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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Answer:
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)
cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)
tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)
∴ sin A = \(\frac{15}{17}\);
cos A = \(\frac{8}{17}\)
tan A = \(\frac{15}{8}\)

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.
Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR² = PQ² + PR²
(25)² = (7)² + PR²
PR² = (25)² – (7)² = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);
tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)
∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{(24)^{2}-(7)^{2}}{168}\)
= \(\frac{576-49}{168}\)
= \(\frac{527}{168}\)

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Answer:
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC² = AB² + BC²
(25)² = AB² + (24)²
AB² = 25² – 24² = 625 – 576
AB² = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)
tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)
Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A.
Answer:
From the identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\left(\frac{12}{13}\right)^{2}\)
= 1 – \(\frac{144}{169}\)
= \(\frac{169-144}{169}\)
= \(\frac{25}{169}\)
∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Answer:
Given 3 tan A = 4
⇒ tan A = \(\frac{4}{3}\)
From the identify sec² A – tan² A = 1
⇒ 1 + tan² A = sec² A

If cos A = \(\frac{3}{5}\) then from
sin² A + cos² A = 1
We can write sin²A = 1 – cos²A
= 1 – \(\left(\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}\)
⇒ sin² A = \(\frac{16}{25}\)
⇒ sin A = \(\frac{4}{5}\)
∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.
Answer:

In the given triangle, cos A = cos X
⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot² θ = (cot θ)²
= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ
So cot θ = \(\frac{7}{8}\)
⇒ tan θ = \(\frac{8}{7}\)
⇒ tan² θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)
From sec² θ – tan² θ = 1
⇒ 1 + tan² θ = sec² θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:
Given, tan A = \(\frac{\sqrt{3}}{1}\)
Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC² = AB² + BC²
(By Pythagoras theorem)
⇒ AC² = (1k)² + (√3k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k²
∴ AC = \(\sqrt{4 k^{2}}\) = 2k

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
Answer:
Given (sin θ + cos θ)² + (sin θ – cos θ)²
= (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ – 2 sin θ cos θ) [∵ (a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin² θ + cos² θ = 1]
= 1 + 1
= 2

iii) (sec² θ – 1) (cosec² θ – 1)
Answer:
Given (sec² θ – 1) (cosec² θ – 1)
= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1]
= tan² θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.
Show that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)²]; (a – b)(a + b) = a² — b²]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan² A
Answer:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec² A – tan² A [∵ sec² A – tan² A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Answer:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A . cosec A) + (cos² A – sec² A + 2 cos A . sec A) [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + cosec² A + 2 sin A . \(\frac{1}{\sin A}\) + sec² A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot² A) + 2 + (1 + tan² A) + 2
[∵ sin² A + cos² A = 1; cosec² A = 1 + cot² A; sec² A = 1 + tan² A]
= 7 + tan² A + cot² A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot² θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ]
= sin² θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec² θ – tan² θ = 1
sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec² θ – cot² θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a² – b² = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)

According to identity cos² θ + sin² θ = 1

Hence proved.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm
We know that ∠OPQ = 90°
Hence in △OPQ
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
12² = 5² + PQ²
∴ PQ² = 144 – 25 .
PQ² = 119
PQ = √119

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
Steps:

1. Draw a circle with some radius.
2. Draw a chord of the circle.
3. Draw a line parallel to the chord intersecting the circle at two distinct points.
4. This is secant of the circle (l).
5. Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:
Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP² + PQ² – OQ²
9² + PQ² = 15²
PQ² = 15² – 9²
PQ² = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.
O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.
Answer:

Let the height of the tower = h mts say
Width of the road be = x m.
Distance between two points of observation = 10 cm.
Angles of elevation from the two points = 60° and 30°.
From the figure
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
⇒ h = √3x …….(1)
Also tan 30° = \(\frac{h}{10+x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)
⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)
From equations (1) and (2) h
h = √3x = \(\frac{10+x}{\sqrt{3}}\)
∴ √3x = \(\frac{10+x}{\sqrt{3}}\)
√3 × √3x = 10 + x
⇒ 3x – x = 10
⇒ 2x = 10
⇒ x = \(\frac{10}{2}\) = 5m
∴ Width of the road = 5 m
Now Height of the tower = √3x = 5√3 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.
Answer:

Height of the temple = 30 m
Height of the man = 1.5 m
Initial distance between the man and temple = d m. say
Let the distance walked = x m.
From the figure
tan 30° = \(\frac{30-1.5}{d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)
∴ d = 28.5 × √3m ………(1)
Also tan 60° = \(\frac{28.5}{d-x}\)
⇒ √3 = \(\frac{28.5}{d-x}\)
⇒ √3(d-x) = 28.5
⇒ √3(28.5 × √3-x) = 28.5
⇒ 28.5 × 3 – √3x = 28.5
⇒ √3x = 3 × 28.5-28.5
⇒ √3x = 2 × 28.5 = 57
∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3
= 19 × 1.732
= 32.908 m.
∴ Distance walked = 32.908 m.

