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AP State Syllabus 10th Class English Study Material Guide Textbook Solutions Pdf Free Download

Unit 1 Personality Development

Unit 2 Wit and Humour

Unit 3 Human Relations

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Unit 5 Bio-Diversity

Unit 6 Nation and Diversity

AP Board 10th Class Maths Study Material Guide Textbook Solutions State Syllabus

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AP 10th Class Maths Textbook SSC Solutions 2020 Chapter 1 Real Numbers

AP State Board Class 10 Maths Solutions Chapter 2 Sets

AP SSC 10th Class Maths Solutions Chapter 3 Polynomials

AP SSC Maths Solutions 10th Class Chapter 4 Pair of Linear Equations in Two Variables

AP SSC Maths Solutions Class 10 Chapter 5 Quadratic Equations

AP 10th Class Maths Solutions Chapter 6 Progressions

AP 10th Class Maths Guide Chapter 7 Coordinate Geometry

AP 10th Maths Solutions SSC Chapter 8 Similar Triangles

Maths Guide for Class 10 SSC Chapter 9 Tangents and Secants to a Circle

AP State 10th Class Maths Textbook Solutions Chapter 10 Mensuration

10th Class Maths Textbook SSC Solutions Telangana Chapter 11 Trigonometry

Andhra Pradesh SSC Class 10 Solutions for Maths Chapter 12 Applications of Trigonometry

AP SSC Class 10 Solutions for Maths Chapter 13 Probability

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AP Board Solutions

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Practice the AP 10th Class Maths Bits with Answers Chapter 11 Trigonometry on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 1st Lesson Chapter 11 Trigonometry with Answers

Question 1.
Find the value of cos 12° – sin 78°
Answer:
0
Explanation:
cos 12°- sin(90°- 12°)
⇒ cos 12° – cos 12° = 0

Question 2.
If x = cosec θ + cot θ and y = cosec θ – cot θ, then write the relation between ‘x’ and ‘y’
Answer:
xy = 1
Explanation:
xy = (cosec θ + cot θ) (cosec θ – cot θ)
⇒ cosec2 θ – cot2 θ = 1

Question 3.
Write a formula to cos (A – B).
Answer:
cos A cos B + sin A sin B

Question 4.
The value of cos (90 – θ).
Answer:
sin θ

Question 5.
In Δ ABC sin C = \(\frac {3}{5}\),then find cos A.
Answer:
\(\frac {3}{5}\)

Question 6.
Complete the value tan2 θ – sec2 θ.
Answer:
-1
Explanation:
-(sec2 θ – tan2 θ) = – 1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 7.
The value of sec (90 – A).
Answer:
cosec A

Question 8.
If cosec θ + cot θ = 5, then cosec θ – cot θ.
Answer:
\(\frac {1}{5}\)

Question 9.
If x = 2 sec θ; y = tan θ,then the value of x2 – y2.
Answer:
4
Explanation:
sec2 θ = \(\left(\frac{x}{2}\right)^{2}\),tan2 θ = \(\left(\frac{y}{2}\right)^{2}\)
⇒ sec2 θ – tan2 θ = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)
⇒ 4 = x2 – y2

Question 10.
If √3 tan θ = 1,then the value of θ.
A.
30°

Question 11.
The value of (sec 60)(cos 60).
Answer:
1

Question 12.
How much the value of sin (60 + 30)?
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 13.
If sec θ + tan θ = \(\frac{1}{2}\) then find sec θ – tan θ.
Answer:
2

Question 14.
The value of cos(90 – θ).
Answer:
sin θ

Question 15.
Simplify: tan 26°. tan 64°
Answer:
1
Explanation:
tan 26° . tan 64°
⇒ tan 26° . tan (90° – 26°)
⇒ tan 26° . cot 26° = 1

Question 16.
How much value of the angle ‘θ’ in the figure?
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 1
Answer:
30°
Explanation:
sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2}{4}=\frac{1}{2}\) = sin 30°
∴θ = 30°

Question 17.
Find the value of \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\).
Answer:
0
Explanation:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-1}{1+1}=\frac{0}{2}\) = 0

Question 18.
If sin x = \(\frac{5}{7}\), then find the value of cosec x.
Answer:
\(\frac{7}{5}\)
Explanation:
sin x = \(\frac{5}{7}\) ⇒ cosec x = \(\frac{7}{5}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 19.
Given ∠A = 75°, ∠B = 30°,then find the value of tan (A – B).
Answer:
1
Explanation:
tan (75° – 30°) = tan 45° = 1

Question 20.
If sec θ + tan θ = \(\frac{1}{3}\),then find the value of sec θ – tan θ.
Answer:
3
Explanation:
⇒ sec θ + tan θ = \(\frac{1}{3}\)
⇒ sec θ – tan θ = 3

Question 21.
If cosec θ + cot θ = 2,then find the value of cos θ.
Answer:
\(\frac{3}{5}\)
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 2

Question 22.
If cos (A + B) = 0, cos B = \(\frac{\sqrt{3}}{2}\),then
how much value of A?
Answer:
60°
Explanation:
cos (A + B) = 0, cos B = \(\frac{\sqrt{3}}{2}\)
⇒ cos B = cos 30°
⇒ A + B = 90°
⇒ A + 30° = 90°
∴ A = 90°- 30° = 60°

Question 23.
If sec θ – tan θ = 3, then find the value of sec θ + tan θ.
Answer:
\(\frac{1}{3}\)

Question 24.
If sin 2θ = cos 3θ, then how the value of θ.
Answer:
18°
Explanation:
sin 2θ = cos 3θ
⇒ 2θ + 3θ = 90°
⇒ 5θ = 90°
θ = 18°

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 25.
If cos θ = \(\frac{3}{5}\), then find the value of sin θ.
Answer:
\(\frac{4}{5}\)
Explanation:
cos θ = \(\frac{3}{5}\) ⇒ sin θ = \(\frac{4}{5}\)

Question 26.
Simplify : cos 60° + sin 30°.
Answer:
1
Explanation:
cos 60° + sin 30° = \(\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\) = 1

Question 27.
If sec A + tan A = \(\frac{1}{5}\), then find the value of sec A – tan A.
Answer:
5

Question 28.
sin (90 – A) = \(\frac{1}{2}\) , then how much the value A?
A.
60°

Question 29.
If cot A = \(\frac{5}{12}\), then find the value of sin A + cos A.
Answer:
\(\frac{17}{13}\)

Question 30.
Write any value which is not possible for sin x?
Answer:
\(\frac{5}{4}\)
Explanation:
sin x = \(\frac{5}{4}\) > 1, so it is not possible.

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 31.
If sin θ = cos θ (0 < θ < 90), then the value of tan θ + cot θ.
Answer:
2
Explanation:
sin θ = cos θ
⇒ 0 = 45°
⇒ tan 45° + cot 45° = 2

Question 32.
If sec θ + tan θ = 3, then find the value of sec θ – tan θ.
Answer:
\(\frac{1}{3}\)

Question 33.
In ΔABC; AB = c, BC = a, AC = b and ∠BAC = 0, then calculate the value of area of ΔABC is ………………(θ is acute)
Answer:
\(\frac{1}{2}\) bc sin θ

Question 34.
Write the value of tan θ in terms of cosec θ.
Answer:
\(\frac{1}{\sqrt{\operatorname{cosec}^{2} \theta-1}}\)

Question 35.
Observe the following:
(I) sin2 20° + sin2 70° = 1
(II) log2 (sin 90°) = 1
Which one is CORRECT?
Answer:
(I) only

Question 36.
Simplify : tan 36°. tan 54° + sin 30°
Answer:
\(\frac{3}{2}\)
Explanation:
tan 36° . tan (90 – 36°) + sin 30°
⇒ tan 36° . cot 36° + \(\frac{1}{2}\)
⇒ 1 + \(\frac{1}{2}=\frac{3}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 37.
If sin A = \(\frac{24}{25}\), then find the value of sec A.
Answer:
\(\frac{25}{7}\)

Question 38.
Which one of the following is NOT defined?
sin 90°, cos 0°, sec 90°, cos 90°.
Answer:
sec 90°
Explanation:
sec 90° is not defined

Question 39.
Simplify:\(\sqrt{\frac{1-\cos ^{2} A}{1+\cot ^{2} A}}\)
Answer:
sin2A
Explanation:
\(\sqrt{\frac{\sin ^{2} A}{\operatorname{cosec}^{2} A}}\) = \(\sqrt{\sin ^{4} A}\) = sin2A

Question 40.
If cot θ – cosec θ = p, then find cot θ + cosec θ.
Answer:
\(\frac{-1}{\mathrm{p}}\)

Question 41.
Express tan θ in terms of cos θ.
Answer:
\(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\)

Question 42.
Who was introduced Trigonometry?
Answer:
Hipparchus

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 43.
\(\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}\) equal to
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 3

Question 44.
sin247° + sin243° equal to
Answer:
1

Question 45.
sin 2A equal to
Answer:
2sin A cos A

Question 46.
sin 30° + cos 60° equal to
Answer:
1

Question 47.
sec4A – sec2A equal to
Answer:
tan4 A – tan2 A
Explanation:
sec4 A – sec2A
⇒1 + tan4 A – (1 + tan2 A)
⇒ tan4 A – tan2 A

Question 48.
Find the value of \(\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}\).
Answer:
sin2 θ

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 49.
2sin 45°. cos 45° equal to
Answer:
1

Question 50.
sin 81° equal to
Answer:
cos 9°
1

Question 51.
tan θ = \(\frac{1}{\sqrt{3}}\), then find cos θ.
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 52.
If cos θ = \(\frac{1}{2}\); then find cos\(\frac{\theta}{2}\)
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
cos θ = \(\frac{1}{2}\)
⇒ cos θ = cos 60° ⇒ θ = 60°
⇒ cos\(\frac{\theta}{2}\) ⇒ cos 30° = \(\frac{\sqrt{3}}{2}\)

Question 53.
\(\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}\) equal to
Answer:
cos θ + sin θ

Question 54.
cos 300° equal to
Answer:
\(\frac{1}{2}\)
Explanation:
cos (270° + 30°) = sin 30° =\(\frac{1}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 55.
\(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\) equal to
Answer:
tan θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 4

Question 56.
(1 + tan2 θ)cos2 θ equal to
Answer:
1

Question 57.
If 3 tan θ = 1; then find θ.
Answer:
30°

Question 58.
\(\frac{\sqrt{\sec ^{2} \theta-1}}{\sec \theta}\) equal to
Answer:
sin θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 5

Question 59.
\(\frac{1}{\sqrt{1+\tan ^{2} \theta}}\) equal to
Answer:
cos θ

Question 60.
Find the value of cosec 60° × cos 90°.
Answer:
0

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 61.
sec2 θ + cosec2 θ equal to
Answer:
sec2 θ.cosec2 θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 6

Question 62.
sec2 33° – cot2 57° equal to
Answer:
1

Question 63.
If sin θ = \(\frac{11}{15}\), then find cos θ.
Answer:
\(\frac{2 \sqrt{26}}{15}\)

Question 64.
equal to \(\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}\)
Answer:
sin θ

Question 65.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\); cos θ = \(\frac{\mathbf{c}}{\mathbf{d}}\); then find tan θ.
Answer:
\(\frac{\mathrm{ad}}{\mathrm{bc}}\)
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 7

Question 66.
tan (A+B) equal to
Answer:
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\)

Question 67.
If sin A = \(\frac{1}{\sqrt{2}}\) ; then find tan A.
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 68.
sin\(\frac{\pi}{6}\) + cos\(\frac{\pi}{3}\) equal to
Answer:
1

Question 69.
Find the value of cos 75°.
Answer:
sin 15°
Explanation:
cos 75° = cos (90° – 15°) = sin 15°

Question 70.
If sin 0 = \(\frac{1}{2}\); then find cos \(\frac{3 \theta}{2}\).
Answer:
\(\frac{1}{\sqrt{2}}\)

Question 71.
\(\frac{1}{\sec ^{2} A}+\frac{1}{\operatorname{cosec}^{2} A}\) equal to
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 8

Question 72.
sin 240° equals to
Answer:
–\(\frac{\sqrt{3}}{2}\)
Explanation:
sin 240° = sin (270° – 30°)
= -cos 30°= \(\frac{-\sqrt{3}}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 73.
If sec θ + tan θ = \(\frac{1}{5}\) ,then find sin θ.
Answer:
\(\frac{12}{13}\)

Question 74.
tan 240° equal to
Answer:
√3

Question 75.
Find the value of
sin 60° cos 30°+ cos 60°. sin 30°.
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 9

Question 76.
\(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) equal to
Answer:
2sec2 θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 10

Question 77.
tan 0° equal to
Answer:
0

Question 78.
\(\frac{\sqrt{1+\tan ^{2} \theta}}{\sqrt{1+\cot ^{2} \theta}}\) equal to
Answer:
tan θ

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 79.
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) equal to
Answer:
1

Question 80.
If sin θ = cos θ, then find θ.
Answer:
45°

Question 81.
In right triangle ΔABC; ∠B= 90°; tan C = \(\frac{5}{12}\) , then find the length of hypotenuse.
Answer:
13
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 11
By Py thagoras theorem,
AC2 = AB2 + BC2
AC2 = 52 + 122
= 25 + 144= 169
⇒ AC = hypotenuse = 13

