SELINA Solution Class 9 Chapter 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] Exercise 23 C

Question 1.1

Solve the following equation for A, if 2 sin A = 1

Sol:

2 sin A = 1
sin A = ${\displaystyle \frac{1}{2}}$
sin A = sin 30°
A = 30°

Question 1.2

Solve the following equation for A, if 2cos2A = 1

Sol:

2 cos 2 A = 1 
cos 2 A = ${\displaystyle \frac{1}{2}}$
cos 2 A = cos 60°
2A = 60°
A = 30°

Question 1.3

Solve the following equation for A, if sin 3 A = ${\displaystyle \frac{\sqrt{3}}{2}}$

Sol:

sin 3A = ${\displaystyle \frac{\sqrt{3}}{2}}$
sin 3A = sin 60°
3A = 60°
A = 20°

Question 1.4

Solve the following equation for A, if sec 2A = 2

Sol:

According to the question,
We have,

sec 2A = 2
sec 2A = sec 60°
2A = 60°
A = 30°.

Question 1.5

Solve the following equations for A, if tan A = 1

Sol:

tan 3 A = 1
tan 3 A = tan45°
3A = 45°
A = 15°

Question 1.6

Solve the following equation for A, if ${\displaystyle \sqrt{3}}$ tan 3 A = 1

Sol:

${\displaystyle \sqrt{3}\tan }$ A = 1
tan A = ${\displaystyle \frac{1}{\sqrt{3}}}$
tan A = tan 30°
A = 30°

Question 1.7

Solve the following equation for A, if 2 sin 3 A = 1

Sol:

2 sin 3 A = 1
sin 3 A = ${\displaystyle \frac{1}{2}}$
sin 3A = sin 30°
3A = 30°
A = 10°

Question 1.8

Solve the following equation for A, if ${\displaystyle \sqrt{3}}$ cot 2 A = 1

Sol:

${\displaystyle \sqrt{3}\cot }$ 2 A = 1
cot 2 A = ${\displaystyle \frac{1}{\sqrt{3}}}$
cot 2 A = cot 60°
2A = 60°
A = 30°

Question 2.1

Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0

Sol:

( sin A – 1) ( 2 cos A – 1) = 0
(sin A – 1) = 0 and 2 cos A – 1 = 0
sin A = 1  and cos A = ${\displaystyle \frac{1}{2}}$
sin A = sin90° and cos A = cos60°
A = 90° and A = 60°

Question 2.2

Calculate the value of A, if (tan A - 1) (cosec 3A - 1) = 0

Sol:

( tan A – 1) ( cosec 3A – 1) = 0
tan A – 1 = 0 and  cosec 3A – 1  = 0
tan A = 1  and cosec 3A = 1
tan A = tan45° and cosec 3A = cosec90°
A = 45° and A = 30°

Question 2.3

Calculate the value of A, if (sec 2A - 1) (cosec 3A - 1) = 0

Sol:

( sec 2A – 1) ( cosec 3A – 1) = 0
sec 2A – 1 = 0 and  cosec 3A – 1  = 0
sec 2A = 1  and cosec 3A = 1
sec 2A = sec0° and cosec 3A = cosec90°
A = 0° and A = 30°

Question 2.4

Calculate the value of A, if cos 3A. (2 sin 2A - 1) = 0

Sol:

cos 3A ( 2 sin 2A – 1 ) = 0
cos 3A = 0 and 2 sin 2A  – 1 = 0
cos 3A = cos90° and 2 sin 2A = ${\displaystyle \frac{1}{2}}$
3A = 90° and sin 2A = sin 30°
A = 30°  2A = 30° ⇒ A = 15°

Question 2.5

Calculate the value of A, if (cosec 2A - 2) (cot 3A - 1) = 0

Sol:

( cosec 2A – 2) (cot 3A – 1) = 0
cosec 2A – 2 = 0  and cot 3A – 1 = 0
cosec 2A = 2 and cot 3A = 1
cosec 2A = cosec 30° and cot 3A = cot 45°
2A = 30° and 3A = 45°
A = 15° and A = 15°

Question 3

If 2 sin x° - 1 = 0 and x is an acute angle;
find :
(i) sin x°
(ii) x° 
(iii) cos x and tan x°.