Question 3.
A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.
Answer:

Height of the pedestal = 2 m.
Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.
Angle of elevation of top of the pedestal = 45°.
Let the distance between the point of observation and foot of the pedestal = x m.
From the figure
tan 45° = \(\frac{2}{x}\)
1 = \(\frac{2}{x}\)
∴ x = 2 m.
Also tan 60° = \(\frac{2+h}{x}\)
⇒ √3 = \(\frac{2+h}{x}\)
⇒ 2√3 = 2 + h
⇒ h = 2√3 – 2
= 2(√3-1)
= 2(1.732 – 1)
= 2 × 0.732
= 1.464 m.

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.
Answer:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Distance between tower and building = 7 m.
Let the height of the building = x m and tower = y m.
From the figure
tan 45° = \(\frac{x}{7}\)
1 = \(\frac{x}{7}\)
∴ x = 7 m.
Also tan 60° = \(\frac{y-x}{7}\)
⇒ √3 = \(\frac{y-x}{7}\)
⇒ 7√3 = y – 7
∴ y = 7 + 7√3
= 7 (√3 + 1)
= 7(1.732 + 1)
= 2.732 × 7
= 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?
Answer:

Length of the wire = 18 m
Let the length of the wire removed = x
Height of the pole be = h
From the figure
sin 30° = \(\frac{h}{18}\)
⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)
⇒ h = \(\frac{18}{2}\) = 9 m
Also sin 60° = \(\frac{h}{18-x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)
√3(18-x) = 9 × 2
18√3 – √3x = 18
√3x = 18√3 – 18
√3x = 18(√3-1)
x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)
= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)
= 6√3(√3-1)
= 6(3-√3)
= 18 – 6√3
= 18 – 10.392
= 7.608 m.

Question 6.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
Answer:

Height of the tower = 30 m
Angle of elevation of the top of the tower = 60°.
Angle of elevation of the top of the building = 30°.
Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
From the figure
tan 60° = \(\frac{30}{d}\)
√3 = \(\frac{30}{d}\)
⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m
Also tan 30° = \(\frac{x}{d}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)
⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m
∴ Height of the building = 10 m

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:

Width of the road = 120 f.
Angle of elevation of the top of the 1st tower = 60°.
Angle of elevation of the top of the 2 tower = 30°.
Let the distance of the point from the 1st pole = x.
Then the distance of the point from
the 2nd pole = 120 – x.
and height of each pole = h say.
From the figure
tan 60° = \(\frac{h}{x}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x ……..(1)
Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ h = \(\frac{120-x}{\sqrt{3}}\)
From (1) and (2)
√3x = \(\frac{120-x}{\sqrt{3}}\)
⇒ √3.√3x = 120-x
⇒ 3x = 120 – x
⇒ 3x + x = 120
⇒ 4x = 120
⇒ x = \(\frac{120}{4}\) = 30 ft
Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet
∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
Height of each pole = 51.96 ft.

Question 8.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
Answer:

Let the height of the tower = h m.
Angles of elevation of the top of the tower from two points = x° and (90° – x)
From the figure
tan x = \(\frac{h}{4}\) ……. (1)
Also tan (90° – x) = \(\frac{h}{9}\)
⇒ cot x = \(\frac{h}{9}\)
⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)
∴ tan x = \(\frac{9}{h}\) …….. (2)
From (1) and (2)
tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)
∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)
h × h = 9 × 4
⇒ h² = 36
⇒ h = 6 m

Question 9.
The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)
Answer:

Height of the plane from the ground PM = RN = 1500√3 m.
Angle of elevation are 30° and 60°.
From the figure
tan 60° = \(\frac{PM}{QM}\)
√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)
QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
Also tan 30° = \(\frac{RN}{QN}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)
QM + MN = 1500√3 × √3
1500 + MN = 1500 × 3
MN = 4500 – 1500
MN = 3000 mts.
∴ Distance travelled in 15 seconds = 3000 mts.
∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s
= 200 × \(\frac{18}{5}\) kmph
= 720 kmph
Speed = 200 m/sec. or 720 kmph.

AP SSC 10th Class Maths Textbook Solutions

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.4

Question 1.
Find the height of the statue from the adjacent diagram.

Answer:
From the given diagram
Height of the statue (BC) = x m
Height of the Pedastal (AB) = 1.9 m
Total Height (AC) = 3.6 m
BC + AB = 3.6 m
⇒ x + 1.9 m = 3.6 m
⇒ x + 1.9 – 1.9 = 3.6 – 1.9 (Subtract 1.9 on both sides)
⇒ x = 1.7 m
∴ Height of the statue = 1.7 m

Question 2.
The sum of twice a number and 4 is 80, find the number.
Answer:
Let the number be x.
Twice the number = 2x
Sum of twice a number and 4 = 80
2x + 4 = 80
⇒ 2x + 4 – 4 = 80 – 4
(Subtract 4 on both sides)
⇒ 2x = 76
⇒ \(\frac{2 x}{2}\) = \(\frac{76}{2}\) (Divide by 2 on both sides)
⇒ x = 38
∴ Number = 38

Question 3.
The difference between a number and one-fourth of itself is 24, find the number.
Answer:
Let the number be a.
One-fourth of a number = \(\frac{1}{4}\) of a = \(\frac{a}{4}\)
Difference of number and one-fourth of itself = 24

⇒ a = 32
∴ The number a = 32

Question 4.

