# AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry InText Questions

## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry InText Questions

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

### 10th Class Maths 11th Lesson Trigonometry InText Questions and Answers

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Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.  (Page No. 271)

Question 1.
For angle R In the △PQR
Opposite side = PQ
Hypotenuse = PR Question 2.
i) For angle X
ii) For angle Y        (Page No. 271) In the △XYZ,
i) For angle X
Opposite side = YZ
Hypotenuse = XY
ii) For angle Y
Opposite side = XZ
Hypotenuse = XY

Question 3.
Find (i) sin C (ii) cos C and (iii) tan C in the given triangle.    (Page No. 274) By Pythagoras theorem
AC2 = AB2 + BC2
132 = AB2 + 52
AB2 = 169 – 25
AB2 = √144
⇒ AB = √144 = 12  Question 4.
In a triangle XYZ, ∠Y is right angle, XZ = 17 cm and YZ = 15 cm, then find (i) sin X (ii) cos Z (iii) tan X.    (Page No. 274)
Given △XYZ, ∠Y is right angle. By Pythagoras theorem
XZ2 = YZ2 + XY2
172 = 152 + XY2
XY2 = 172 – 152 = 289 – 225
XY2 = 64
XY = √64 = 8 Question 5.
In a triangle PQR with right angle at Q, the value of ∠P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.    (Page No. 274)
Given right angled triangle is PQR with right angle at Q. The value of ∠P is x. By Pythagoras theorem
PR2 = PQ2 + QR2
PR2 = 72 + 242
PR2 = 49 + 576
PR2 = 625
PR2 = √625 = 25
sin x = $$\frac{QR}{PR}$$ = $$\frac{24}{25}$$
cos x = $$\frac{PQ}{PR}$$ = $$\frac{7}{25}$$ Try This

Question 1.
Write lengths of “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.
1. For angle C
2. For angle A          (Page No. 271) By Pythagoras theorem
AC2 = AB2 + BC2
(5)2 = AB2 + 42
25 = AB2 + 16
AB2 = 25 – 16
AB2 = 9
AB = √9 = 3
For angle C:
Opposite side = AB = 3 cm
Adjacent side = BC = 4 cm
Hypotenuse = AC = 5 cm
For angle A:
Opposite side = BC = 4 cm
Adjacent side = AB = 3 cm
Hypotenuse = AC = 5 cm

Question 2.
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC – CA = 7 cm, then find sin A and tan B.    (Page No. 274)
In a right angle triangle ABC, right angle is at C. BC = $$\frac{30}{2}$$ = 15
BC = 15
Substituting BC = 15 in equation (1)
BC + CA = 23
CA = 23 – BC = 23 – 15
CA = 8
By Pythagoras theorem
AB2 = AC2 + BC2
= 82 + 152
= 64 + 225 = 289
= √289 = 17
sin A = $$\frac{BC}{AB}$$ = $$\frac{15}{17}$$
cos B = $$\frac{AC}{BC}$$ = $$\frac{8}{15}$$

Question 3.
What will be the ratios of sides for sec A and cot A?    (Page No. 275)
sec A = $$\frac{\text { Hypotenuse }}{\text { Adjacent side of the angle } \mathrm{A}}$$
cot A = $$\frac{\text { Adjacent side of the angle } A}{\text { Opposite side of the angle } A}$$ Think & Discuss

Question 1.
i) sin x = $$\frac{4}{3}$$ does exist for some value of angle x?
ii) The value of sin A and cos A is always less than 1. Why?
iii) tan A is product of tan and A.        (Page No. 274)
i) The value of sin 0 always lies between 0 and 1. Here sin x = $$\frac{4}{3}$$ which is greater than 1. So, it does not exist.
ii) We observe in above sin A, cos A, hypotenuse is in denominator which is greater than other two sides
∴ sin A = $$\frac{\text { Opposite side }}{\text { Hypotenuse }}$$ = $$\frac{Nr}{Dr}$$
here denominator is more than numerator. Hence its value will be less than 1.
cos A = $$\frac{\text { Adjacent side }}{\text { Hypotenuse }}$$
here also adjacent side is always less than hypotenuse. Hence its value is also less than or equal to 1.

iii) The symbol tan A is used as an abbreviation for “the tan of the angle A”.
tan A is not the product of “tan” and A. “tan” separated from A’ has no meaning.