Sol:

(i) 2 sin x° – 1 = 0

sin x° = ${\displaystyle \frac{1}{2}}$

(ii) sin x° = ${\displaystyle \frac{1}{2}}$

sin x° = sin 30°
x° = 30°

(iii) cos x° = cos 30° = ${\displaystyle \left(\sqrt{3}\right)}$

tax x° = tan 30° = ${\displaystyle \frac{1}{\sqrt{3}}}$

Question 4

If 4 cos² x° - 1 = 0 and 0 ∠ x° ∠ 90°,
find:(i) x°
(ii) sin² x° + cos² x°
(iii) ${\displaystyle \frac{1}{{\cos }^2\times °}–({\tan }^2\times °)}$

Sol:

(i) 4 cos²x° – 1 = 0
4 cos²x° = 1

co^s2x° = ${\displaystyle {\left(\frac{1}{2}\right)}^2}$

cosx° = ${\displaystyle \frac{1}{2}}$

cosx° = cos60°

x° = 60°

(ii) sin² x° + cos²x° = sin²60° + cos²60°

= ${\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$

= ${\displaystyle \frac{3}{4}+\frac{1}{4}}$

= 1

(iii) ${\displaystyle \frac{1}{{\cos }^2\times °}–{\tan }^2\times °=\frac{1}{{\cos }^{260}°}–{\tan }^{2}60°}$

= ${\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}^2}–{\left(\sqrt{3}\right)}^2}$

= 4 – 3

= 1

Question 5

If 4 sin² θ - 1= 0 and angle θ is less than 90°, find the value of θ and hence the value of cos² θ + tan²θ.

Sol:

4 sin² θ – 1 = 0

sin² θ = ${\displaystyle \frac{1}{4}}$

sin θ = ${\displaystyle \frac{1}{2}}$

sin θ = sin30°

θ = 30°

cos² θ + tan² θ= cos²30° + tan²30° 

= ${\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}$

= ${\displaystyle \frac{3}{4}+\frac{1}{3}}$

= ${\displaystyle \frac{9+4}{12}}$

= ${\displaystyle \frac{13}{12}}$

Question 6.1

If sin 3A = 1 and 0 < A < 90^°, find sin A

Sol:

sin 3A = 1 
sin 3A = sin90°
3A = 90°
A = 30°

sin A = sin30°
sin A = ${\displaystyle \frac{1}{2}}$

Question 6.2

If sin 3A = 1 and 0 < A < 90^°, find cos 2A

Sol:

sin 3A = 1 
sin 3A = sin90°
3A = 90°
A = 30°

cos 2A = cos 2(30°)
= cos 60°
= ${\displaystyle \frac{1}{2}}$

Question 6.3

If sin 3A = 1 and 0 < A < 90^°, find ${\displaystyle {\tan }^{2}A-\frac{1}{{\cos }^2\text{A}}}$

Sol:

sin 3A = 1 
sin 3A = sin90°
3A = 90°
A = 30°

${\displaystyle {\tan }^{2}A–\frac{1}{{\cos }^2\text{A}}={\tan }^{2}30°–\frac{1}{{\cos }^{2}30°}}$

= ${\displaystyle {\left(\frac{1}{\sqrt{3}}\right)}^2–\frac{1}{{\left(\frac{\sqrt{3}}{2}\right)}^2}}$

= ${\displaystyle \frac{1}{3}–\frac{4}{3}}$

= ${\displaystyle \frac{–3}{3}}$

= – 1

Question 7

If 2 cos 2A =  ${\displaystyle \sqrt{3}}$ and A is acute,
find:
(i) A 
(ii) sin 3A
(iii) sin² (75° - A) + cos² (45° +A)

Sol:

(i) 2 cos 2A = ${\displaystyle \sqrt{3}}$

cos 2A = ${\displaystyle \frac{\sqrt{3}}{2}}$

cos 2A = cos 30°

2A = 30°

A = 15°

(ii) sin 3A = sin 3(15°)
= sin 45°
= ${\displaystyle \frac{1}{\sqrt{2}}}$

(iii) sin²(75° – A ) + cos² (45 + A) = sin² ( 75° –15°) + (cos² ( 45° + 15°) 

= sin² 60° + cos² 60° 

= ${\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$

= ${\displaystyle \frac{3}{4}+\frac{1}{4}}$

= 1

Question 8.1

If sin x + cos y = 1 and x = 30°, find the value of y

Sol:

Given that x = 30°
sin x + cos y = 1

sin 30° + cos y = 1

cos y = 1 – sin 30°

cos y = 1 –${\displaystyle \frac{1}{2}}$

cos y = ${\displaystyle \frac{1}{2}}$

cos y = cos 60°

y = 60°

Question 8.2

If 3 tan A - 5 cos B = ${\displaystyle \sqrt{3}}$ and B = 90°, find the value of A

Sol:

Given that B = 90°
3 tan A – 5 cos B = ${\displaystyle \sqrt{3}}$

3 tan A – 5 cos 90° = ${\displaystyle \sqrt{3}}$

3 tan A – 0 = ${\displaystyle \sqrt{3}}$

tan A = ${\displaystyle \frac{\sqrt{3}}{3}}$

tan A = ${\displaystyle \frac{1}{\sqrt{3}}}$

tan A = tan 30°

A = 30°

Question 9

From the given figure,
find:
(i) cos x°
(ii) x°
(iii) ${\displaystyle \frac{1}{{\tan }^2\times °}–\frac{1}{{\sin }^2\times °}}$ 
(iv) Use tan x^o, to find the value of y.

Sol:

(i) cos x° = ${\displaystyle \frac{10}{20}}$

cos x° = ${\displaystyle \frac{1}{2}}$

 

(ii) cos x° = ${\displaystyle \frac{1}{2}}$

cos x° = cos 60°

x° = 60°

 

(iii)  ${\displaystyle \frac{1}{{\tan }^{2}x°}–\frac{1}{{\sin }^{2}x°}=\frac{1}{{\tan }^{2}60°}–\frac{1}{{\sin }^{2}60°}}$

= ${\displaystyle \frac{1}{{\left(\sqrt{3}\right)}^2}–\frac{1}{{\left(\frac{\sqrt{3}}{2}\right)}^2}}$

= ${\displaystyle \frac{1}{3}–\frac{4}{3}}$

= – 1

(iv) tan x° = tan 60°
= ${\displaystyle \sqrt{3}}$

We know that tan x° = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

⇒ tan x° = ${\displaystyle \frac{\text{y}}{10}}$

⇒ y = 10 tan x°
⇒ y = 10 tan 60°
⇒ y = 10${\displaystyle \sqrt{3}}$

Question 10

Use the given figure to find:
(i) tan θ°
(ii)  θ°
(iii) sin²θ° - cos²θ°
(iv) Use sin θ° to find the value of x.

Sol:

(i) tan θ° = ${\displaystyle \frac{5}{5}=1}$

(ii) tan θ° = 1
tan θ° = tan 45°
θ° = 45°

(iii) sin²θ° – cos²θ° = sin²45° – cos² 45°

= ${\displaystyle {\left(\frac{1}{\sqrt{2}}\right)}^2–{\left(\frac{1}{\sqrt{2}}\right)}^2}$

= 0

(iv) sinθ° = ${\displaystyle \frac{5}{x}}$

sin 45° = ${\displaystyle \frac{5}{x}}$

x = ${\displaystyle \frac{5}{\sin 45°}}$

x = ${\displaystyle \frac{5}{\frac{1}{\sqrt{2}}}}$

x = 5${\displaystyle \sqrt{2}}$

Question 11.1

Find the magnitude of angle A, if 2 sin A cos A - cos A - 2 sin A + 1 = 0

Sol:

2 sin A cos A – cos A – 2 sin A + 1 = 0
2 sin A cos A – cos A = 2 sin A – 1
(2 sin A – 1) cos A – (2 sin A – 1) = 0
(2 sin A – 1) = 0 and cos A = 1
sin A =${\displaystyle \frac{1}{2}}$ and cos A = cos 0°
A = 30°  and A = 0°

Question 11.2

Find the magnitude of angle A, if tan A - 2 cos A tan A + 2 cos A - 1 = 0

Sol:

tan A – 2 cos A tan A + 2 cos A – 1 = 0
tan A – 2 cos A tan A = 1 – 2 cos A
tan A ( 1 –  2 cos A ) –  (1 – 2 cos A )= 0
(1 –  2 cos A) (tan A –  1) = 0
1 –  2 cos A = 0 and tan A –  1 = 0
cos A = ${\displaystyle \frac{1}{2}}$ and tan A = 1
A = 60° and A = 45°

Question 11.3

Find the magnitude of angle A, if 2 cos² A - 3 cos A + 1 = 0

Sol:

2 cos2 A –  3 cos A + 1 = 0
2 cos2 A –  cos A - 2cosA +1 = 0
cos A(2cos A –  1) –  (2 cos A – 1) = 0
( 2 cosA – 1) (cos A – 1) = 0
2 cos A –  1 = 0 aand cos A –  1 = 0
cos A = ${\displaystyle \frac{1}{2}}$ and cos A = 1
A = 60° and A = 0°

Question 11.4

Find the magnitude of angle A, if 2 tan 3A cos 3A - tan 3A + 1 = 2 cos 3A

Sol:

2tan 3A cos 3A –  tan 3A + 1 = 2 cos 3A
2 tan 3A cos 3A –  tan 3A = 2 cos 3A –  1
tan 3A (2 cos 3A –  1) = 2 cos 3A –  1
(2 cos 3A –  1)(tan 3A –  1) = 0
 2 cos 3A –  1 = 0 and tan 3A –  1 = 0
cos 3A = ${\displaystyle \frac{1}{2}}$ and tan 3A = 1
3A = 60° and 3A = 45°
A = 20° and A = 15°

Question 12.01

Solve for x : 2 cos 3x - 1 = 0

Sol:

2 cos 3x – 1 = 0
cos 3x = ${\displaystyle \frac{1}{2}}$
3x = 60°
x = 20°

Question 12.02

Solve for x : cos  ${\displaystyle \frac{x}{3} –1}$ = 0

Sol:

cos${\displaystyle \frac{x}{3}}$ – 1 = 0

cos${\displaystyle \frac{x}{3}}$ = 1

${\displaystyle \frac{x}{3}}$ = 0°

x =0°

Question 12.03

Solve for x : sin (x + 10°) = ${\displaystyle \frac{1}{2}}$ 

Sol:

sin (x + 10°) = ${\displaystyle \frac{1}{2}}$
sin ( x + 10°) = sin 30°
x + 10° = 30°
x = 20°

Question 12.04

Solve for x : cos (2x - 30°) = 0

Sol:

cos (2x – 30°) = 0
cos ( 2x – 30°) = cos 90°
2x – 30° = 90°
2x = 120°
x = 60°

Question 12.05

Solve for x : 2 cos (3x - 15°) = 1

Sol:

2 cos(3x –15°) = 1
cos (3x – 15°) = ${\displaystyle \frac{1}{2}}$
cos (3x – 15°) = cos 60°
3x – 15° = 60°
3x = 75°
x = 25°

Question 12.06

Solve for x : tan² (x - 5°) = 3

Sol:

tan² (x – 5°) = 3
tan (x – 5°) = ${\displaystyle \sqrt{3}}$
tan (x – 5°) =  tan 60°
x – 5° = 60°
x = 65°

Question 12.07

Solve for x : 3 tan² (2x - 20°) = 1

Sol:

3 tan² ( 2x – 20°) = 1
tan ( 2x – 20°) = ${\displaystyle \frac{1}{\sqrt{3}}}$
tan ( 2x – 20°) = tan 30°
2x  –20° = 30°
2x = 50°
x = 25°