Question 82.
If A, B are acute angles ;
sin(A – B)= \(\frac{1}{2}\); cos (A + B) = \(\frac{1}{2}\), then find B.
Answer:
15° (or) \(\frac{\pi}{12}\)
Explanation:
sin (A – B) = \(\frac{1}{2}\) = sin30°
A – B = 30°
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 12

Question 83.
cos (270° – θ) equal to
Answer:
-sin θ

Question 84.
Find the value of
cos 0° + sin 90° + √2sin 45°.
Answer:
3
Explanation:
1 + 1 + √2.\(\frac{1}{\sqrt{2}}\) = 1 + 1 + 1 = 3

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 85.
If tan θ = \(\frac{1}{\sqrt{3}}\); then find cos θ.
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
⇒ θ = 30°
∴ cos θ = cos 30° = \(\frac{\sqrt{3}}{2}\)

Question 86.
Equal to cosec (90 + θ).
Answer:
sec θ

Question 87.
sin θ. sec θ equals to
Answer:
tan θ

Question 88.
Find the value of 3sin2 45°+2cos2 60°.
Answer:
2
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 13

Question 89.
Find the value of tan2 30° + 2 cot2 60°.
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 14

Question 90.
Find the value of secA.\(\sqrt{1-\sin ^{2} A}\)
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 91.
If 5 sin A= 3; then find sec2 A – tan2 A.
Answer:
1
Explanation:
sin A = \(\frac{3}{5}\) ⇒ sec2A = tan2A = 1

Question 92.
Find the value of cos 240°.
Answer:
–\(\frac{1}{2}\)

Question 93.
If sin θ. cosec θ = x; then find x.
Answer:
1

Question 94.
sin(45°+ θ) – cos(45°- θ).
Answer:
0

Question 95.
Find the value of cos2 17° – sin2 73°.
Answer:
0
Explanation:
cos2 17°- sin2 73°
= cos2 (90 – 73) – sin2 73°
= sin2 73° – sin2 73° = 0

Question 96.
Find the value of sin2 60° – sin2 30°.
Answer:
\(\frac{1}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 97.
ten θ is not defined when θ is equal to
Answer:
90°

Question 98.
Find the value of sin 45° + cos 45°.
Answer:
√2
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 15

Question 99.
Simplify: \(\frac{1-\sec ^{2} A}{\operatorname{cosec}^{2} A-1}\)
Answer:
– tan4A

Question 100.
Find the value of sin θ. cosec θ+ cos θ. sec θ + tan θ. cot θ.
Answer:
3
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 16
= 1 + 1 + 1 = 3

Question 101.
If A = 30°, then sin 2A equals to
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 102.
If sec θ = 3k and tan θ = \(\frac{3}{\mathbf{k}}\), then find the value of (k2 – \(\frac{1}{\mathbf{k}^{2}}\))
Answer:
\(\frac{1}{9}\)
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 17

Question 103.
If tan θ + sec θ = 8, then find sec θ – tan θ.
Answer:
\(\frac{1}{8}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 104.
If sin θ = \(\frac{12}{13}\), then find tan θ.
Answer:
\(\frac{12}{5}\)

Question 105.
In a right angled ΔABC, right angle at C if tan A = \(\frac{8}{15}\), then find the value of cosec2 A – 1.
Answer:
\(\frac{225}{64}\)
Explanation:
tan A = \(\frac{8}{15}\),
cosec 2 A – 1 = cot2A = \(\left(\frac{15}{8}\right)^{2}=\frac{225}{64}\)

Question 106.
Find the value of \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)
Answer:
sin 60°

Question 107.
If sin A = ;\(\frac{1}{\sqrt{2}}\) then find tan A
Answer:
1

Question 108.
In ΔABC, sin \(\left(\frac{B+C}{2}\right)\) equal to .
Answer:
A. cos\(\frac{A}{2}\)

Question 109.
tan 26°. tan 64° equal to
Answer:
1
Explanation:
tan 26° . tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° = 1

Question 110.
cos2 θ equal to
Answer:
1 – sin2 θ

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 111.
If tan A = \(\frac{3}{4}\) then find sec2 A – tan2 A.
Answer:
1

Question 112.
sin4 θ – cos4 θ equal to
Answer:
2sec2 θ – 1

Question 113.
\(\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\) equal to
Answer:
tan θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 18

Question 114.
sin(90 – Φ) equal to
Answer:
cos Φ

Question 115.
sec θ – tan θ = \(\frac{1}{n}\), then find sec θ + tan θ.
Answer:
n

Question 116.
x = 2 cosec θ, y = 2 cot θ; find x2 – y2.
Answer:
4

Question 117.
tan θ is not defined if θ.
Answer:
90°

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 118.
sec θ is not defined if θ.
Answer:
90°

Question 119.
tan2 Φ – sec2 Φ equal to
Answer:
-1

Question 120.
\(\left|\begin{array}{ll}
\tan \theta & \sec \theta \\
\sec \theta & \tan \theta
\end{array}\right|\)
Answer:
1
Explanation:
\(\left|\tan ^{2} \theta-\sec ^{2} \theta\right|=|-1|\) = 1

Question 121.
sin 225° equal to
Answer:
\(\frac{-1}{\sqrt{2}}\)

Question 122.
cos (x – y) equal to
Answer:
cos x cos y + sin x sin y

Question 123.
\(\frac{\sec 35^{\circ}}{\operatorname{cosec} 55^{\circ}}\) equal to
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 19

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 124.
\(\frac{1}{\sec ^{2} A}+\frac{1}{\operatorname{cosec}^{2} A}\) equal to
Answer:
1

Question 125.
sin (-θ) equal to
Answer:
-sin θ

Question 126.
cosec (270 – θ) equal to
Answer:
-sec θ

Question 127.
sec (90 + θ) equal to
Answer:
-cosec θ

Question 128.
tan (360 – θ) equal to
Answer:
-tan θ

Question 129.
cos (-θ) equal to
Answer:
cos θ

Question 130.
sin (180 – θ) equal to
Answer:
sin θ

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 131.
cos (270 – θ) equal to
Answer:
-sin θ

Question 132.
Find the maximum value of cos θ.
Answer:
1

Question 133.
Find the minimum and maximum values of tan θ.
Answer:
(- ∞ ,∞ )

Question 134.
sin 420° equal to
Answer:
\(\frac{\sqrt{3}}{2}\)

Question 135.
sec 240° equal to
Answer:
-2

Question 136.
cos 0° + sin 90° + √3 cosec 60° equal to
Answer:
4
Explanation:
1 + 1 + 3 . \(\frac{2}{\sqrt{3}}\) = 4

Question 137.
sec θ + tan θ = \(\frac{1}{2}\); then find sin θ.
Answer:
\(\frac{12}{13}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 138.
sin 45°.cos 45° + √3 sin 60° equal to
Answer:
2

Question 139.
tan 30° + cot 30° equal to
Answer:
\(\frac{4}{\sqrt{3}}\)

Question 140.
tan (A – B) equal to
Answer:
\(\frac{\tan A-\tan B}{1+\tan A \tan B}\)

Question 141.
If sin A = \(\frac{3}{5}\); then find sin (90 + A).
Answer:
\(\frac{4}{5}\)
Explanation:
sin A = \(\frac{3}{5}\), sin (90 + A) = cos A = \(\frac{4}{5}\)

Question 142.
Find the value of cos 1°.cos 2°.cos 3°……………, cos 180°.
Answer:
0
Explanation:
cos 1° × cos 2° × cos 3° × …………….. × cos 90° × ……….. × cos 180°
cos 1° × cos 2° × cos 3° × …………. × 0 × …………. × (- 1) = 0

Question 143.
Find the value of cos217° – sin2 73°.
Answer:
0

Question 144.
If cosec θ + cot θ = 2; then find cosec θ – cot θ.
Answer:
\(\frac{1}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 145.
cosec 60°. sec 60° equal to
Answer:
\(\frac{4}{\sqrt{3}}\)

Question 146.
sin (A – B) equal to
Answer:
sin A cos B – cos A sin B

Question 147.
If tan (15°+ B)= √3 ; then find B.
Answer:
45°
Explanation:
tan (15° + B) = √3 = tan 60°
15 + B = 60 ⇒B = 60°- 15° = 45°

Question 148.
Simplify: \(\frac{\sin (90-\theta) \sin \theta}{\tan \theta}\) – 1
Answer:
-sin2 θ

Question 149.
sin 450° equal to
Answer:
1

Question 150.
cos 150° equal to
Answer:
–\(\frac{\sqrt{3}}{2}\)

Question 151.
If sin θ = \(\frac{1}{2}\) ; then find cot θ.
Answer:
√3

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 152.
Find the value of sin 29° – cos 61°
Answer:
0

Question 153.
If tan θ = 1; then find cos θ.
Answer:
\(\frac{1}{\sqrt{2}}\)

Question 154.
cos (A + B) equal to
Answer:
cos A cos B – sin A sin B

Question 155.
Express tan θ, in terms of sin θ.
Answer:
\(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\)

Question 156.
sin θ + sin2 θ = 1, then find cos2 θ + cos4 θ.
Answer:
1

Question 157.
in ΔABC,write tan(\(\frac{B+C}{2}\))equal to
Answer:
Cot (\(\frac{\mathrm{A}}{2},\))
Explanation:
A + B + C = 180°
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 20

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 158.
(cos A + sin A)2 + (cos A – sin A)2 equal to
Answer:
2

Question 159.
cos(180 – θ) equal to
Answer:
– cos θ

Question 160.
Find the value of (cosec θ – cot θ).
Answer:
\(\frac{1-\cos \theta}{\sin \theta}\)
Explanation:
cosec θ – cot θ
= \(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\frac{1-\cos \theta}{\sin \theta}\)

Question 161.
sin2 75° + cos2 75° equal to
Answer:
1

Question 162.
If cos θ. sin θ = \(\frac{1}{2}\) ; then find tan θ.
Answer:
1
Explanation:
cos θ . sin θ = cos 45° . sin 45°
= \(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}\)
Then tan 45° = 1.

Question 163.
For which value of cosine equal to sin 81°.
Answer:
cos 9°.

Question 164.
sin 750° equal to
Answer:
\(\frac{1}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 165.
\(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) equal to
Answer:
tan θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 21

Question 166.
If sin(A+B) = 1; sin B = \(\frac{1}{2}\) ; then find A.
Answer:
60°
Explanation:
sin (A + B) = 1 = sin 90°
A + B = 90°
sin B = sin 30° ⇒ B = 30°
A + 30° = 90° ⇒ A = 60°

Question 167.
If tan θ = √3 , then find sec θ.
Answer:
2
Explanation:
tan θ = √3 = tan 60° ⇒ θ = 60°
sec 60° = 2

Question 168.
Find the value of
cos 0°+ sin 90° + √3 cosec 60°.
Answer:
4

Question 169.
\(\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}\) equal to
Answer:
sec θ . cosec θ
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 22

Question 170.
Find the value of
cos 60°. cos 30° – sin 60°. sin30°.
Answer:
0

Question 171.
cot2θ – \(\frac{1}{\sin ^{2} \theta}\) equal to
Answer:
-1
Explanation:
cot2 θ – cosec2 θ = – 1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 172.
π radians equal into degrees.
Answer:
180°

Question 173.
sin (A + B). cos(A – B) + sin (A – B). cos (A + B) equal to
Answer:
sin 2A

Question 174.
If cos (A+B) = 0, cos B = \(\frac{\sqrt{3}}{2}\) ;then find A.
Answer:
60°

Question 175.
cos6 θ + sin6 θ equal to .
Answer:
1 – 3 sin2 θ.cos2 θ
Explanation:
cos6 θ + sin6 θ = (cos2 θ)3 + (sin2 θ)3
a3 + b3 = (a + b) – 3ab (a + b)
= (sin2 θ + cos2 θ)3
– 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)
= 1 – 3 sin2 θ cos2 θ (1)
= 1 – 3 sin2 θ cos2 θ

Question 176.
sin 225° equal to
Answer:
–\(\frac{1}{\sqrt{2}}\)

Question 177.
sin 180° equal to
Answer:
0

Question 178.
If x = 2 cosec θ; y = 2 cot θ; then find x2 – y2.
Answer:
4
Explanation:
\(\frac{\mathrm{x}}{2}\) = cosec θ, \(\frac{\mathrm{y}}{2}\) = cot θ
cosec2 θ – cot2 θ = \(\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2}\right)^{2}\)
1 = \(\frac{x^{2}}{4}-\frac{y^{2}}{4}\)
x2 – y2 = 4

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 179.
cos θ. tan θ equal to
Answer:
sin θ
Explanation:
cos θ = \(\frac{\sin \theta}{\cos \theta}\) = sin θ

Question 180.
If cot2 θ = 3; then find cosec θ.
Answer:
2
Explanation:
cot θ = √3 = cot 30° ⇒ θ = 30°
∴ cosec 30° = 2.