Find the value of x from the adjacent diagram.
Answer:

From the above diagram
AB = 12 cm; CD = x cm
EF = 5 cm; PQ = 24 cm
We know,
AB + CD + EF = PQ
⇒ 12 + x + 5 = 24
⇒ x + 17 = 24
⇒ x + 17 – 17 = 24 – 17
(Subtract 17 on both sides)
⇒ x = 7 cm

Question 5.
To convert temperature from Fahrenheit to Centigrade, we use the formula (F – 32) = \(\frac{9}{5}\) × C. If C = – 40° C, then find F.
Answer:
Given formula (F – 32) = \(\frac{9}{5}\) × C and C = – 40° C

⇒ F – 32 = 9(- 8)
⇒ F – 32 = – 72
⇒ F – 32 + 32 = – 72 + 32 (Add 32 on both sides)
⇒ F = – 40° F
Therefore we conclude that
– 40° C = – 40° F.

Question 6.
Rahim has ₹x, from which he spends ₹6. If twice of the money left with him is ₹86, then find x.
Answer:
Rahim has money = ₹x
Money after spends ₹6 = x – 6
Twice the money left with Rahim
= 2(x – 6) = ₹86
⇒ – 2(x – 6) = 86
⇒ \(\frac{2(x-6)}{2}\) = \(\frac{86}{2}\) (Divide by 2 on both sides)
⇒ x – 6 = 43
⇒ x – 6 + 6 = 43 + 6 (Add 6 on both sides)
⇒ x = 49
Therefore money at Rahim = ₹49.

Question 7.
The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
Answer:
Let the smaller number is x.
Larger number = x + 7
Six times the smaller = 6.x
Six times smaller plus the larger = 77
⇒ 6x + x + 7 = 77
⇒ 7x + 7 = 77

⇒ 7x = 70
⇒ \(\frac{7 \mathrm{x}}{7}\) = \(\frac{70}{7}\) (Divide by 7 on both sides) ⇒x = 10
∴ Smaller number x = 10
Larger number = x + 7 = 10 + 7 = 17
∴ The numbers are 10 and 17.

Question 8.
The sum of three consecutive even numbers is 54. Find the numbers.
Answer:
Let the three consecutive even numbers are 2x, 2(x +1), 2(x + 2).
Sum of three consecutive even numbers = 54
⇒ 2x + 2(x + 1) + 2(x + 2) = 54
⇒ 2x + 2x + 2 + 2x + 4 = 54
⇒ 6x + 6 = 54

⇒ 6x = 48
⇒ \(\frac{6 x}{6}\) = \(\frac{48}{6}\) (Divide by 6 on both sides)
⇒ x = 8
Numbers are, 2x, 2(x + 1), 2(x + 2)
2(8), 2(8 + 1), 2(8 + 2)
∴ The numbers are 16, 18, 20.

Question 9.
In a class of 48 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.
Answer:
Let the number of boys = x
The number of girls = \(\frac{1}{3}\) of boys
= \(\frac{1}{3}\) of x = \(\frac{x}{3}\)
Boys + Girls = Total students
⇒ \(\frac{x}{1}+\frac{x}{3}\) = 48
⇒ \(\frac{3 x+x}{3}\) = 48
⇒ \(\frac{4 x}{3}\) = 48
⇒ \(\frac{4 x}{3}\) × 3 = 48 × 3
(Multiply by 3 on both sides)
⇒ 4x = 144
⇒ \(\frac{4 x}{4}\) = \(\frac{144}{4}\) (Divide by 4 on both sides)
⇒ x = 36
∴ Number of boys = 36
Number of girls = \(\frac{x}{3}\) = \(\frac{36}{3}\) = 12
Boys = 36 and girls = 12

Question 10.
The present ages of Mary and Joseph are in the ratio 5 : 3. After 3 years sum of their ages will be 38. Find their present ages.
Answer:
Ratio of present ages of Mary and Joseph = 5:3 = 5x : 3x
After 3 years their ages
= 5x + 3 : 3x + 3
Sum of their ages = 38
⇒ 5x + 3 + 3x + 3 = 38
⇒ 8x + 6 = 38
⇒ 8x + 6 – 6 = 38 – 6 (Subtract 6 on both sides)
⇒ 8x = 32
⇒ \(\frac{8 x}{8}\) = \(\frac{32}{8}\)
⇒ x = 4
⇒ Ratio of present ages = 5x : 3x
= 5(4) : 3(4) = 20 : 12
Present ages of Mary and Joseph are 20 years and 12 years.

Question 11.
A sum of ₹500 is in the form of notes of denominations of ₹5 and ₹10. If the total number of notes is 90, then find the number of notes of each type.

Answer:
Let, number of f5 notes = x
Given sum of number of ₹5 notes and ₹10 notes is 90.
So, number of ₹10 notes = 90 – x
Sum of ₹5 notes = ₹5 × x = 5x
Sum of ₹ 10 notes = ₹ 10 × (90 – x)
= 900 – 10x
⇒ 5x + 900 – 10x = 500
⇒ – 5x + 900 = 500
⇒ – 5x + 900 – 900 = 500 – 900 (Subtract 900 on both sides)
⇒ – 5x = – 400
⇒ \(\frac{-5 x}{-5}\) = \(\frac{-400}{-5}\)
(Divide by – 5 on both sides)
⇒ x = 80
∴ Number of ₹5 notes x = 80
Number of ₹10 notes
= 90 – x = 90 – 80 = 10.