Question 2.
Is $$\frac{\sin \mathrm{A}}{\cos \mathrm{A}}$$ equal to tan A?      (Page No. 275)
Yes, $$\frac{\sin \mathrm{A}}{\cos \mathrm{A}}$$ = tan A
Proof:  Question 3.
Is $$\frac{\cos \mathrm{A}}{\sin \mathrm{A}}$$ equal to cot A?     (Page No. 275)
Yes, $$\frac{\cos \mathrm{A}}{\sin \mathrm{A}}$$ = tan A
Proof: cot A = $$\frac{\text { Adjacent side }}{\text { Opposite side }}$$

Do this

Question 1.
Find cosec 60°, sec 60° and cot 60°.       (Page No. 279) Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = $$\frac{1}{2}$$ BC = $$\frac{2a}{2}$$ = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD2 = AB2 – BD2
(By Pythagoras theorem)
= (2a)2 – (a)2 = 3a2
From definitions of trigonometric ratios,  Try this

Question 1.
Find sin 30°, cos 30°, tan 30°, cosec 30°, sec 30° and cot 30° by using the ratio concepts.     (Page No. 279) Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the given figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.
Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.
BD = $$\frac{1}{2}$$ BC = $$\frac{2a}{2}$$ = a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD2 = AB2 – BD2
(By Pythagoras theorem)
= (2a)2 – (a)2 = 3a2
Therefore, AD = $$\sqrt{3 a^{2}}$$ = √3a
BD = a, AD = √3a and hypotenuse = AB = 2a and ∠DAB = 30°.
sin 30° = $$\frac{BD}{AB}$$ = $$\frac{a}{2a}$$ = $$\frac{1}{2}$$ Question 2.
Find the values for tan 90°, cosec 90°, sec 90° and cot 90°.     (Page No. 281)
From the adjacent figure, the trigonometric ratios of ∠A, gets larger and larger in △ABC till it becomes 90°. As ∠A get larger and larger, ∠C gets smaller and smaller. Therefore, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC. ∴ AB = 0 and AC = BC = r
From trigonometric ratios
sin A = $$\frac{BC}{AC}$$
sin A = $$\frac{AB}{AC}$$
If A = 90°, then AB = 0 and AC = BC = r,  Think & Discuss

Question 1.
What can you say about cosec 0° = $$\frac{1}{\sin 0^{\circ}}$$? Is it defined? Why?    (Page No. 280)
sin 0° = 0
cosec 0° = $$\frac{1}{\sin 0^{\circ}}$$ = $$\frac{1}{0}$$ = not defined.
It is not defined.
Reason:
Division by ‘0’ is not allowed, hence $$\frac{1}{0}$$ is indeterminate.

Question 2.
What can you say about cot 0° = $$\frac{1}{\tan 0^{\circ}}$$. Is it defined? Why?    (Page No. 281)
Is it defined? Why?
tan 0° = 0
cot 0° = $$\frac{1}{\tan 0^{\circ}}$$ = $$\frac{1}{0}$$ = undefined.
Reason:
Division by ‘0’ is not allowed, hence $$\frac{1}{0}$$ is indeterminate.

Question 3.
sec 0° = 1. Why?      (Page No. 281)
sec 0° = $$\frac{1}{\cos 0^{\circ}}$$ [∵ cos 0° = 1]
= $$\frac{1}{1}$$ = 1

Question 4.
What can you say about the values of sin A and cos A, as the value of angle A increases from 0° to 90°? (Observe the above table)
i) If A ≥ B, then sin A ≥ sin B. Is it true?
ii) If A ≥ B, then cos A ≥ cos B. Is it true? Discuss.      (Page No. 282)
i) Given statement
“If A ≥ B, then sin A ≥ sin B”
Yes, this statement is true.
Because, it is clear from the table below that the sin A increases as A increases. ii) Given statement
“If A ≥ B, then cos A ≥ cos B”
No, this statement is not true.
Because, it is clear from the table below that cos A decreases as A increases.  Think & Discuss

Question 1.
For which value of acute angle
(i) $$\frac{\cos \theta}{1-\sin \theta}$$ + $$\frac{\cos \theta}{1+\sin \theta}$$ = 4 is true?
For which value of 0° ≤ θ ≤ 90°, above equation is not defined?    (Page No. 285)  ⇒ cos θ = cos 60° [from trigonometric ratios table]
⇒ θ = 60°
Given statement is true for the acute angle i.e., θ = 60°.