Question 181.
If sec θ = cosec θ; then find the value of θ.
Answer:
\(\frac{\pi}{4}\)

Question 182.
\(\frac{\tan \theta \cdot \sqrt{1-\sin ^{2} \theta}}{\sqrt{1-\cos ^{2} \theta}}\) equal to
Answer:
1
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 23

Question 183.
cos(x – y) equal to
Answer:
cos x cos y + sin x sin y

Question 184.
Simplify : cosec 31° – sec 59°
Answer:
0
Explanation:
cosec 31° – sec (90° – 31°)
[∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31°
= 0

Question 185.
(sec 45° + tan 45°) (sec 45° – tan 45°) equal to
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 186.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\); cos θ = \(\frac{\mathbf{c}}{\mathbf{d}}\), then find cot θ.
Answer:
\(\frac{\mathrm{bc}}{\mathrm{ad}}\)

Question 187.
\(\sqrt{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}\) equal to
Answer:
1

Question 188.
sin2 45° + cos2 45° + tan2 45° equal to
Answer:
2
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 24

Question 189.
sec (360° – θ) equal to
Answer:
sec θ

Question 190.
tan θ. cot θ = sec θ. x ; then find x.
Answer:
cos θ
Explanation:
tan θ . cot θ = sec θ . x
\(\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}=\frac{x}{\cos \theta}\)
cos θ = x

Question 191.
If 4 sin 30°. sec 60° = x tan 45°; then find x.
Answer:
4
Explanation:
4 . sin 30° – sec 60° = x . tan 45°
4 . \(\frac{1}{2}\) . 2 = x . 1
⇒ x = 4

Question 192.
In the following which are in geometric progression?
A) sin 30°, sin 45°, sin 60°
B) sec 30°, sec 45°, sec 60°
C) tan 30°, tan 45°, tan 60°
D) cos 45°, cos 60°, cos 90°
Answer:
C) tan 30°, tan 45°, tan 60°

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 193.
(1 + tan2 A) (1 – sin2 A)
equal to
Answer:
1
Explanation:
sec2 A × cos2 A = 1

Question 194.
Find the value of
cos 60° cos 30° + sin 60°. sin 30°.
Answer:
\(\frac{\sqrt{3}}{2}\)
Explanation:
\(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}\)

Question 195.
\(\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) equal to
Answer:
\(\frac{1}{2}\)

Question 196.
cos(\(\frac{3 \pi}{2}\) + θ) equal to
Answer:
sin θ
Explanation:
cos (270 + θ) = sin θ

Question 197.
\(\sqrt{1+\cot ^{2} \theta}\) equal to
Answer:
cosec θ

Question 198.
If tan θ + cot θ = 2; then find tan2 θ + cot2 θ.
Answer:
2
Explanation:
tan θ + cot θ = 2
⇒ tan2 θ + cot2 θ + 2 . tan θ . cot θ = 4
⇒ tan2 θ + cot2 θ = 4 – 2 = 2

Question 199.
If sec θ = \(\frac{13}{12}\), then find sin θ.
Answer:
\(\frac{5}{13}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 200.
How much the value of
(sin θ + cos θ)2 + (sin θ – cos θ)2 ?
Answer:
2
Explanation:
(a + b)2 + (a – b)2 = 2(a2 + b2)
= 2 (sin2 θ + cos2 θ) = 2

Question 201.
(1 + tan θ)2 equal to
Answer:
sec2 θ + 2 tan θ

Question 202.
sin(A – B) = \(\frac{1}{2}\); cos (A+B) = \(\frac{1}{2}\). So
find A.
Answer:
45°
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 25
⇒ A = 45°

Question 203.
Find the value of tan 135°.
Answer:
-1

Question 204.
Find the value of \(\sqrt{1+\sin A} \cdot \sqrt{1-\sin A}\)
Answer:
cos A
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 26

Question 205.
Find the value of tan 60° – tan 30°.
Answer:
\(\frac{2 \sqrt{3}}{3}\)
Explanation:
tan 60° – tan 30°
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 27

Question 206.
In ΔABC, a = 3; b = 4; c = 5, then find cos A.
Answer:
4/5
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 28

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 207.
sin3 θ cos θ.cos3 θ sin θ equals to
Answer:
sin θ cos θ

Question 208.
If tan θ \(\frac{1}{\sqrt{3}}\) , then find the value of 7 sin2 θ + 3 cos2 θ.
Answer:
4
Explanation:
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 29

Question 209.
cot (270° – θ) equal to
Answer:
tan θ

Question 210.
(1 + tan2 60°)2 equals to
Answer:
16
Explanation:
[1 + (√3)2]2 = (1 + 3)2 = 42 = 16

Question 211.
sin (270° + θ) equal to
Answer:
– cos θ

Question 212.
If tan2 60° + 2 tan2 45° = x tan 45°, then find x.
Answer:
5
Explanation:
(√3)2 + 2(1)2 = x . 1
⇒ 3 + 2 = x ⇒ x = 5

Question 213.
cos2 0° + cos2 60° equal to
Answer:
5/4

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 214.
Simplify : sin4 θ – cos4 θ
Answer:
2sin2 θ – 1

Question 215.
If α + β = 90° and α = 2β, then find cos2 β + sin2 β.
Answer:
1
Explanation:
α = 90 – β
⇒ 90 – β = 2β ⇒ 3β = 90° ⇒ β = 30°
∴ cos2 30° + sin2 30° = (\(\frac{\sqrt{3}}{2}\))2 + (\(\frac{1}{2}\))2
= \(\frac{3}{4}+\frac{1}{4}=\frac{4}{4}\) = 1

Question 216.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\) , then find cos θ.
Answer:
\(\frac{\sqrt{b^{2}-a^{2}}}{b}\)

Question 217.
2sin θ = sin2 θ is true for the value of θ is
Answer:

Question 218.
If sin θ = \(\frac{\mathbf{a}}{\mathbf{b}}\) , then find tan θ.
Answer:
\(\frac{a}{\sqrt{b^{2}-a^{2}}}\)

Question 219.
\(\frac{\sin \theta}{1+\cos \theta}\) is equal to
Answer:
\(\frac{1-\cos \theta}{\sin \theta}\)

Question 220.
If sin θ = cos θ, then find the value of 2 tan θ + cos2 θ.
Answer:
\(\frac{5}{2}\)
Explanation:
sin θ = cos θ ⇒ θ = 45°
2 tan 45° + cos2 45° = 2 + \(\frac{1}{2}=\frac{5}{2}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 221.
If sin x = cos x, 0 ≤ x ≤ 90°, then find x.
Answer:
45°

Question 222.
How much the maximum value of sin θ?
Answer:
1

Question 223.
In the figure find tan x.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 30
Answer:
\(\frac{15}{8}\)

Question 224.
sin A = cos B, then find A + B.
Answer:
90°

Question 225.
If cosec θ + cot θ = 3, then find cosec θ – cot θ.
Answer:
\(\frac{1}{3}\)

Question 226.
If tan 2A = cot (A – 18°) where 2A is an acute angle, then find A.
Answer:
36°
Explanation:
90 – 2A = A – 18°
⇒ 3A = 108° ⇒ A = 36°

Question 227.
sec 0° equal to
Answer:
1

Question 228.
cosec 300° equal to
Answer:
\(\frac{-2}{\sqrt{3}}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 229.
cos 240° equal to
Answer:
\(\frac{-1}{2}\)

Question 230.
\(\frac{\operatorname{cosec}^{2} \theta}{\cot \theta}\) – cot θ equal to
Answer:
tan θ

Question 231.
Find the minimum value of cos θ.
Answer:
-3

Question 232.
If sec θ = cosec θ; then find the value of θ in radians.
Answer:
\(\frac{\pi^{\mathrm{c}}}{4}\)
Explanation:
sec θ = cosec θ ⇒ θ = 45° = \(\frac{\pi^{\mathrm{c}}}{4}\)

Question 233.
Reciprocal of cot A.
Answer:
tan A

Question 234.
sin \(\frac{\pi^{\mathrm{c}}}{4}\) + cos 45° equal to
Answer:
√2

Question 235.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 31
Answer:
\(\frac{\mathrm{x}}{\mathrm{z}}\)

Question 236.
If cosec θ = \(\frac{25}{7}\), then find cot θ.
Answer:
\(\frac{24}{7}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 237.
If sin (A + B) = \(\frac{\sqrt{3}}{2}\); cos B = \(\frac{\sqrt{3}}{2}\) , then find A.
Answer:
30°

Question 238.
tan 750° equal to
Answer:
\(\frac{1}{\sqrt{3}}\)
Question 239.
(1 – sec2 θ) (1 – cosec2 θ) equal to
Answer:
1

Question 240.
If cosec θ – cot θ = 4, then find cosec θ + cot θ.
Answer:
\(\frac{1}{4}\)

Question 241.
\(\sqrt{\tan ^{2} \theta+\cot ^{2} \theta+2}\) equal to
Answer:
tan θ + cot θ
Explanation:
= \(\sqrt{1+\tan ^{2} \theta+1+\cot ^{2} \theta}\)
= \(\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}\)
= tan θ + cot θ

Question 242.
If cos θ = \(\frac{3}{5}\); then cos (-θ) equal to
Answer:
\(\frac{3}{5}\)

Question 243.
\(\frac{\sqrt{\operatorname{cosec}^{2} \theta-1}}{\operatorname{cosec} \theta}\)
equal to
Answer:
cos θ

Question 244.
If 5 sin A = 3, then find sec2 A – tan2 A.
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 245.
In the figure, find AB.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 32
Answer:
20√3 (or) \(\frac{60}{\sqrt{3}}\)

Question 246.
If cot θ = x; then find cosec θ.
Answer:
\(\sqrt{\mathrm{x}_{1}^{2}+1}\)
Explanation:
cot θ = x ⇒ cosec θ = \(\sqrt{x^{2}+1}\)

Question 247.
\(\sqrt{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta-\cos ^{2} \theta}\) equal to
Answer:
cot θ

Question 248.
Find the value of \(\frac{1}{\sec \theta-\tan \theta}\)
Answer:
sec θ + tan θ

Question 249.
sin (A + B) equal to
Answer:
sin A cos B + cos A sin B

Question 250.
\(\sqrt{(\sec \theta+1)(\sec \theta-1)}\) equal to
Answer:
tan θ

Question 251.
Find the value of tan 5° × tan 30° × 4 tan 85°.
Answer:
\(\frac{4}{\sqrt{3}}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 252.
cos 110°.cos 70° – sin 110°.sin 70° equal to ,
Answer:
-1
Explanation:
cos A . cos B – sin A . sin B
= cos (A + B)
= cos (110 + 70) = cos 180° = – 1

Question 253.
Find the value of tan 1°.tan 2°.tan 3°………….tan 89°.
Answer:
1

Question 254.
If cos θ = -cos θ; then Write θ in radian measure.
Answer:
\(\frac{\pi^{\mathrm{c}}}{3}\)

Question 255.
sec A = cosec B, then write A and B are ……….. angles.
Answer:
Complementary.

Question 256.
Find the value of tan 75°.
Answer:
2 + √3

Question 257.
\(\sqrt{\sec ^{2} \theta-\tan ^{2} \theta+\cot ^{2} \theta}\) equal to
Answer:
cosec θ

Question 258.
sin 240° + sin 120° equal to
Answer:
0

Question 259.
Find the value of
sec 70°. sin 20° + cos 20°. cosec 70°.
Answer:
2

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 260.
If sec A + tan A = \(\frac{1}{3}\); then find sec A – tan A.
Answer:
3

Question 261.
(sec2 θ – 1)(1 – cosec2 θ)equal to
Answer:
-1

Question 262.
cos θ equal to
Answer:
\(\frac{\cot \theta}{\operatorname{cosec} \theta}\)

Question 263.
The radius of a circle is ‘r’; an arc of length ‘L’ is making an angle θ, at the centre of the circle, then find θ.
Answer:
L/r

Question 264.
cos (A – B) = \(\frac{1}{2}\); sin B = \(\frac{1}{\sqrt{2}}\), find measure of A.
Answer:
105°

Question 265.
If sec θ + tan θ = 4; then find cos θ.
Answer:
\(\frac{8}{17}\)

Question 266.
sec (360° – θ) equals to
Answer:
sec θ

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 267.
If A is acute and tan A = \(\frac{1}{\sqrt{3}}\); then find sin ‘A’.
Answer:
\(\frac{1}{2}\)

Question 268.
(1 +cot2 45°)2 equal to
Answer:
4

Question 269.
\(\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}\) equal to
Answer:
2\(\frac{1}{2}\)

Question 270.
\(\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}\) equal to
Answer:
sec x + tan x

Question 271.
\(\frac{\sin ^{4} A-\cos ^{4} A}{\sin ^{2} A-\cos ^{2} A}\) equal to
Answer:
1

Question 272.
If the angle in a triangle are in the ratio of 1:2:3, then find the smallest angle in radins.
Answer:
π/6

Question 273.
If sin θ + cos θ = √2; then find the value of ‘θ’.
Answer:
45°

Question 274.
If cosec θ = 2 and cot θ = √3 P; where θ is an acute angle, then find the value of ‘P’.
Answer:
1

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 275.
If cos 2θ = sin 4θ; here 2θ and 4θ are acute angles, then find the value of θ.
Answer:
15°
Explanation:
2θ + 4θ = 90°
⇒ 6θ = 90°⇒ θ = \(\frac{90^{\circ}}{6}\) = 15°

Question 276.
If sin 45°.cos 45°+cos 60° = tan θ, then find the value of θ.
Answer:
45°

Question 277.
If P, Q and R are interior angles of a ΔPQR, then tan \(\left(\frac{\mathbf{P}+\mathbf{Q}}{2}\right)\) equals
Answer:
cot (\(\frac{\mathrm{R}}{2}\))