Question 12.
John and Ismail donated some money to the Relief Fund. The amount paid by Ismail is ₹85 more than twice that of John. If the total money paid by them is ₹4000, then find the amount of money donated by John.
Answer:
Let the amount donated by John = ₹x
The amount donated by Ismail is ₹85 more than twice of John.
So, the amount donated by Ismail = ₹(2x + 85)
Total amount = John amount + Ismail amount = ₹4000
⇒ x + 2x + 85 = 4000
⇒ 3x + 85 = 4000
⇒ 3x + 85 – 85 = 4000 – 85 (Subtract 85 on both sides)
⇒ 3x = 3915
⇒ \(\frac{3 x}{3}\) = \(\frac{3915}{3}\) (Divide by 3 on both sides)
⇒ x = 1305
∴ Money donated by John = x = ₹ 1305
∴ Money donated by Ismail = 2x + 85
= 2(1305) + 85
= ₹2695

Question 13.
Length of a rectangle is 4 m less than 3 times its breadth. If the perimeter of rectangle is 32 m. Then find its length and breadth.
Answer:
Let breadth of a rectangle (b) = a m
Length of rectangle (l) = 4 m less than 3 times of breadth
= (3 ∙ a – 4) m
Given perimeter = 2(l + b) = 32 m
⇒ 2[3a – 4 + a] = 32
⇒ 2[4a – 4] = 32
⇒ (8a – 8) = 32 (Distributive property)
⇒ 8a – 8 + 8 = 32 + 8 (Add 8 on both sides)
⇒ 8a = 40
⇒ \(\frac{8 \mathrm{a}}{8}\) = \(\frac{40}{8}\) (Divide by 8 on both sides) o o
⇒ a = 5
∴ breadth (b) = 5 m
length (l) = 3a – 4
= 3(5) – 4 = 15 – 4 = 11 m
∴ length = 11 m and breadth = 5 m.

Question 14.
A bag contains some number of white balls, twice the number of white balls are blue balls, thrice the number of blue balls are the red balls. If the total number of balls in the bag are 27. Then calculate the number of balls of each colour present in the bag.
Answer:
Let the number of white balls = x
Number of blue balls = twice the number of white balls
= 2(x) = 2x
Number of red balls
= thrice the number of blue balls = 3(2x) = 6x
Total balls = White + Blue + Red = 27
⇒ x + 2x + 6x = 27
⇒ 9x = 27
⇒ \(\frac{9 x}{9}\) = \(\frac{27}{9}\) (Divide by 9 on both sides) 9 9 .
⇒ x = 3
Therefore white balls x = 3
Blue balls = 2x = 2(3) = 6
Red balls = 6x = 6(3) = 18

Question 15.

Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60°
b) 30°
c) 45°
d) 90°
Answer: [ d ]
If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
a) 7 cm
b) 12 cm
c) 15 cm
d) 24.5 cm
Answer: [ a ]

O – centre of the circle
OP – a circle radius = ?
OQ = 25 cm
PQ = 24 cm
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
25² = OP² + 24²
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°
b) 70°
c) 80°
d) 90°
Answer: [ b ]
In □ OPAQ,
∠OPA = ∠OQA = 90°
∠POQ = 110°
∴ ∠O + ∠P + ∠A + ∠Q = 360°
⇒ 90° + 90° + 110° + ∠PAQ – 360°
⇒ ∠PAQ = 360° – 290° = 70°

iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Answer: [None]

If ∠APB = 80°
then ∠AOB = 180° – 80° = 100°
[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°

Answer: [ c ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction: Join OP and OB.
In △OPB ;
∠OPB = 90°
[radius is perpendicular to the tangent]
OP = 3cm OB = 5 cm
Now, OB² = OP² + PB²
[hypotenuse² = Adj. side² + Opp. side², Pythagoras theorem]
5² = 3² + PB²
PB² = 25 – 9 = 16
∴ PB = √l6 = 4cm.
Now, AB = 2 × PB
[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]
AB = 2 × 4 = 8 cm.
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:

Given: A circle with centre ‘O’.
A parallelogram ABCD, circumscribing the given circle.
Let P, Q, R, S be the points of contact.
Required to prove: □ ABCD is a rhombus.
Proof: AP = AS …….. (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See below figure). Find the sides AB and AC.

Answer:
The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.
∴ BF = BD = 9 cm
CD = CE = 3 cm
AF = AE = x cm say
∴ The sides of die triangle are
12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
or S = 12 + x
S – a = 12 + x – 12 = x
S – b = 12 + x – 3 – x = 9
S – c = 12 + x – 9 – x = 3
∴ Area of the triangle

Squaring on both sides we get,
27 (x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x- 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
⇒ x (x + 12) – 6 (x + 12) = 0
⇒ (x – 6) (x + 12) = 0
⇒ x = 6 or – 12
But ‘x’ can’t be negative hence, x = 6
∴ AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.
Answer:
Steps of construction:

1. Draw a circle with centre ‘O’ and radius 6 cm.
2. Take a point P outside the circle such that OP =10 cm. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP
OA² + AP² = 6² + 8²
= 36 + 64 = 100
OP² = 10² = 100
∴ OA² + AP² = OP²
Hence AP is a tangent.
Similarly BP is a tangent.