Question 2.
Check and discuss the above relations in the case of angles between 0° and 90°, whether they hold for these angles or not? So,          (Page No. 286)
i) sin (90° – A) = cos A
ii) cos (90° – A) = sin A
iii) tan (90° – A) = cot A and
iv) cot (90° – A) = tan A
v) sec (90° – A) = cosc A
vi) cosec (90° – A) = sec A
Let A = 30°
i) sin (90° – A) = cos A
⇒ sin (90° – 30°) = cos 30°
⇒ sin 60° = cos 30° = $$\frac{\sqrt{3}}{2}$$

ii) cos (90° – A) = sin A
⇒ cos (90° – 30°) = sin 30°
⇒ cos 60° = sin 30° = $$\frac{1}{2}$$

iii) tan (90° – A) = cot A
⇒ tan (90° – 30°) = cot 30°
⇒ tan 60° = cot 30° = √3

iv) cot (90° – A) = tan A
⇒ cot (90° – 30°) = tan 30°
⇒ cot 60° = tan 30° = $$\frac{1}{\sqrt{3}}$$

v) sec (90° – A) = cosec A
⇒ sec (90° – 30°) = cosec 30°
⇒ sec 60° = cosec 30° = 2

vi) cosec (90° – A) = sec A
⇒ cosec (90° – 30°) = sec 30°
⇒ cosec 60° = sec 30° = $$\frac{2}{\sqrt{3}}$$

So, the above relations hold for all the angles between 0° and 90°. Do this

(Page No. 290)

Question 1.
i) If sin A = $$\frac{15}{17}$$ then find cos A.
Given sin A = $$\frac{15}{17}$$
cos A = $$\sqrt{1-\sin ^{2} A}$$ [From Identity -I] ii) If tan x = $$\frac{5}{12}$$, then find sec x. (AS j)
Given tan x = $$\frac{5}{12}$$
We know that sec2 x – tan2 x = 1
sec2 x = 1 + tan2 x iii) If cosec θ = $$\frac{25}{7}$$, then find cot θ.
Given cosec θ = $$\frac{25}{7}$$
We know that cosec2 θ – cot2 θ = 1
cot2 θ = cosec2 θ – 1 Try This

(Page No. 290)

Question 1.

i) $$\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}$$
Given $$\frac{\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} 36^{\circ}+\cos ^{2} 54^{\circ}}$$ [∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= $$\frac{1}{1}$$ = 1 [By sin2 θ + cos2 θ = 1]

ii) sin 5° cos 85° + cos 5° sin 85°
Given sin 5° cos 85° + cos 5° sin 85°
= sin 5° . cos (90° – 5°) + cos 5° . sin (90° – 5°)
= sin 5° . sin 5° + cos 5° . cos 5°
[∵ sin (90° – θ) = cos θ; cos (90° – θ) = sin θ]
= sin2 5° + cos2
= 1 [∵ sin2 θ + cos2 θ = 1]

iii) sec 16° cosec 74° – cot 74° tan 16°.
Given sec 16° cosec 74° – cot 74° tan 16°
= sec 16° . cosec (90° – 16°) – cot (90° – 16°) . tan 16°
= sec 16°. sec 16° – tan 16° . tan 16° [∵ cosec (90° – θ) = sec θ; cot (90° – θ) — tan θ]
= sec2 16° – tan2 16°
= 1 [∵ sec2 θ – tan2 θ = 1] Think & Discuss

(Page No. 290)

Question 1.
Are these identities true for 0° ≤ A ≤ 90°? If not for which values of A they are true?
i) sec2 A – tan2 A = 1
Given identity is sec2 A – tan2 A = 1
Let A = 0°
L.H.S. = sec2 0° – tan2
= 1 – 0 = 1 = R.H.S.
Let A = 90°
tan A and sec A are not defined.
So it is true.
∴ For all given values of ‘A’ such that 0° ≤ A ≤ 90° this trigonometric identity is true.

ii) cosec2 A – cot2 A = 1