Question 278.
If tan θ = 1, then find the value of \(\frac{5 \sin \theta+4 \cos \theta}{5 \sin \theta-4 \cos \theta}\)
Answer:
9
Explanation:
tan θ = 1 = tan 45° ⇒ θ = 45°
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 33

Question 279.
If sec 2A = cosec (A – 27°), where 2A is an acute angle, then find the measure of ∠A.
Answer:
39°
Explanation:
90 – 2A = A – 27°
⇒ 117° = 3A ⇒ A = \(\frac{117^{\circ}}{3}\) = 39°

Question 280.
If sin C = \(\frac{3}{5}\); then find cos A.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 34
Answer:
3/5

Question 281.
Expressing tan θ, interms of sec θ.
Answer:
\(\sqrt{\sec ^{2} \theta-1}\)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 282.
If sin θ. cos θ = k; then find sin θ + cos θ.
Answer:
\(\sqrt{1+2 \mathrm{k}}\)

Question 283.
If \(\frac{1}{2}\) tan2 45° = sin2 A and ’A’ is acute, then find the value of ‘A’.
Answer:
45°

Question 284.
Find the value of (\(\frac{11}{\cot ^{2} \theta}-\frac{11}{\cos ^{2} \theta}\))
Answer:
-11
Explanation:
11 (tan2 θ – sec2 θ) = 11 (-1) = -11

Question 285.
Find the maximum value of \(\frac{1}{\sec \theta}\)
0° ≤ θ ≤ 90°.
Answer:
1

Question 286.
If π < θ < \(\frac{3 \pi}{2}\), then θ lies in which quadrant?
Answer:
Third quadrant

Question 287.
If cos θ = \(\frac{\sqrt{3}}{2}\) and’θ’is acute, then find the value of 4sin2 θ + tan2 θ.
Answer:
4/3

Question 288.
If tan θ = \(\frac{7}{8}\), then find the value of \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) ,
Answer:
\(\frac{64}{49}\)

Question 289.
When 0° ≤ 0 ≤ 90°; find the maximum value of sin θ + cos θ.
Answer:
√2

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 290.
In ΔABC, ∠B = 90° ; ∠C = θ. From the figure find tan θ.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 35
Answer:
\(\frac{15}{8}\)

Question 291.
If sin (x – 20°) = cos(3x – 10°), then find x’.
Answer:
15°

Question 292.
If sin (A – B)= \(\frac{1}{2}\); cos (A + B)= \(\frac{1}{2}\),
then find ‘B’.
Answer:
15°

Question 293.
If 5 tan θ = 4, then find die value of
\(\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta}\)
Answer:
\(\frac{1}{7}\)

Question 294.
If 4 cos2 θ – 3 = 0, then find the value of sin θ.
Answer:
\(\frac{1}{2}\)

Choose the correct answer satisfying the following statements.
Question 295.
Statement (A): sin2 67° + cos2 67° = 1
Statement (B) : For any value of θ,
sin2 θ + cos2 θ = 1
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true
Explanation:
sin2 θ + cos2 θ = 1
⇒ sin2 67° + cos2 67° = 1
Hence, (i) is the correct option.

Question 296.
Statement (A): If cos A + cos2 A = 1,
then sin2 A + sin4 A = 2
Statement (B): 1 – sin2 A = cos2 A, for any value of A.
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
iii) A is false, B is true
Explanation:
cos A + cos2 A = 1
cos A = 1 – cos2 A = sin2 A
∴ sin2 A + sin4 A = cos A + cos2 A = 1
⇒ sin2 A + sin4 A = 1
Hence, (iii) is the correct option.

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 297.
Statement (A):
The value of sec2 10° – cot2 80° is 1.
Statement (B):
The value of sin 30° = \(\frac{1}{2}\)
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true
Explanation:
We have sec2 10° – cot2 80°
= sec2 10°-cot2 (90°- 10°)
= sec2 10° – tan2 10° = 1
Also, sin 30° = \(\frac{1}{2}\).
Hence, (i) is the correct option.

Question 298.
Statement (A) : The value of sin θ cos (90 – θ) + cos θ sin(90 – θ) ‘ equal to 1.
Statement (B): tan θ = sec(90 – θ)
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
ii) A is true, B is false
Explanation:
sin θ cos (90 – θ) + Cos θ sin (90 – θ)
= sin θ . sin θ + cos θ – cos θ
= sin2 θ + cos2 θ = 1 and tan θ = cot (90 – θ)
Hence, (ii) is the correct option.

Question 299.
Statement (A) : The value of sin θ = \(\frac{4}{3}\) is not possible.
Statement (B): Hypotenuse is the largest side in any right angled triangle.
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true
Explanation:
sin 2 = \(\frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{3}\)
Here, perpendicular is greater than the hypotenuse which is not possible in any right triangle.
Hence, (i) is the correct option.

Question 300.
Statement (A) : In a right angled triangle, if tan θ = \(\frac{3}{4}\), the greatest side of
the triangle is 5 units.
Statement (B) : (Greatest side hypotenuse)2 = (perpendicular)2 – (base)2
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true
Explanation:
Both A and B are correct and B is the correct explanation of the A.
Greatest side = \(\sqrt{(3)^{2}+(4)^{2}}\) = 5 units.
Hence, (i) is the correct option.

Question 301.
Statement (A) : In a right angled triangle, if cos θ \(\frac{1}{2}\) = and sin θ = \(\frac{\sqrt{3}}{2}\) then tan θ = √3
‘Statement (B) : tan θ = \(\frac{\sin \theta}{\cos \theta}\)
i) Both A and B are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
i) Both A and B are true
Explanation:
Both A and B are correct and B is the correct explanation of the A.
tan θ = \(\frac{\sqrt{3}}{2}\) × 2 = √3
Hence, (i) is the correct option.

Question 302.
Statement (A) : sin 47° cos 43°.
Statement (B) : sin θ = cos(90 + θ),
where θ is an acute angle.
i) Both A and B .are true
ii) A is true, B is false
iii) A is false, B is true
iv) Both A and B are false.
Answer:
ii) A is true, B is false
Explanation:
A is correct, but B is not correct,
sin θ = cos (90 – θ)
sin 47° = cos (90 – 47)
= cos 43°
Hence, (ii) is the correct option.

❖ Study the given information and answer to the following questions.
In ΔABC, right angled at B.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 36
AB + AC = 9 cm and BC = 3 cm

Question 303.
The value of cot C is
Answer:
\(\frac{3}{4}\)
Explanation:
cot C = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}\)
[∴ In ΔABC, By Pythagoras theorem,
AC2 = AB2 + BC2
AB = 4 cm, AC = 5 cm]

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 304.
The value of sec C is
Answer:
\(\frac{5}{3}\)
Explanation:
sec C = \(\frac{A C}{B C}=\frac{5}{3}\)

Question 305.
sin2 C + cos2 C is equal to
Answer:
1
In figure, ΔABC has a right angle at B. If AB = BC = 1cm and AC = √2 cm.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 37
Explanation:
sin C= \(\frac{4}{5}\) ;cosC = \(\frac{3}{5}\)
L.H.S. = sin2 C + cos2 C
= \(\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}\)
= \(\frac{16+9}{25}\) = 1 = R.H.S.

Question 306.
Find sin C.
Answer:
\(\frac{1}{\sqrt{2}}\)
Explanation:
sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}\)

Question 307.
Find cos C.
Answer:
\(\frac{1}{\sqrt{2}}\)
Explanation:
cos C =\(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}\)

Question 308.
Find tan C.
Answer:
1
The length of a pendulum is 80 cm. Its end describes an arc of length 16 cm.
Explanation:
tan C = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{1}\) = 1

Question 309.
To find the length of the arc which formula is useful?
Answer:
l = r0

Question 310.
Calculate the angle of arc makes at centre.
Answer:
θ = \(\frac{1}{r}=\frac{16}{80}=\frac{1}{5}\)
In ΔPQR, right angled at Q,
PR + QR = 25 cm and PQ = 5 cm

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 311.
Determine the value of “QR”.
Answer:
QR = 12 cm

Question 312.
Determine the value of “PR”.
Answer:
PR = 13 cm

Question 313.
Find the value of sin P.
Answer:
\(\frac{12}{13}\) cm

Question 314.
Find the value of cos P.
Answer:
\(\frac{5}{13}\) cm

Question 315.
Find the value of tan P.
Answer:
\(\frac{12}{5}\) cm

Question 316.
In ΔABC, ∠B = 90°, AB = 3 cm and BC = 4 cm, then match the column.
A) sin C [ ] i) 3/5
B) tanA [ ] ii) 4/5
C) cos C [ ] iii) 5/3
D) sec A [ ] iv) 4/3
Answer:
A – (i), B – (iv), C – (ii), D – (iii)

Question 317.
Match the following.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 38
Answer:
A – (iv), B- (ii), C – (iii), D – (i)

AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers

Question 318.
Match the following.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 39
Answer:
A – (ii), B- (i), C – (iii), D – (iv)

Question 319.
If sin θ = \(\frac{7}{25}\), then
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 40
Answer:
A – (i), B- (iii), C – (ii), D – (iv)

Question 320.
Match the following.
AP 10th Class Maths Bits Chapter 11 Trigonometry with Answers 41
Answer:
A – (iv), B – (iii), C – (v), D – (ii), E – (i)

Question 321.
What is the value of
sec 16°- cosec 74° – cot 74° • tan 16° ?
Answer:
1 (one)

Question 322.
If x = 2019°, then what is the value of sin2 x + cos2 x ?
Solution:
If x = 2019°, then
sin2x + cos2x = sin22019° + cos22019° = 1 [∵ sin2θ + cos2θ = 1]

Question 323.
If x is in first quadrant and sin x = cos x, then what is the value of x?
Solution:
Given, sin x = cos x
We know, sin (90°- θ) = cos θ
So, cos x = sin(90° – x)
⇒ sin x = sin(90° – x)
[note : If sin A = sin B, then A = B]
⇒ x = 90° – x
⇒2x = 90°
∴ x = 45°

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Practice the AP 10th Class Maths Bits with Answers Chapter 7 Coordinate Geometry on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 7th Lesson Coordinate Geometry with Answers

Question 1.
Write the nearest point to origin,
i) (2, – 3)
ii) (5, 0)
iii) (0, – 5)
iv) (1, 3)
Answer:
(1,3)

Question 2.
The distance of a point (3, 4) from the origin is how many units ?
Answer:
5 units.

Question 3.
Write the formula to find the area of a triangle.
Answer:
Δ = \(\frac { 1 }{ 2 }\) bh and
Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Question 4.
Find the mid point of (2, 3) and (-2,3).
Answer:
(0, 3)

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 5.
Find the distance to X – axis from the point (3, – 4).
Answer:
4 units

Question 6.
Find the centroid of the triangle formed by these points (0, 3); (3, 0) and (0, 0).
Answer:
(1, 1)

Question 7.
Where do the points lie on co-ordinate axis ?
(- 4, 0), (2, 0), (6, 0), (- 8, 0)
Answer:
On X-axis.

Question 8.
The graph of y = 5 represents.
Answer:
Parallel to X – axis.

Question 9.
Find sum of the distances from A(3, 4) to X – axis and from B(5, 7) to Y – axis.
Answer:
9 units.

Question 10.
Find the distance from origin to (2,3).
Answer:
\(\sqrt{13}\) units.

Question 11.
Find slope of the line passing through the points (0, sin 60°) and (cos 30°, 0).
Answer:
m = – 1

Question 12.
If the mid point of (x – y, 8) and (2, x + y) is (5, 10), then find (x, y).
Answer:
(10,2)

Question 13.
Where the point (0, 5) lies ?
Answer:
On Y – axis.

Question 14.
Find the area of a triangle whose verti-ces (points) Eire (0, 0), (3, 0) and (0, 4).
Answer:
6 sq. units.

Question 15.
Write the slope of Y – axis.
Answer:
Not defined.

Question 16.
Find the mid point of line segment joined by (4, 5) and (- 6, 3).
Answer:
(-1,4)

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 17.
(x, y), (2,0), (3,2) and (1,2) are vertices of a parallelogram, then find (x, y).
Answer:
(0, 0)

Question 18.
Find centroid of a triangle, whose ver-tices are (- a, 0), (0, b) and (a, 0).
Answer:
(0, \(\frac { b }{ 3 }\))

Question 19.
Find the distance between two points A (a cos 0, 0), B (0, a sin 0).
Answer:
a units.

Question 20.
Find the distance between (0, 0), (x1, y1) points.
Answer:
\(\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}}\)

Question 21.
If A(log2 8, log5 25) and B(log10 10, log10 100), then find the mid-point of AB.
Answer:
(2,2)

Question 22.
Find the distance between (0, 7) and (- 7, 0).
Answer:
7\(\sqrt{2}\) units.

Question 23.
Find slope of the line passing through the points (- 1, 1) and (1, 1).
Answer:
0

Question 24.
Find the slope of the line passing through the points (4, 6) and (2, – 5).
Answer:
\(\frac { 11 }{ 2 }\)

Question 25.
In the given figure, find the area of ΔOAB.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 1
Answer:
6 sq. units.

Question 26.
A line makes 45° with X – axis, then find its slope.
Answer:
m = tan θ = tan 45° = 1

Question 27.
If a line is passing through (2, 3) and (2, – 3), then write the nature of that line.
Answer:
The line is parallel to Y – axis and The slope of the line is not defined.