Question 6.
Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:
Steps of construction:

1. Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
2. Take a point ‘P’ on larger circle and join O, P.
3. Draw the perpendicular bisector of OP which intersects it at M.
4. Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
5. Join PQ, which is a tangent to the smaller circle.

Question 7.
Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.
Answer:
Steps of construction:

1. Draw a circle with the help of a bangle.
2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
3. Taking an outside point P, join OP.
4. Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Answer:
Let ABC be a right triangle right angled at P.
Consider a circle with diametere AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are two tangents to the circle from the same point P.
QB = QP …….. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC (2)
From (1) and (2);
QB = QC Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.
Answer:
Only two tangents can be drawn from a given point outside the circle.

Andhra Pradesh AP Board 4th Class Maths Solutions 3rd Lesson Addition Textbook Exercise Questions and Answers.

AP State Syllabus 4th Class Maths Solutions Chapter 3 Addition

Play and Win :

• Divide the students into 2 groups.
• Supply each group 2 sets of number cards from 0 to 9.
• Each student will take one set.
• Flip the cards to hide the numbers.
• Jumble the cards to change the order.
• One student from the first group selects 4 cards from second set of numbers and makes another four digit number.
• Then they add two numbers and write down the sum in the note book.
• Every group does the same. .
• The group which gets the largest sum will get 1 point.
• After 10 rounds who got more points they will be declared as winners.
  

Addition Machine:

This is an Addition machine. It has a fixed 4-digit number 4,728. If you place any 4 digit number in the machine, it is added to 4728 and the sum if displayed on the screen. You should place a 4-digit number of your choice and find the number displayed by the machine.

Question 1.

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Textbook Page No. 34

Do this

1. Add

a)

Answer:

b)

Answer:

c)

Answer:

d)

Answer:

Question 2.
Add 4789 and 2946
Answer:

Question 3.
Find 7645 + 5895
Answer:

Example – 2:

In another shop there are two types of bicycles namely gear bicycles and small bicycles.

Now you find out the two bicycles whose cost would be less than 10,000 and fill the table. The cost of each bicycle is given inboxes.

Answer:

Textbook Page No. 34

Do this

Question 1.
Estimate the sum in each case and tick (✓) on the correct box which is nearest to its sum.

Answer:

Question 2.
Estimate the sum.

There is a kids train. All the compartments have their numbers on them. Find the pairs of compartments whose sum is greater than 8000.
Answer:
(2536, 7326), (5329, 7326), (7326, 861), (2536, 6297), (5329, 4932), (5329, 3210), (5329, 6297), (1923, 7326), (1923, 6297), (7326, 4932), (7326, 3210), (7326, 6297), (7326, 2134) (4932, 3210), (4932, 6297), (3210, 6297), (6297, 2134),

Textbook Page No. 37

Do this

Question 1.
A water tank supplies drinking water to two villages. One village gets 3870 buckets of water. The other gets 5,295 buckets. How many buckets of water in total are supplied to both the villages ?

Solution:
Number of buckets supplied to first village = 3,870
Number of buckets supplied to second village = 5,295
Number of buckets supplied to

Totally = 9,165

Question 2.
On an independent day, 7,305 saplings wree planted in schools and 2859 saplings were planted in offices. How many saplings were planted in total ?
Solution:
Number of saplings planted in schools = 7,365
Number of saplings planted in Offices = 2,895
Number of saplings planted

Totally = 10,260

Textbook Page No. 38, 40

Try this

Make word problems for the following:

a) 6,854 + 3,521
Solution:
Latha had 6,854 bangles and Prasanna had 3,521 bangles. How many bangles were there in all ?

b) 5,340 + 3,564
Solution:
A fruit seller sold 5,340 apples and 3,564 oranges in a day.How many fruits did he sell?

c) 4,563 + 8,520
Solution:
Sirisha has 4,563 pencils and Sohan has 8,520 pencils. How many pencils are there altogether ?

1. Find the missing numerals.

a)

Answer:

b)

Answer:

2. Find the sum by suitable re-grouping.

a) 740 + 320 + 260 + 2,680
Solution:
We have to re-group the given in the following way.
= 740 + 260+ 320 + 2,680
= 1000 + 3,000
= 4,000

b) 5,986 + 2,976 + 14 + 24
Solution:
Re-group the given numbers.
= 5,986+ 14 + 2,976 + 24
= 6,000 + 3,000
= 9,000

c) 4,893 + 894 + 106 + 107
Solution:
Re-group the given numbers.
= 4,893 + 107 + 894 + 106
= 5,000 + 1,000
= 6,000

Exercise – 3.1

1. Add

a)

Answer:

b)

Answer:

c)

Answer:

d)

Answer:

2. Identify whether the following additions are correct or not. Justify your answer.

a)

Answer:
Given

is wrong
Justification

b)

Answer:
Given

is wrong
Justification

c)

Answer:
Given

is wrong
Justification

3. Write word problems for the following additions:

a) 3,268 + 5,634 = ?
Solution:
The balloon seller had 3,268 blue balloons and 5,634 red balloons. How many balloons did he have in all ?

b) 6,240 + 5,425 = ?
Solution:
A gift box has 6,240 chocolates and 5,425 lollipops. How many things are there in the box ?