Question 28.
Find area of the triangle formed by the points A(0, 0), B(1, 0) and C(0, 1).
Answer:
\(\frac { 1 }{ 2 }\) sq. units.

Question 29.
Find the distance from X – axis to (- 4, 3) is units.
Answer:
3 units.

Question 30.
Find the area of the triangle BOA is …………… sq. units.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 2
Answer:
3 sq. units.

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 31.
Find the slope of the line that passes through the points P (x1, y1) and Q(x2, y2) and making an angle ‘θ’ with X – axis.
Answer:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Question 32.
The area of given triangle is 60 sq. units, then find x = …………units.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 3
Answer:
12 units.

Question 33.
A line makes 45° with X – axis, then find its slope.
Answer:
1

Question 34.
Find the distance between the points (x1, y1) and (x2, y2) which are on the line parallel to Y – axis.
Answer:
|y – y2| or |y2 – y1 |

Question 35.
If the co-ordinates of the vertices of a rectangle are (0, 0), (4, 0), (4, 3) and (0, 3), then find the length of its di¬agonal.
Answer:
5 units.

Question 36.
Find the distance from Y-axis to (4, 0) is ……………. units.
Answer:
4 units.

Question 37.
Draw the graph represented by y = x.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 4

Question 38.
If origin is the centroid of a triangle, whose vertices are (3, 2), (- 6, y) and (3, – 2), then calculate ‘y’.
Answer:
y = 0

Question 39.
In a coordinate plane, if line segment AB is parallel to X – axis, then write about points A and B.
Answer:
X – coordinates of points A and B are equal.

Question 40.
Find the distance between the points (0, 7) and (- 7, 0).
Answer:
7\(\sqrt{2}\) units.

Question 41.
Find the distance of the point (- 8, 3) from the origin.
Answer:
\(\sqrt{73}\) units

Question 42.
If points (x, 0), (0, y) and (1, 1) are collinear, then find \(\frac{1}{x}+\frac{1}{y}\).
Answer:
1

Question 43.
Write a point on the X – axis is of the form.
Answer:
(x, 0)

Question 44.
Find the points (- 3, 0), (0, 5) and (3, 0) are the vertices of which type of triangle ?
Answer:
Isosceles triangle.

Question 45.
Find the area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b).
Answer:
0

Question 46.
Write a point on the Y – axis is of the form.
Answer:
(0, y)

Question 47.
The point which divides the line segment joining the points (3, 4) and (7, – 6) internally in the ratio 1 : 2 lies in the quadrant.
Answer:
Q4

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 48.
Find the distance between the points (- 2, 3) and (2, – 3).
Answer:
\(\sqrt{52}\) units.

Question 49.
AOBC is a rectangle whose three ver-tices are A(4, 0), B(0, 3) and O (0, 0), then find its diagonal ?
Answer:
5 units.

Question 50.
Find the distance of the point (-8, -7) from Y – axis.
Answer:
8 units.

Question 51.
A circle is drawn with origin as centre and passing through (2, 3), then find its radius.
Answer:
\(\sqrt{13}\) units.

Question 52.
Find the perimeter of a triangle whose vertices are A(12, 0), 0(0,.0) and B(0, 5).
Answer:
30 units.

Question 53.
Find the distance of the point (- 4, 3) from X – axis.
Answer:
3 units.

Question 54.
If the distance between the points (4, y) and (1, 0) is 5, then find ‘y’.
Answer:
y = ± 4.

Question 55.
Write the distance of (x, y) from X-axis.
Answer:
y units.

Question 56.
Find the distance of the point (- 9, 40) from the origin.
Answer:
41 units

Question 57.
If (0, 0), (a, 0) and (0, b) are collinear, then write the relation between ‘a’ and b’.
Answer:
ab = 0

Question 58.
Which ratio the centroid divides each median ?
Answer:
2:1

Question 59.
Find the value of ‘p’ if the distance be-tween (2, 3) and (p, 3) is 5. ,
Answer:
p = 7

Question 60.
Find the angle between X – axis and Y – axis.
Answer:
90°

Question 61. Find the distance between the points (a cos θ, 0) and (0, a sin θ).
Answer:
‘a’ units

Question 62.
(- 2, 8) belongs to which quadrant ?
Answer:
Q2

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 63.
Find the centroid of the triangle whose vertices are (2, – 3), (4, 6), (- 2, 8).
Answer:
(\(\frac{4}{3}, \frac{11}{3}\))

Question 64.
Guess shape of the closed figure formed by the points (- 2, 0), (2, 0), (2, 2), (0, 4) and (-2,-2).
Answer:
Pentagon

Question 65.
Find the midpoint of the line joining of (2, 3) and (- 2, 3).
Answer:
(0, 0)

Question 66.
If the centroid of the triangle formed with (a, b); (b, c) and (c, a) is O (0, 0), then the value of a3 + b3 + c3.
Answer:
3 abc

Question 67.
If the points (a, 2a), (3a, 3a) and (3,1) are collinear, then find k.
Answer:
k = \(\frac { -1 }{ 3 }\)

Question 68.
Write the coordinates of the midpoint joining P(x1, y1) and Q(x2, y2).
Answer:
\(\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)\)

Question 69.
Find the slope of line joining of (5, -1), (0, 8).
Answer:
\(-\frac{9}{5}\) =m

Question 70.
If the distance between the points (3, k) and (4, 1) is \(\sqrt{10}\), then find the value of k.
Answer:
4 (or)-2.

Question 71.
If (- 2, – 1), (a, 0), (4, b) and (1,2) are the vertices of a parallelogram, then find a’.
Answer:
a = 1

Question 72.
Write the slope of X – axis.
Answer:
0

Question 73.
P(2, 2), Q(- 4, 4) and R(5, – 8) are the vertices of a ΔPQR, then find length of median from ‘R’.
Answer:
\(\sqrt{157}\) units.

Question 74.
Find the value of ‘k’ if the distance be-tween (2, 8) and (2, k) is 3.
Answer:
k = 5.

Question 75.
Find the distance of a point (α, β) from the origin.
Answer:
\(\sqrt{\alpha^{2}+\beta^{2}}\)

Question 76.
If the points (1, 2), (-1, x) and (2, 3) are collinear, then find the value of x.
Answer:
x = 0.

Question 77.
If (- 2,8) and (6, – 4) are the end points of the diameter of a circle, then find the centre of the circle.
Answer:
(2, 2) = centre.

Question 78.
A(0, -1), B(2, 1) and C(0, 3) are the vertices of AABC, then find median
through ‘B’ has a length . units.
Answer:
2

Question 79.
Two vertices of a triangle are (3, 5) and (- 4, – 5). If the centroid of the triangle is (4, 3), find the third vertex.
Answer:
(13, 9).

Question 80.
If A, B, C are collinear, then find the area of AABC.
Answer:
Δ = 0

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 81.
A circle drawn with origin as centre 13
passes through (\(\frac { 13 }{ 2 }\),0). Find the point which doesn’t lie in the interior of the circle.
Answer:
(-6,3)

Question 82.
Find area of triangle formed by (-4, 0), (0, 0) and (0, 5) is ……………… sq. units.
Answer:
10 sq.units.

Question 83.
Find the ratio in which the point (4, 8) divide the line segment joining the points (8, 6) and (0, 10).
Answer:
1:1

Question 84.
Write a formula to the coordinates of the point which divides the line join¬ing (x1, y1) and (x2, y2) in the ratio m: n internally.
Answer:
P = \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)

Question 82.
If A(4, 0), B(8, 0), then find \(\overline{\mathbf{A B}}\).
Answer:
4 units.

Question 83.
Find the slope of the line \(\frac{\mathbf{x}}{\mathbf{a}}+\frac{\mathbf{y}}{\mathbf{b}}\) = 1.
Answer:
m = \(\frac{-\mathrm{b}}{\mathrm{a}}\)

Question 87.
Find .the radius of the circle whose centre is (3, 2) and passes through (- 5, 6) is……………..units.
Answer:
4\(\sqrt{5}\) units.

Question 88.
In Heron’s formula ‘s’ represents.
Answer:
s = \(\frac{a+b+c}{2}\) = Semi perimeter

Question 89.
Slope of the line joining the points (2, 5) and (k, 3) is 2, then find k.
Answer:
k = 1.

Question 90.
If A(4, 5), B(7, 6), then find \(\overline{\mathbf{A B}}\).
Answer:
\(\sqrt{10}=\overline{\mathrm{AB}}\)

Question 91.
Write the distance of (x, y) from Y-axis.
Answer:
‘x’units.

Question 92.
A(2, 0), B(l, 2), C(l, 6), then find ∆ABC.
Answer:
∆ = 0, so the points are collinear.

Question 93.
Find the mid point of the line joining the points (1,1) and (0, 0).
Answer:
( \(\frac{1}{2}, \frac{1}{2}\) )

Question 94.
How much the slope of vertical line ?
Answer:
Not defined.

Question 95.
A(1, – 1), B(0, 6) and C(- 3, 0), then find G (centroid).
Answer:
G = (\(\frac{-2}{3}, \frac{5}{3}\))

Question 96.
A(a, b) and B(- a, – b), then find \(\overline{\mathbf{B A}}\).
Answer:
\(\overline{\mathrm{BA}}=2 \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\)

Question 97.
Find the centroid of the triangle formed with the line x + y = 6 with the coordinate axes.
Answer:
G .= (2, 2)

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 98.
Find the area of triangle formed with (- 5,-1), (3,-5) and (5, 2).
Answer:
32 sq. units.

Question 99.
How much the slope of horizontal line?
Answer:
0 = m

Question 100.
Find the angle between the lines x = 2 and y = 3.
Answer:
θ = 90°.

Question 101.
Write the slope of the line y = mx.
Answer:
‘m’

Question 102.
The midpoint of the line joining the points (1, 2) and (1, p) is (1, – 1), then find p.
Answer:
p = – 4.

Question 103.
Name the point of concurrence of me-dians of a triangle is called
Answer:
Centroid

Question 104.
If AC = AB + BC, then the points A, B, C are called points.
Answer:
Collinear

Question 105.
ax + by + c = 0, represents a
Answer:
Straight line

Question 106.
If the points (k, k), (2, 3) and (4, – 1) are collinear, then find k.
Answer:
\(\frac{7}{3}\) = k

Question 107.
Write other name for x-coordinate of a points.
Answer:
Abscissa

Question 108.
Find the slope of the line joining the points A(-1.4, – 3.7) and B(-2.4, 1.3).
Answer:
m = – 5

Question 109.
If a < 0, then (- a, – a) belongs to which quadrant ?
Answer:
Q1

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 110.
If θ is the angle made by a line with X – axis, then find slope’m’.
Answer:
m = tan θ

Question 111.
Find the area of square formed with the vertices (0, – 1), (2, 1), (0, 3) and (-2, 1) taken in order as vertices.
Answer:
∆ = 8 sq. units.

Question 112.
Name the person who was introduced coordinate geometry.
Answer:
Rene Descartes

Question 113.
Find the coordinates of centroid of the triangle formed with the vertices (-1,3), (6, -3) and (-3, 6).

Question 114.
In quadrilateral ABCD, AB = BC = CD = AD and \(\overline{\mathbf{A C}} \neq \overline{\mathbf{B D}}\), then it is Answer:……………..type of quadrilateral.
Answer:
Rhombus

Question 115.
Write the slope of the line joining the points (2a, 3b) and (a, – b).
Answer:
m = \(\frac{4 \mathrm{~b}}{\mathrm{a}}\)

Question 116.
Write a formula to distance of (x, y) from origin.
Answer:
\(\sqrt{x^{2}+y^{2}}\)

Question 117.
In rhombus all sides are……………….
Answer:
Equal in length.

Question 118.
If the slope of a line passing through (- 2, 3) and (4, a) is \(\frac { -5 }{ 3 }\), then find Answer:
Answer:
a = – 7.

119.
A(2a, 4a), B(2a, 6a), C(2a+ \(\sqrt{3}\), 5), then write ΔABC is a type of tri¬
angle.
Answer:
Equilateral triangle.

Question 120.
How much each angle in equilateral triangle ?
Answer:
60° = each angle.

Question 121.
In the below figure, G is the centroid then AG : GD = ………………
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 5
Answer:
2:1

Question 122.
In the below figure AD : GD =
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 6
Answer:
3 : 1

Question 123.
Write the midpoint of a line segment divides it in the ratio.
im
Answer:
1 : 1

Question 124.
If the distance between the points (x1, y1) and (x2, y2) is |x1 – x2|, then they are parallel to ……………..
Answer:
x-axis.

Question 125.
Find slope of the line joining the points A(0,0), B(1/2,1/2)
Answer:
1 = m

Question 126.
Write the equation of X – axis.
Answer:
Y = 0

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 127.
Write the point of concurrence of alti-tudes of a triangle is called ……………..
Answer:
Orthocentre

Question 128.
P(cos θ, – cos θ), Q (sin θ, sin θ), then find \(\overline{\mathbf{P Q}}\).
Answer:
\(\sqrt{2}\) units.

Question 129.
Diagonals in a parallelogram …………….. to each other.
Answer:
Bisect

Question 130.
Find slope of the line 3x – 2 = 0.
Answer:
Not defined = (\(\frac{0}{3}\))

Question 131.
If A(p, q), B(m, n) and C(p – m, q – n) are collinear, then find pn.
Answer:
qm

Question 132.
Write Y-axis can be represented as.
Answer:
X = 0

Question 133.
Write number of medians of a triangle.
Answer:
3

Question 134.
A(cot θ, 1), B(0, 0), then find \(\overline{\mathbf{B A}}\).
Answer:
cosec θ

Question 135.
If the point (4 – p) lie on X – axis, then find the value of p2 + 2p – 1.
Answer:
– 1

Question 136.
A(t, 2t), B(- 2, 6), C(3, 1) and ΔABC = 5 sq.units, then find ‘t’.
Answer:
t = 2

Question 137.
y-intercept of the line x – 2y + 1 = 0 is …………..
Answer:
b = \(\frac { 1 }{ 2 }\)

Question 138.
In the below figure find ‘x’.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 7
Answer:
-9 = x

Question 139.
In the below figure find ‘y’.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 8
Answer:
3 = y

Question 140.
Where the X and Y axes will inter¬sects ?
Answer:
(0, 0)

Question 141.
If the point (a, 5) lies on Y – axis find the value of ‘a’.
Answer:
a = 0.

Question 142.
Write (3, 0), (8, 0), (1/2, 0) points lie on ………….
Answer:
X – axis.

Question 143.
Nature of the line that does not pass through origin and having a zero slope is
Answer:
Parallel to X – axis.

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 144.
Y-intercept of the line y = mx + c is ….
Answer:
‘c’

Question 145.
In ΔABC, all the side are different, then it is called type of triangle.
Answer:
Scalene

Question 146.
A = (\(\frac{1}{2}, \frac{3}{2}\)) , B = (\(\frac{3}{2}, \frac{-1}{2}\)) then find \(\overline{\mathbf{B A}}\)
Answer:
\(\sqrt{5}\)

Question 147.
Find the area of below parallelogram, if ΔABC = 5 sq. units.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 9
Answer:
10 sq. units

Question 148.
Find x-intercept of the line x – y +1 =0.
Answer:
– 1

Question 149.
In ΔPQR, PQ = QR, then it is called ……………… triangle.
Answer:
Isosceles

Question 150.
If (1, x) is at \(\sqrt{10}\) units from origin, then find the value of ‘x’.
Answer:
x = ± 3

Question 151.
A(1, – 1), B(2 1/2, 0), C(4, 1), then find area of ΔABC.
Answer:
Δ = 0.

Question 152.
Name the line joining the mid point of one side of a triangle from opposite vertex is called …………….
Answer:
Median

Question 153.
Find angle made by the line y = x with the positive direction of X – axis.
Answer:
45°.

Choose the correct answer satistying the following statements.

Question 154.
Statement (A): The point (0, 4) lies on Y – axis.
Statement (B) : The X co-ordinate of the point on Y – axis zero.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 155.
Statement (A): The value of y is 6, for which the distance between the points P(2, – 3) and Q(10, y) is 10.
Statement (B): Distance between two given points A(x1, y1) and B(x2, y2) is given by
AB = \(\sqrt{\left(x_{2}+x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 156.
Statement (A) : The point (- 1, 6) di¬vides the line segment joining the points (- 3, 10) and (6, – 8) in the ratio 2 : 7 internally.
Statement (B): Three points A, B and C are collinear if area of AABC = 0.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 157.
Statement (A) : Centroid of a triangle formed by the points (a, b), (b, c) and (c, a) is at origin, then a + b + c = 0.
Statement (B) : Centroid of a AABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is given by
\(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 158.
Statement (A): The area of the triangle with vertices (- 5,-1), (3, – 5), (5, 2) is 32 square units.
Statement (B): The point (x, y) divides the line segment joining the points A(xj, y2) and B(x2, y2) in the ratio k : 1 externally, then
\(x=\frac{k x_{2}+x_{1}}{k+1}, y=\frac{k y_{2}+y_{1}}{k+1}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 159.
Statement (A): The ratio in which the segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6) is 2 : 7.
Statement (B) : If A(x1, y1), B(x2, y2) are two points. Then the point C(x, y) such that C divides AB internally in the ratio k : 1 is given by
x = \(\frac{\mathrm{kx}_{2}+\mathrm{x}_{1}}{\mathrm{k}+1}, \mathrm{y}=\frac{\mathrm{ky}_{2}+\mathrm{y}_{1}}{\mathrm{k}+1}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 160.
Statement (A) : If three vertices of a parallelogram taken in order are (- 1, – 6), (2, – 5) and (7, 2), then its fourth vertex is (4, 1).
Statement (B) : Diagonals of a paral-lelogram bisect each other.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 161.
Statement (A) : The points (k -I- 1, 1), (2k + 1, 3) and (2k + 2, 2k) are col- linear, then k = 4.
Statement (B) : Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear if and only if
x1(y2 – y3) + x2(y3-y1) + x3(y1-y2) = 0
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 162.
Statement (A) : Let the vertices of a ΔABC are A(- 5, – 2), B(7, 6) and C(5, – 4), then coordinate of circum- centre is (1, 2).
Statement (B) : In a right angle tri¬angle, mid-point of hypotenuse is the circumcentre of the triangle.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 163.
Statement (A): If A(2a, 4a) and B(2a, 6a) are two vertices of a equilat¬eral triangle ABC, then the vertex C is given by (2a + a\(\sqrt{3}\) , 5a).
Statement (B): In equilaterahtriangle all the coordinates of three vertices can be rational.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 164.
Statement (A) : The equation of the straight line which passes through the point (2,-3) and the point of the inter-section of the lines x + y + 4 = 0 and 3x – y – 8 = 0 is 2x – y – 7 = 0.
Statement (B) : Product of slopes of two perpendicular straight lines is – 1.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Read Che below passages and answer to the following questions.

Let there be two points (4, 1) and (5,-2) in a two dimensional coordi¬nate system. A line which passes through the above give points and intersects the coordinate axes forms a triangle.

Question 165.
Write the equation of the line passing through the above given points.
Answer:
3x + y – 13 = 0.

Question 166.
Find the point of intersection of the above line with both the coordinate axes.
Answer:
(\(\frac { 13 }{ 3 }\),0) and (0, 13).

Question 167.
Find the area of the triangle so formed.
Answer:
\(\frac { 169 }{ 6 }\) sq. units.

In the diagram on a Lunar eclipse, if the position of Sun, Earth and Moon are shown by (- 4, 6) (k, – 2) and (5, – 6) respectively.

Question 168.
In Lunar eclipse what is the positions of Sun, Earth and Moon ?
Answer:
All are in same line, i.e., collinear.

Question 169.
To solve the above problem which mathematical concept is used ?
Answer:
Co-ordinate Geometry.

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 170.
Which formula is used to find the value of k ?
Δ = \(\frac { 1 }{ 2 }\) |x1(y2 – y3) + x2(y3-y1) + x3(y1 – y2) | = 0

Manowbhiram calculated the dis¬tance between T(5, 2) and R(-4, -1) to the nearest length is 9.5 units.

Question 171.
Do you agree with Manowbhiram ?
Answer:
Yes, I agree with him.

Question 172.
Which mathematical concept is used to you support Manowbhiram ?
Answer:
Co-ordinate Geometry (or) (Distance formula).

Question 173.
Column – II gives distance between pair of points given in column -I, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 10
Answer:
A – (iv), B – (i)

Question 174.
Column – II gives distance between pair of points given in column -I, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 11
Answer:
A – (ii), B – (iii)

Question 175.
Column – II gives the coordinates of the point ’p’ that divides the line segment join¬ing the points given in column -I, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 12
Answer:
A – (iv), B – (ii)

Question 176.
Column – II gives the coordinates of the point ‘p’ that divides the line segment join¬ing the points given in column -I, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 13
Answer:
A – (iii), B – (i).

Question 177.
Column – II gives the area of triangles whose vertices are given in column -1, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 14
Answer:
A – (iv), B – (iii).

Question 178.
Column – II gives the area of triangles whose vertices are given in column -1, match them correctly.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 15
Answer:
A – (ii), B – (i).

Question 179.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 16
Answer:
A – (iv), B – (iii).

Question 180.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 17
Answer:
A – (iii), B – (i).

AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers

Question 181.
Name the quadrilateral, which satis¬fies both the conditions given below.
Statement (A) : Diagonals are equal
Statement (B) : All sides are equal
a) Rhombus
b) Parallelogram
c) Rectangle
d) Square
Answer:
(d)

Question 182.
Find the area of the shaded triangle, in the figure given below.
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 18
Answer:
6 sq. units

Question 183.
What is the slope of the line joining the points (2, 0) and (- 2, 0).
Solution:
AP 10th Class Maths Bits Chapter 7 Coordinate Geometry with Answers 19

Question 184.
Name the point which is the point of intersection of medians of a triangle.
Answer:
Centroid of a triangle.

AP 10th Class Maths Bits Chapter 6 Progressions with Answers

Practice the AP 10th Class Maths Bits with Answers Chapter 6 Progressions on a regular basis so that you can attempt exams with utmost confidence.

AP SSC 10th Class Maths Bits 6th Lesson Progressions with Answers

Question 1.
Which term of Answer:P., 18, 15, 12, ………….. equal to ‘0’ ?
Answer:
7
Explanation :
a = 18, d = 15 – 18 = -3
an = 0 ⇒ a + (n – 1)d = 0
18 + (n-1)(-3) = 0
(n-1)(-3) = -18
n – 1 = 6 ⇒ n = 7

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 2.
Find the 21st term of an Answer:P., whose first two terms are – 3 and 4.
Answer:
137
Explanation :
a1 = -3, a2 = 4
⇒ d = 4 – (-3) = 7
a21 = a + 20d
= (-3) + 20(7)
= -3 + 140 = 137

Question 3.
Which term of G.P., 3, 3\(\sqrt{3}\) , 9,
equals to 243 ?
Answer:
9
Explanation :
AP 10th Class Maths Bits Chapter 6 Progressions Bits 8

Question 4.
If a, b, c are in G.P., then find b’.
Answer:
\(\sqrt{\mathrm{ac}}\)

Question 5.
Find the sum of 10 terms of the progression log 2 + log 4 + log 8 + log 16 + ………………..
Answer:
55 log 2

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 6.
Find nth term of a progression a, ar, ar2 ……………
Answer:
arn – 1

Question 7.
Find the sum of first 100 natural num-bers.
Answer:
5050

Question 8.
Find the common difference of Answer:P. log2 2, log2 4, log2 8.
Answer:
1
Explanation :
log22,log222 , log223
log22, 2 log22, 3 log22
1, 2, 3, ….
⇒ d = 1

Question 9.
In a GP, a1 = 20 and a4 = 540, then ‘ find ‘r’.
Answer:
3
Explanation :
a = 20, a4 = a – r3 = 540
⇒ 20.r3 = 540
⇒ r3 = \(\frac{540}{20}\) = 27 ⇒ r3 = 33 ⇒ r = 3

Question 10.
In an Answer:P., if a = 1, an = 20 and Sn = 399, then find ‘n’.
Answer:
38
Explanation :
AP 10th Class Maths Bits Chapter 6 Progressions Bits 9

Question 11.
Find the common difference of the AP x – y, x, x + y.
Answer:
y

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 12.
Find the common difference of the AP 1,-1, -3.
Answer:
– 2

Question 13.
If k, 2k + 1, 2k + 3 are three consecutive terms in Answer:P., then find the value of’k’.
Answer:
1
Explanation :
k + 1 – k = 2k + 3 – (2k + 1)
⇒ k + 1 = 2 ⇒ k = 1

Question 14.
Find the common difference in the AP 2a – b, 4a – 3b, 6a – 5b.
Answer:
2a – 2b.

Question 15.
Which term of the arithmetic progres-sion 24,21,18, is the first negative term ?
Answer:
10th term
Explanation :
a = 24, d = 21-24 = -3
an = a + (n – 1)d = 0
⇒ 24 + (n- 1) (-3) = 0
⇒ 24 – 3n + 3 = 0
⇒ 27 – 3n = 0
⇒ n = 9
∴ First negative term is ’10’.

Question 16.
Find the next term in Answer:P. \(\sqrt{3}, \sqrt{12}, \sqrt{27}\)
Answer:
\(\sqrt{48}\)

Question 17.
Find the common difference of an arithmetic progression in which
a25 – a12 = -52
Answer:
-4
Explanation :
a + 24d – (a + 11d) = – 52
an + 24d – a – 11d = – 52
⇒ 13d = – 52 ⇒ d = \(-\frac{52}{13}\) = – 4.