4. Fill in the blanks.

a) 632 + 984 = 984 + _________
Solution:
632

b) 2,735 + __________ = 2,569 + 2,735
Solution:
2.569

Question 5.
A number exceeds 6897 by 5,478. What is that number?
Solution:
Sum of 6897 + 5478
∴ That number = 12,375

Question 6.
Veerayya sold maize for ₹ 5,397 and pearl millets for ₹ 3,849 in a village fair. How much amount did he get ?
Solution:
Selling price of maize = ₹ 5,397
Selling price of millets = ₹ 3,849

Total amount he got = ₹ 9,246

Question 7.
Madhav produced 3,985 watermelons in his field. Vijender produced 854 more than Madhava’s. What is the total number of watermelons produced by him?
Solution:
Watermelons produced by Madhav = 3,895
Vijendar produced 854 more than Madhav.
∴ Watermelons produced by Vijendar
= 3,985 + 854
= 4,839.

Question 8.
The number of visistors to Arasavalli temple on three consecutive days in Karthika masam is 3842, 2642 and 1,958 respectively. What is the total number of visitors during these days ?
Solution:
Number of visitors visited on First day = 3,842
Number of visitors visited on Second day = 2,642
Number of visitors visited on Third day = 1,958

Total number of visitors during these days = 8,442

Multiple Choice Questions

Question 1.
Nearest to thousand to the number 7,250 is
A) 7,000
B) 7,500
C) 8,000
D) 7,750
Answer:
A) 7,000

Question 2.
The sum of 1,984 + 2,020 to nearest thousand number
A) 2,000
B) 3,000
C) 4,000
D) 5,000
Answer:
C) 4,000

Question 3.
3,265 + 2,678 = ……………… + 3,265
A) 2,678
B) 3,265
C) 5,943
D) None
Answer:
A) 2,678

Question 4.
There are 2,475 girls and 3,950 boys in a school. How many students are there in all ?
A) 6,425
B) 6,000
C) 6,500
D) 7,000
Answer:
A) 6,425

Question 5.
Sum of 2,896 and 4,728
A) 4,267
B) 7,624
C) 6,724
D) 7,642
Answer:
B) 7,624

AP State Syllabus AP Board 8th Class Social Studies Important Questions Chapter 24 Disaster Management.

AP State Syllabus 8th Class Social Studies Important Questions 24th Lesson Disaster Management

Question 1.
What do you know about Tsunami?
Answer:

1. A tsunami consists of a series of waves and the first wave may not be the largest. The danger from subsequent tsunami waves can last for several hours after the arrival of the first wave.
2. Tsunami can move at 50 km per hour on coastal plain, faster than a person can run.

Question 2.
What is IWMP?
Answer:
The government is implementing Integrated Watershed Management Programme (IWMP) in drought prone areas to reduce the impact of droughts.

Question 3.
How was the disaster of ‘Budameru’ floods managed?
Answer:
The local government officials shifted the people to the schools and community halls in the area. They supplied food packets and drinking water. Some people donated bedsheets, old clothes to the needy. Thus the disaster was managed.

Question 4.
Have you ever observed any disaster in your locality? Explain.
Answer:
There is one ‘sponge dusters company’ in our area. There was a fire accident on some day evening. All the stock was burned within minutes. The workers also had small burns.

Question 5.
Write about the work of coastal tidal gauges.
Answer:
Coastal tidal gauges can stop tsunamis close to the shore, but they are useless in deep oceans.

Question 6.
What is the work of Tsunameters?
Answer:
They transmit warnings of buoys on the sea surface, which relay it to satellites.

Question 7.
Write about ‘Drought’.
Answer:
Drought is basically a disaster situation caused by lack of rainfall. The deficiency in rainfall is defined as meteorological drought.

Question 8.
Explain the types of disasters.
Answer:
Disasters can be categorised into various types based on the speed and origin/ cause.
A. Based on speed, a disaster can be termed as slow or rapid.

1. Slow onset disaster: A disaster that prevails for many days, months or even years like drought, environmental degradation, pest infection, famine are some examples of a slow onset disaster.
2. Rapid onset disaster: A disaster that is triggered by an instance causes shock. The impact of this disaster may be short lived or long-term. Earthquake, cyclone, flash floods, volcanic eruptions are some examples of rapid onset disasters.

B. Based on the cause, disaster can be natural or human induced.

1. Natural disaster: A natural disaster is an event that is caused by nature and leads to human, material, economic and environmental losses. The types of natural disasters:
   1. Earthquakes
   2. Cyclones
   3. Floods
   4. Droughts
   5. Tsunamis
   6. Land slides
   7. Volcanoes etc.
2. Human induced disasters: A serious disruption of normal life triggered by human- induced hazard causing human, material, economic and environmental losses, which exceed the ability of those affected to cope. Some examples are the 1984 Bhopal Gas tragedy, the 1997 Uphaar Cinema fire in Delhi, Rajdhani Express train derailment in 2002, Kumbakonam school fire tragedy in 2003, Jaipur serial blasts in 2008 etc.