Question 18.
Find the sum of first ‘n’ odd natural numbers.
Answer:
n2

Question 19.
Find the common difference of an Answer:P. for which a18 – a14 = 32.
Answer:
8

Question 20.
Write a formula for sum of first ‘n’ terms in an AP.
Answer:
Sn = \(\frac{n}{2}\) [2a + (n – 1 )d] (or)
Sn = \(\frac{n}{2}\)[a + l]

Question 21.
Which term of the G.P. \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \ldots\) is \(\frac{1}{2187}\) ?
Answer:
7th

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 22.
If an = \(\frac{\mathbf{n}}{\mathbf{n}+\mathbf{1}}\) then find a2017
Answer:
\(\frac{2017}{2018}\)

Question 23.
In an arithmetic progression, 4th term is 11 and 7th term is 17, then find its common difference.
Answer:
2
Explanation :
a4 = a + 3d = 11 and a7 = a + 6d = 17
AP 10th Class Maths Bits Chapter 6 Progressions Bits 10

Question 24.
If x, x + 2, x + 6 are three consecutive terms in G.P. Find the value of’x’.
Answer:
2

Question 25.
The ’nth’ term of an Answer:P. is an = 3 + 2n, then find the common difference.
Answer:
2

Question 26.
If an = \(\frac{n(n+3)}{n+2}\)then find a17.
Answer:
\(\frac{340}{19}\)

Question 27.
In an AP an = \(\frac{5 n-3}{4}\), then find a7.
Answer:
8
Explanation :
a7 = \(\frac{5 \times 7-3}{4}=\frac{32}{4}\) = 8

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 28.
Find the common difference of an arithmetic progression, whose 3rd term is 5 and 7th</sup? term is 9.
Answer:
1

Question 29.
If 4, a, 9 are in G.P., then find ‘a’,
Answer:
±6

Question 30.
If \(-\frac{2}{7}\) x , \(-\frac{7}{2}\) are in Geometric Progression, then find the value of x.
Answer:
1

Question 31.
If (i) – 1.0, – 1.5, – 2.0, – 2.5,…… and
(ii)- 1,-3, -9, -27, ………
ate two progressions, then which of them is a geometric progression ?
A) (i) only
B) (ii) only
C) (i) and (ii) both
D) None of them
Answer:
B) (ii) only

Question 32.
In a G.P., a = 81, r = \(-\frac{1}{3}\), then find a3.
Answer:
9
Explanation :
a3 = Answer:r2 = 81. (\(\left(\frac{-1}{3}\right)^{2}\)) ⇒ a3 = 9

Question 33.
Write a G.P. with r = 2 and a = 7.
Answer:
7, 14, 28 ………………..

Question 34.
If a, b, c are in AP, then find ’b’.
Answer:
\(\frac{a+c}{2}\) = b.

Question 35.
3, \(\frac{3}{2}\) , \(\frac{3}{4}\), …………. then find ‘r’.
Answer:
r = \(\frac{t_{2}}{t_{1}}=\frac{\frac{3}{2}}{3}=\frac{3}{2} \times \frac{1}{3}=\frac{1}{2}\)

Question 36.
Find the sum of first 1000 positive integers.
Answer:
500500

Question 37.
In the AP – 9, – 14, – 19, – 24, …………….. then find the value of a30 – a20.
Answer:
-50
Explanation :
a = – 9, d = -14 + 9 = -5
a30 = a + 29d
= – 9 + 29 (- 5)
= -9-145 = -154
a20 = a + 19d
= -9 + 19 (-5)
= -9-95 = -104
∴ a30 – a20 = – 154 + 104 = – 50

Question 38.
If 4, x, 9 are in G.P., then find ‘x’.
Answer:
6

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 39.
Find 15th term of the AP x – 7, x – 2, x + 3, …………………….
Answer:
x + 63

Question 40.
\(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\) find a7.
Answer:
\(\frac{1}{2187}\)

Question 41.
Write a formula 12 + 22 + 32 + … + n2.
Answer:
Σn2 = \(\frac{n(n+1)(2 n+1)}{6}\)

Question 42.
1 + \(\frac{1}{2}+\frac{1}{2^{2}}\) + ………….. then find ‘r’.
Answer:
\(\frac { 1 }{ 2 }\)

Question 43.
Find the common ratio of the G.P. 2, \(\sqrt{8}\), 4.
Answer:
\(\sqrt{2}\)

Question 44.
1,4,7,10, ……………… find ‘d’.
Answer:
3

Question 45.
If an = \(\frac{\mathbf{n}}{\mathbf{n}+\mathbf{2}}\), then find a3.
Answer:
\(\frac{3}{5}\)

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 46.
Write an in G.P.
Answer:
arn-1 = an

Question 47.
2, \(\frac{5}{2}\), 3, find S25.
Answer:
S25 = 200
Explanation :
a = 2, d = \(\frac{5}{2}\) – 2 = \(\frac{1}{2}\)
S25 = \(\frac{n}{2}\)[2a + (n- 1) d]
= \(\frac{25}{2}\) [2(2) + 24 (1/2)]
= \(\frac{25}{2}\)[4 + 12]
= \(\frac{25}{2}\) x 16 = 25 x 8 = 200

Question 48.
Write G.M of a and b.
Answer:
\(\sqrt{\mathrm{ab}}\) = G.M.

Question 49.
In an Answer:P. a1 = -4, a6 = 6, then find a2
Answer:
-2.

Question 50.
If a, b, c are in Answer:P., then find ‘b’.
Answer:
\(\frac{a+c}{2}\) = b.

Question 51.
Which term of the G.P. 2, 6, 18, 54,…. is 2 x 310 ?
Answer:
11th term
Explanation :
a = 2, r = 3, an = 2 x 310
Answer:rn-1 = 2x 310
2 x 3n-1 = 2 x 310
⇒ n – 1 = 10 ⇒ n = 11

Question 52.
Find the sum of 15 terms of the Answer:P. 4, 7, 10, ………………
Answer:
375

Question 53.
Which term of the Answer:P. 100, 90, 80, …………… is zero ?
Answer:
11th term
Explanation :
a = 100, d = 90 – 100 = -10
an = a + (n – 1)d = 0
⇒ 100 + (n- 1)(- 10) = 0
⇒ (n – 1) (- 10) = – 100
⇒ n- 1 = 10
⇒ n = 11

Question 54.
Is the numbers, – 15, -11,-7, – 3, are in Answer:P. ? If so, find’d’.
Answer:
Yes, it is in Answer:P., with d = 4.

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 55.
\(\frac{1}{\sqrt{2}},-2, \frac{8}{\sqrt{2}}\) find a5
Answer:
\(32 \sqrt{2}\)

Question 56.
Find AM of 24 and 16.
Answer:
20

Question 57.
In the Answer:P. – 11,-9, – 7, find d’.
Answer:
2

Question 58.
Which term of Answer:P. 21, 18, 15, …………… is -81 ?
Answer:
35th term

Question 59.
Write a G.P. your own with r = – 2.
Answer:
5,- 10, 20,-40, ……………

Question 60.
Find the sum of first ’n’ natural num-bers.
Answer:
1 + 2 + 3 + 4 ……………+ n = Σ \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Question 61.
In AP a12 = 37, d = 3, then find S12.
Answer:
246
Explanation :
a12 = 37, d = 3
⇒ a + 11d = 37
⇒ a + 11 x 3 = 37
⇒ a = 37 – 33 = 4
S12 = \(\frac { 12 }{ 2 }\)[2 x 4 + 11 x 3]
= 6[8 + 33] = 6 x 41 = 246

Question 62.
If 3, x, 11 are in Answer:P., then find ‘x’.
Answer:
7 = x
Explanation :
x – 3 = 11- x ⇒ 2x = 14 ⇒ x = 7

Question 63.
In an AP 7a7 = 11a11, then find a18.
Answer:
0 = a18

Question 64.
Find number of terms of the Answer:P.
-5 + (-8) + (- 11) + + (-230).
Answer:
76
Explanation :
a = -5, d = -8 + 5 = -3, an ⇒ a + (n – 1) d = – 230
⇒ – 5 + (n – 1) (- 3) = – 230
⇒ (n – 1)(-3) = -225
⇒ n – 1 = \(\frac{-225}{-3}\) = 75
⇒ n = 75 + 1 ⇒ n = 76

Question 65.
– 8, – 6, – 4, find a7.
Answer:
6 = a7.

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 66.
1, – 1, 1, – 1, 1, – 1, ………………upto 131 terms, then find S131.
Answer:
1 = S131.

Question 67.
ind the next term of the Answer:P. 51, 59, 67,75.
Answer:
83

Question 68.
an = 9 – 5n, find a4.
Answer:
-11
Explanation :
Explanation :
an = 9 – 5n
⇒ a4 = 9- 5×4 = 9-20 = -11

Question 69.
3, 6, 12, then find ‘r’.
Answer:
r = \(\frac{6}{3}=\frac{12}{6}\) = 2

Question 70.
Which term of Answer:P. 7 + 4 + 1 + is – 56 ?
Answer:
22

Question 71.
n – 1, n – 2, n – 3, ….. find a10.
Answer:
n – 10 = a10

Question 72.
Write Answer:M. of M, P, C.
Answer:
\(\frac{M+P+C}{3}\)

Question 73.
In the formula an = 3.6, a = – 18.9, d = 2.5, then find ‘n’.
Answer:
10
Explanation :
an = 3.6, a = – 18.9, d = 2.5
⇒ a + (n – 1)d = 3.6
⇒ – 18.9 + (n – 1)(2.5) = 3.6
⇒ (n – 1)(2.5) = 3.6 + 18.9 = 22.5
⇒ n-1 = \(\frac{22.5}{2.5}\) = 9
⇒ n = 9 + 1 = 10

Question 74.
Write G.M. of a’ arid \(\frac{1}{a}\).
Answer:
G.M. = 1

Question 75.
In a series an = \(\frac{n(n+1)}{3}\), find a2.
Answer:
a2 = 2.

Question 76.
If a, b, c are in Answer:P., then find b – Answer:
Answer:
c – b

Question 77.
\(\frac{1}{4}, \frac{-1}{4}, \frac{-3}{4}, \frac{-5}{4}\) ………….. find ‘d’
Answer:
d = \(\frac{-1}{2}\)

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 78.
Find 1 + 1 + 1 + + n terms.
Answer:
Sn = n

Question 79.
Find G.M. of x3 and \(\frac{1}{x^{3}}\)
Answer:
G.M. = 1

Question 80.
Write Answer:M. of x2 + y2 and x2 – y2.
Answer:
x2
Explanation :
Answer:M = \(\frac{x^{2}+y^{2}+x^{2}-y^{2}}{2}=\frac{2 x^{2}}{2}\) = x2

Question 81.
a, a2, a3, …… then find r.
Answer:
r = a .

Question 82.
Reciprocals of terms of G.P. are in which progression ?
Answer:
G.P.

Question 83.
2, – 6, 18, – 54,………….find r.
Answer:
-3.

Question 84.
Find the value of – 5 + (- 8) + (- 11) + ………….. + (-230).
Answer:
-8930

Question 85.
In a G.P. 25, – 5,1, \(-\frac{1}{5}\),…. then find ‘r’.
Answer:
\(-\frac{1}{5}\) = r

Question 86.
If 2, x, 6 are in G.P., then find ‘x’.
Answer:
\(2 \sqrt{3}\) = x.

Question 87.
In a G.P. a8 = 192, r = 2, then find a12.
Answer:
3072 = a12.

Question 88.
Which term of G.P., 2,8,32, is 512?
Answer:
5

Question 89.
1, 2, 3, ……….. find sum to ’10’ terms.
Answer:
S10 = 55
Explanation :
a = 1,d = 1
S10 = \(\frac{10}{2}\)[21 + 9 x 1]
= 5[2 + 9] = 5 x 11 = 55

Question 90.
\(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}\), …………. find an.
Answer:
\(\frac{5}{2^{n}}\) = an

Question 91.
4,16, , 256, ……….. then find
A,
64

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 92.
In the Answer:P. 100, 103, 106, find d’.
Answer:
d = 3

Question 93.
Find the Answer:P. with first term is 8 and common difference is 2\(\frac { 1 }{ 2 }\).
Answer:
8, 10\(\frac { 1 }{ 2 }\), 13, …………….

Question 94.
How many terms of Answer:P. – 6, \(\frac { -11 }{ 2 }\), – 5, ……………. are needed to obtain a sum – 25 ?
Answer:
5 or 20 terms.
Explanation :
a = -6,d = \(\frac{-11}{2}\) + 6 = \(\frac{1}{2}\) Sn = -25
Sn = \(\frac{n}{2}\)[2a + (n – 1)d] = -25
AP 10th Class Maths Bits Chapter 6 Progressions Bits 11
⇒ n (n – 25) – 100
⇒ n2 – 25n + 100 = 0
⇒ n2 – 20n – 5n + 100 = 0
⇒ n (n – 20) – 5 (n – 20) = 0
⇒ (n – 5) (n – 20) = 0
∴ n = 5 or 20

Question 95.
(a + 3d), (a + d), (a – d), ……….. find the next term of the Answer:P.
Answer:
a – 3d

Question 96.
Find the 103rd term of 1, -1,1,- 1, ….
Answer:
– 1

Question 97.
Find the sum of first 50 even numbers.
Answer:
2550
Explanation :
Sum of first 50 even numbers
= n(n + 1)
= 50(51) = 2550

Question 98.
Find the common ratio of the G.P.192, 36, 9, …………
Answer:
1/4

Question 99.
Find the 25th term of
– 300, – 290, – 280,
Answer:
-60.

Question 100.
a, b, c are in AP, then write, \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in which progression ?
Answer:
Harmonic progression (H.P)

Question 101.
In Answer:P. ap = q, aq = p, then find ap + q.
Answer:
0 = ap + q

Question 102.
a, b, c are in Answer:P., then 3a, 3b, 3c fire in which series ?
Answer:
In geometric Progression (G.P).

Question 103.
22, 32, 42, find a7.
Answer:
a7 = 82

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 104.
an = 2n, then find a5.
Answer:
32 = a5.

Question 105.
Write G.M. of 5 and 125.
Answer:
25

Question 106.
Σn = 10, then Σn3.
Answer:
1000
Explanation :
(Σn)3 = (10)3 = 1000

Question 107.
Write G.M. of x, y, z.
Answer:
G.M. = \(\sqrt[3]{x y z}\)

Question 108.
Find the value of 16 + 11 + 6 + …. 23 terms.
Answer:
S23 = – 897.

Question 109.
Write a formula to 13 + 23 + 33 +….+ n3
Answer:
\(\frac{n^{2}(n+1)^{2}}{4}=\Sigma n^{3}\)

Question 110.
an = (n – 1) (n – 2), then find a2.
Answer:
a2 = 1

Question 111.
If a, b, c are in G.P., then find \(\frac{\mathbf{b}}{\mathbf{a}}\)
Answer:
\(\frac{b}{a}=\frac{c}{b}\)

Question 112.
-1, \(\frac{1}{4}, \frac{3}{2}\), ……….. find sum to 10 terms.
Answer:
46.25

Question 113.
Find the sum of first 40 positive integers which are divisible by 6.
Answer:
4920
Explanation :
Sn = 6 + 12 + 18 + 24 + …. + 240
= \(6\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]=6\left[\frac{40 \times 41}{2}\right]\)
= 6 x 20 x 41
= 4920

Question 114.
If a, b, c are in G.P., then find b2.
Answer:
ac = b2

Question 115.
Calculate the common ratio of the G.P. 4, 20, 100, 500, ……………
Answer:
r = 5

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 116.
In the Answer:P., first term is 4 and common difference is – 1, then find Answer:P.
Answer:
4,3, 2, Answer:P.

Question 117.
Find the sum of first 20 odd numbers..
Answer:
400 = S20.

Question 118.
How many numbers are divisible by ‘4’ lying between 101 and 250 ?
Answer:
37

Question 119.
If a7 – a3 = 32, then the common dif-ference of the Answer:P.
Answer:
8 = d

Question 120.
Which term of the Answer:P. 125, 120, 115, ………… is the first negative ?
Answer:
7th term.

Question 121.
If a7 ÷ a4 of a G.P is 343, then find the common ratio.
Answer:
r = 7.

Question 122.
If x, xy, xy2, xy3, …. forms a G.P, then find its 15th term.
Answer:
xy14

Question 123.
Find the nth term of a, a + d, a + 2d,…
Answer:
a + (n – 1) d = an

Question 124.
In a G.P. write a6.
Answer:
ar5 = a6.

Question 125.
Calculate the 16th term of 4, – 4, 4, – 4, ……..
Answer:
4 = a6.

Question 126.
In Answer:P. aa12. = 37, d = 3, then find ‘a’.
Answer:
4 = Answer:

Question 127.
3, – 32, 33, find a6.
Answer:
– 729 = a6.

Question 128.
If there are’n’AM’s between’a’and h’, then find d.
Answer:
d = \(\frac{b-a}{n+1}\)

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 129.
7, 10, 13, find a5.
Answer:
19 = a5.

Question 130.
Find AM of 5 and 95.
Answer:
50 = AM

Question 131.
f 4, x, 16 are in G.P., then find ‘x’.
Answer:
8
Explanation :
b2 = ac ⇒ x = \(\sqrt{4 \times 16}=\sqrt{64}\) = 8

Question 132.
5, 1, – 3, – 7, find a10.
Answer:
– 31 = a10.

Question 133.
Write product of ‘n’ GM’s between a and b.
Answer:
(ab)n/2 = G.M.

Question 134.
If a, b, c are in GP, then b is called
Answer:
Geometric mean.

Question 135.
How many 3-digit numbers are divisible by 7 ?
Answer:
128

Question 136.
If a = 3 and a7 = 33, then find a11.
Answer:
53 = a11
Explanation :
a = 3,
a7 = a + (n – 1)d = 33
⇒ 3 + (7 – 1) d = 33
⇒ 6d = 30 ⇒ d = 5
a11 = a + 10d
= 3 + 10×5 = 3 + 50 = 53

Question 137.
Find the 10th term of the
AP:3, 11, 19, …………..
Answer:
75 = a10

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 138.
aa28 – aa23 = 15, then find the common difference of the Answer:P.
Answer:
3 = d

Question 139.
If x – 1, x + 3, 3x – 1 are in Answer:P., then find ‘x’.
Answer:
4 = x

Question 140.
Find the common ratio of the G.P.
3, 6, 12, 24, …………..
Answer:
2

Question 141.
Find the 8th term from the end of the Answer:P., 7, 10, 13, …………. 1814.
Answer:
163

Question 142.
\(\frac{\mathbf{b}+\mathbf{c}-\mathbf{a}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}-\mathbf{b}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}-\mathbf{c}}{\mathbf{c}}\) are in AP, then write \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in ,
Answer:
Arithmetic Progression.

Question 143.
Which term of the Answer:P, 24, 21, 18, …. is the first negative ?
Answer:
a10 is first negative term.

Question 144.
Find AM of 10 and 20.
Answer:
15 = AM .

Question 145.
Write the next term of the
Answer:P. \(\sqrt{48}, \sqrt{75}, \sqrt{147}, \ldots \ldots \ldots\)
Answer:
\(\sqrt{192}\)

Question 146.
– 20, – 18, – 16, ………… which term of this Answer:P. is a first positive term ?
Answer:
12

Question 147.
Find the 17th term of 1.1, 2.2, 3.3, 4.4 …………….
Answer:
18.7 = a17

Question 148.
If \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) is the AM of ‘a’and ‘b’, then find ‘n’.
Answer:
n = 0.

Choose the correct answer satisfying the following statements.

Question 149.
Statement (A): Common difference of the AP : – 5, – 1, 3, 7, is 4.
Statement (B): Common difference of the a, a + d, a + 2d, is given by
d = 2nd term – 1st term.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
Common difference, d = – 1 – (- 5)
= 4
So, both A and B are correct and B explains Answer: Hence, (i) is the correct option.

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 150.
Statement (A) : an – an-1 is not independent of n, then the given sequence is an AP.
Statement (B) : Common difference d = an – an-1 is constant or independent of n.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
We have, common difference of an Answer:P. d = an – an-1 is independent of ‘n’ or constant. So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 151.
Statement (A): The sum of the first ‘n’ terms of an AP is given by Sn = 3n2 -4n. Then its nth term an = 6n – 7.
Statement (B) : nth term of an AP, whose sum to ’n’ terms is Sn, is given
by an = Sn – Sn-1
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
nth term of an AP be an = Sn – Sn-1
⇒ an = 3n2 – 4n – 3(n – 1)2 + 4 (n – 1)
⇒ an = 6n – 7
So, both A and B are correct and B explains Answer: Hence, (i) is the correct option.

Question 152.
Statement (A) : Common difference of an AP in which a21 – a7 = 84 is 14. Statement (B) : nth term of an AP is given by an = a + (n – 1) d.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
We have, an= a + (n – 1)d
a21 – a7 = {a + (21 – 1)}d – {a + (7-l)d} = 84
⇒ a + 20d – a – 6d = 84
⇒ 14d = 84
⇒ d = 84/14 = 6
⇒ d = 6
So, A is incorrect but B is correct. Hence, (iii) is the correct option.

Question 153.
Statement (A) : Three consecutive terms 2k + 1, 3k + 3 and 5k – 1 from an AP, then k is equal to 6.
Statement (B) : In an AP
a, a + d, a + 2d, the sum to n terms of the AP be Sn = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d].
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
For 2k + 1, 3k + 3 and 5k- 1 to form an Answer:P.
(3k + 3)-(2k + 1) = (5k- l)-(3k + 3)
⇒ k + 2 = 2k-4
⇒ 2 + 4 = 2k – k = k
⇒ k = 6
So, both A and B are correct but B does not explain Answer:
Hence, (i) is the correct option.

Question 154.
Statement (A) : 10th term from the end of AP : 100, 95, 90, 85,……………. 10 is 55.
Statement (B): The nth term from the end of an AP having last term L and common difference d is L – (n – 1) d.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 155.
Statement (A) : If the sum of first ‘n’ terms of an AP is an2 + bn, then its common difference is 2Answer:
Statement (B): In an AP with first term a and last term l, sum of n terms is
given by Sn = \(\frac{n}{2}\)(a + l).
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 156.
Statement (A) : The sum of all natural numbers between 100 and 1000 which are multiple of 7 is 70336.
Statement (B) : The 10th term of an AP is 31 and 20th term is 71. Then t30 = 111
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 157.
Statement (A): 1, 2, 4, 8,……………. is a G.P., 4, 8, 16, 32 is a G.P. and 1 + 4, 2 + 8, 4 + 16, 8 + 32, …. is also a G.P.
Statement (B) : Let general term of a G.P. with common ratio ‘r’ be Tk + 1 and general term of another G.P., with common ratio ‘r’ be Tk + v then the series
whose general term Tk + 1 = Tk + 1 + Tk + 1 is also a G.P. with common ratio ‘r’.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation :
Let Tk + 1 = ark and Tk + 1 = brk
Since T”k + 1 = ark + brk = (a + b)rk ,
∴ T”k + 1 is general term of a G.P.
Option (i) is correct.

Question 158.
Statement (A) : 1111 …………. 1 (upto 91 terms) is a prime number.
Statement (B) : If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}, \frac{a+b-c}{c}\) are in Answer:P., then \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are also in Answer:P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation :
Since 11 11 ………..,
= \(\frac{10^{91}-1}{10-1}\) = divisible by 9.
The given number is not prime. So, A is false, but B is true.
∴ Option (iii) is correct.

Question 159.
Statement (A) : Let the positive numbers a, b, c be in Answer:P, then \(\frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ac}}, \frac{1}{\mathrm{ab}}\) are also in Answer:P.
Statement (B): If each term of an Answer:P. is divided by abc, then the resulting sequence is also in Answer:P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 160.
Statement (A) : Let three distinct positive real numbers a, b, c are in G.P., then a2, b2, c2 are in G.P.
Statement (B) : If we square each term of a G.P., then the resulting sequence is also in G.P.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 161.
Statement (A) : The sum of the series with the nth term, tn = (9 – 5n) is 465, when number of terms n = 15.
Statement (B) : Given series is in Answer:P. and sum of ‘n’ terms of an Answer:P. is
Sn = \(\frac{\mathrm{n}}{2}\)[2a + (n- 1) d]
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Read Che below passages and answer to the following questions.

Following given series are in Answer:P.
2, 4, 6, 8, …………..
3,6,9,12, ……….
First series contains 30 terms, while the second series contains 20 terms. Both of the above given series contains some terms, which are common to both of them.

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 162.
Find the last term of both the above given Answer:P. are
Answer:
(60, 60)
Explanation :
For 2, 4, 6, 8, …………..
Last term, t30 = 2 + (30 – 1)2
= 2 + 2 (29) = 60
3,6,9,12,
Last term, t20 = 3 + (20 – 1)3 x = 3 + 57 = 60

Question 163.
Find the sum of both the above given Answer:P. are
Answer:
(930, 630).
Explanation :
For 2, 4, 6, 8,
S30 = \(\frac { 30 }{ 2 }\) (2 + 60) = 930
For 3, 6, 9, 12,
S20 = \(\frac { 20 }{ 2 }\) (3 + 60) = 630

Question 164.
Write number of terms identical to both the above given Answer:P. are
Answer:
10
Explanation :
Let mth term of the first series is common with the nth term of the second series.
tm = tn
2 + (m – 1)2 = 3 + (n – 1) 3
2 + 2m -2 = 3 + 3n – 3
2m = 3n
\(\frac{m}{3}=\frac{n}{2}=k \text { (let) }\)
m = 3k, n = 2k.
Hence, k = 1, 2, 3, , 10.
[ ∵ 1 ≤ m ≤ 30, 1 ≤ n ≤ 20
1 ≤ 3k ≤ 30, 1 ≤ 2k ≤ 20
\(\frac{1}{2}\) ≤ k ≤ 10, \(\frac{1}{2}\) ≤ k ≤ 10]
For each value of k, we get one identical term.
Thus, number of identical terms =10.

There are 25 trees at equal distances of 5 m in a line with a well, the distance of the well from the nearest tree being 10 m. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next.

Question 165.
Where the well located in the garden?
Answer:
Obviously the well must be on one side of the trees.

Question 166.
How much the distance between the trees ?
Answer:
5 m.

Question 167.
Which mathematical concept is used to find the total distance the gardener will cover in order to water all the trees?
Answer:
Arithmetic Progression.

Question 168.
Column – II give common difference for Answer:P. given column -1, match them cor-rectly.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 1
Answer:
A – (iv), B – (iii).

Question 169.
Column – II give common difference for Answer:P. given column -1, match them cor. rectly.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 2
Answer:
A – (ii), B – (i).

AP 10th Class Maths Bits Chapter 6 Progressions Bits

Question 170.
Column – II give nth term for Answer:P. given column -I, match them correctly.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 3
Answer:
A – (iv), B – (iii).

Question 171.
Column – II give n,h term for Answer:P. given column -I, match them correctly.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 4
Answer:
A – (ii), B – (i).

Question 172.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 5
Answer:
A – (i), B – (ii).

Question 173.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 6
Answer:
A – (iv), B – (iii).

Question 174.
In which progression are the perimeters of triangles formed by joining the midpoints of sides of triangles succes-sively in the given figure.
AP 10th Class Maths Bits Chapter 6 Progressions Bits 7
Answer:
Geometric Progression.