Question 9.
What is disaster management?
Answer:
Disaster management covers the range of activities designed to maintain control over disasters/ emergency situations and to provide a framework for helping people to avoid, reduce the effects of, or recover from the impact of a disaster. These activities may be related to preparedness, mitigation, emergency response, relief and recovery (reconstruction and rehabilitation) and may be conducted before, during or after a disaster.

Question 10.
What measures should be taken before Tsunami?
Answer:

1. Find out if your home, school, work place, or other frequently visited locations are in tsunami hazard prone areas.
2. Plan evacuation routes from your home, school, work place or any other place you could be in where tsunamis present a risk.
3. Practice your evacuation routes
4. Have disaster supplies ready at hand.
5. Discuss tsunamis with your family.

Question 11.
What do you know about Tsunami?
Answer:

1. A tsunami consists of a series of waves and the first wave may not be the largest. The danger from subsequent tsunami waves can last for several hours after the arrival of the first wave.
2. Tsunami can move at 50 km per hour on coastal plain, faster than a person can run.
3. Tsunamis can occur at any time of day or night.

Question 12.
Write about ‘Drought’.
Answer:
Drought is basically a disaster situation caused by lack of rainfall. The deficiency in rainfall is defined as meteorological drought. While in a year, there may be normal rainfall, there might be a wide gap separating two consecutive spells of rain, resulting in crop failure which is termed as agricultural drought. Thus, the quantum as well as the distribution of rainfall are important.
Excess or deficient rainfall is determined by the percent variation from the average rainfall (of 70-100 years) as follows:
Excess + 20 percent or more of the average rainfall Normal + 19 percent to -19 percent of the average rainfall Deficient – 20 percent to -59 percent of the average rainfall Scanty – 60 percent or less of the average rainfall
Certain regions due to their geographical location are more likely to receive less rainfall. These are called ‘drought prone areas’.

Question 13.
What is the impact of drought?
Answer:
There is a sequential impact of drought:

1. Scarcity of drinking water; fall in water-table.
2. Decline in crop acreage.
3. Fall in employment in the agricultural sector due to slowing down of agricultural activity.
4. Fall in purchasing power of those engaged in agriculture.
5. Scarcity of food grains.
6. Scarcity of fodder.
7. Loss of cattle life.
8. Malnutrition, especially among children.
9.  ill health and spread of diseases like diarrhoea, dysentery or cholera and opthalmia caused by starvation.
10. Distress sale and mortgage of land, jewellery and personal property.
11. Migration of people in search of employment.

Question 14.
What is IWMP? What is its main objective?
Answer:
The government is implementing Integrated Watershed Management Programme (IWMP) in drought prone areas to reduce the impact of droughts. The main objective is to strengthen the community and enable them to plan for proper utilisation of natural resources. Land use based on its capability helps in optimum use of land and water and can prevent misuse. The main activities include harvesting rain water in the fields, afforestation, promotion of crops/ trees that require less water and alternative livelihoods.

Question 15.
How can we cope with drought?
Answer:
Unlike sudden disasters drought being a slow onset disaster, gives us ample time for preparedness, response and mitigation. Monitoring and early warning enables timely action by decision makers at all levels. In areas that are normally affected by drought Government, Non-Governmental Organizations (NGOs), local officials and other key players have taken the initiative to bring in awareness on water conservation strategies etc.

Question 16.
How should we harvest rainwater in urban areas?
Answer:
Rainwater harvesting: In urban areas all rainwater as it falls over roofs of houses should be harvested. The easiest thing is to divert it into soak pits for recharging of groundwater. The rainwater may also be stored in sumps/ tanks which are built for this purpose. In certain places, with simple filtering, rain water can be the best source of drinking water.

Question 17.
Do you suggest any precautions to the people?
Answer:
Precautions:

1. We should plan easy escape earlier.
2. Our daily necessaries should be maintained in a bag.
3. If the disaster is related to water, we should reach the highest and safest place in time.
4. Dry food should be stored.
5. Common and emergency medicines should be packed.
6. We should help the needy people.

Question 18.
Have you ever observed any disaster in your locality? Explain.
Answer:
There is one ‘sponge dusters company’ in our area. There was a fire accident on one evening. All the stock was burned within minutes. The workers also had small burns. They met a loss of Rs. 3,50,000/- due to this accident.

Question 19.
Find whether you are a water saver or spender with the help of the following questionnaire. Check how much water you can save and whether you are a water hero or villain!

Total the water you use and check your ranking:

• Eco Hero: <200 It.
• Water saver: 201 – 400 It.
• Water spender: 400 – 600 It.
• Water villain: >601 It.

Answer:
I am in the rank of an Eco Hero.

Question 19.
Read the following passage and answer the given questions.

Detecting Tsunamis:

With the use of satellite technology, it is possible to provide nearly immediate warning of potentially tsuna-migenic earthquakes. Warning time depends upon the distance of the epicenter from the coast line. The warning includes predicted times at selected coastal communities where the tsunami could travel in a few hours.
Coastal tidal gauges can stop tsunamis close to the shore, but they are useless in deep oceans. Tsunami detectors, linked to land by submarine cables, are deployed 50 odd kms out at sea. ‘Tsunameters’ transmit warnings of buoys on the sea surface, which relay it to satellites.

1. What is the technology mentioned here?
Answer:
Satellite technology.

2. On what does the warning time depend?
Answer:
Warning time depends upon the distance of the epicenter from the coast line.

3. What does the warning include?
Answer:
The warning includes predicted times at selected coastal communities where the tsunami could travel in a few hours.

4. Write about the work of coastal tidal gauges.
Answer:
Coastal tidal gauges can stop tsunamis close to the shore, but they are useless in deep oceans.

5. What is the work of Tsunameters?
Answer:
They transmit warnings of buoys on the sea surface, which relay it to satellites.

Question 20.
Read the following para and answer the questions.
The Teachers and students are an integral part of the community and have an important role to play in being prepared for disasters. Students are effective carrier of messages to educate their parents and the community. Teachers have an important responsibility to guide the students in this regard.

1. What is the importance of teachers and students?
Answer:
The teachers and students are an integral part of the community and have an important role to play in being prepared for disasters.

2. How are the students called effective carriers?
Answer:
Students are effective carriers of messages to educate their parents and the community.

3. What is the responsibility of teachers?
Answer:
Teachers have an important responsibility to guide the students in this regard.

Question 21.
Read the paragraph ‘Watershed development’ (in text page no : 259) and then prepare two questions.
Answer:

1. Who implements IWMP?
2. Write any two alternative livelihoods.

Question 22.
Observe the following map and answer the given questions.

1. Write the names of Tsunami affected areas.
Answer:
Alappuzha, Kollam, Kanyakurnari, Cuddlore, Nagapatnam, Chennai, Prakasam, Guntur, Krishna, West and East Godavari, Visakhapatnam, Vizianagaram, West Bengal, Coastal re-gion and Andaman Nicobar islands.

2. On which coast are these areas located?
Answer:
Many areas are in the east coast and some are on the south coast.

AP State Syllabus AP Board 8th Class Social Studies Important Questions Chapter 5 Forests: Using and Protecting Them.

AP State Syllabus 8th Class Social Studies Important Questions 5th Lesson Forests: Using and Protecting Them

Question 1.
What is a forest?
Answer:
A large tract of land covered by trees is called a forest.

Question 2.
Name some conservative centres of medicinal plants.
Answer:

1. Vali Sugriva Medicinal Plant Conservative Centre.
2. Koringa Medicinal Plant Conservative Centre.
3. Karthika Vanam.

These three are in East Godavari District.

Question 3.
“Every year about one hundred sq. kms of forests are lost in our state” – Is this a satisfactory situation?
Answer:
No, this is not a satisfactory situation. This affects our biodiversity, rainfall and soils etc. We can say this as critical condition.

Question 4.
Do you think it is possible for people to take care of forests and use them as well?
Answer:
It is possible. In present society many people are cultivating fruits and vegetables. They are protecting their gardens by selling the fruits and vegetables.

Question 5.
What would they have done if someone had tempted them to cut trees and sell them in the markets?
Answer:
They definitely oppose and reject the suggestion. Eg: All their festivals are related to trees. Mamidi panduga, Gongura panduga, Mutyalamma panduga etc. They may cut a branch or some else but not the trees. They don’t cut their home.

Question 6.
List all the causes for decline of forests in the last 200 years.
Answer:

1. Agriculture
2. Cattle rearing
3. Mega projects
4. Fires
5. Logging
6. Mangroves and shrimp farming
7. Mining oil and gas.

Question 7.
How can you say that the ‘Podu’ cultivation is responsible for decline of forests?
Answer:
We can say that ‘Podu’ is responsible for decline of forests. The tribals left their traditional food, and depended on crops. So the land in forests is changed as agriculture land.

Question 8.
Why do you think the government thought that forest was not important for tribal people’s development ?
Answer:
I think that the government had lost its foresight and thought that forest was not important for tribal people’s development.

Question 9.
How is the implementation of Forest laws in your area?
Answer:
The implementation of Forest laws are somehow nominal.

Question 10.
A few children in a school participated in Vanamahotsavam programme and they planted some saplings. How do you respond to this ?
Answer:
I appreciate this action very much. The children’s participation in this brightens the future of the country. But they should take care of their plants till they grow.

Question 11.
Name some conservative centres of medicinal plants.
Answer:

1. Vali Sugriva Medicinal Plant Conservative Centre.
2. Koringa Medicinal Plant Conservative Centre.
3. Karthika Vanam.

These three are in East Godavari District.

Question 12.
How is the implementation of Forest laws in your area? Though the laws are in vogue, the forests are disappearing day by day. What might be the reasons?
Answer:
The implementation of Forest laws are somehow nominal.
Reasons:
Mining people, rural people and others are misusing the forests. Lack of knowledge regarding forest protection might be one reason. The impurity in the politicians is also another cause. Some more reasons are also there.

Question 13.
Locate the following on the given India map.

1. Alpines of Himalayas
2. Kerala forests
3. Srikakulam
4. Anaimudi forests in western ghats
5. Dandakaranyam

Answer:

Question 14.
Observe the given map and answer the following questions.

1. Where/On which side are more forests located?
Answer:
North side.

2. Which area belongs to Pondicherry?
Answer:
Yanam

3. Where are thorny forests?
Answer:
In Rayalaseema (more)

4. Is there any district without forests?
Answer:
No, each and every district has more or less forests.