Question 12.08

Solve for x : cos ${\displaystyle (\frac{x}{2}+10°)=\frac{\sqrt{3}}{2}}$

Sol:

cos ${\displaystyle (\frac{x}{2}+10°)=\frac{\sqrt{3}}{2}}$

cos ${\displaystyle (\frac{x}{2}+10°)}$ = cos 30°

${\displaystyle \frac{x}{2}}$ + 10° = 30°

x = 40°

Question 12.09

Solve for x : sin² x + sin² 30° = 1

Sol:

sin²x + sin²30° = 1
sin²x  = 1 –sin² 30°

sin²x = 1 – ${\displaystyle \frac{1}{4}}$

sin²x = ${\displaystyle \frac{\sqrt{3}}{2}}$

x = 60°

Question 12.1

Solve for x : cos² 30° + cos² x = 1

Sol:

cos²30° + cos² x = 1
cos² x = 1 – cos² 30°

cos² x = 1 – ${\displaystyle \frac{3}{4}}$

cos x = ${\displaystyle \frac{1}{2}}$

x = 60°

Question 12.11

Solve for x : cos² 30° + sin² 2x = 1

Sol:

cos² 30° + sin² 2x = 1
sin² 2x = 1 – cos² 30°

sin² 2x = ${\displaystyle \frac{1}{2}}$

2x = 30°
x =15°

Question 12.12

Solve for x : sin² 60° + cos² (3x- 9°) = 1

Sol:

sin²60° + cos²(3x – 9°) = 1
cos²(3x – 9°) = 1 – sin²60°
cos²(3x – 9°) = 1 – ${\displaystyle \frac{3}{4}}$
cos²(3x – 9°) = ${\displaystyle \frac{1}{4}}$
cos²(3x – 9°) = ${\displaystyle \frac{1}{2}}$
3x – 9° = 60°
3x = 69°
x = 23°

Question 13

If 4 cos² x = 3 and x is an acute angle;
find the value of :
(i) x 
(ii) cos² x + cot² x
(iii) cos 3x (iv) sin 2x

Sol:

(i) 4 cos²x = 3

cos²x = ${\displaystyle \frac{3}{4}}$

cos x = ${\displaystyle \frac{\sqrt{3}}{2}}$

x = 30°

(ii) cos²x + cot²x = cos²30° + cot²30°

 = ${\displaystyle \frac{3}{4}+3}$

= ${\displaystyle \frac{15}{4}}$

= 3${\displaystyle \frac{3}{4}}$

(iii) cos 3x = cos3(30°) = cos 90° = 0

(iv) sin 2x = sin 2(30°) = sin60° = ${\displaystyle \frac{\sqrt{3}}{2}}$ 

Question 14

In ΔABC, ∠B = 90° , AB = y units, BC = ${\displaystyle \left(\sqrt{3}\right)}$ units, AC = 2 units and angle A = x°,
find:
(i) sin x°
(ii) x°
(iii) tan x° 
(iv) use cos x°  to find the value of y.

Sol:

(i) FromΔABC

sin x° = ${\displaystyle \frac{\sqrt{3}}{2}}$

 

(ii) sin x° = ${\displaystyle \frac{\sqrt{3}}{2}}$

sin x° = sin 60°
x = 60°

 

(iii) tan x° = tan 60°

= ${\displaystyle \left(\sqrt{3}\right)}$

 

(iv) cos x° = ${\displaystyle (y\frac{0}{2}}$

cos 60° = ${\displaystyle \frac{y}{2}}$

${\displaystyle \frac{1}{2}=\frac{y}{2}}$

y = 6

Question 15

If 2 cos (A + B) = 2 sin (A - B) = 1;
find the values of A and B.

Sol:

2 cos (A + B) = 1

cos (A + B) = ${\displaystyle \frac{1}{2}}$

cos (A+B) = cos 60°
A + B = 60°            ........( 1)

2 sin (A – B) = 1

2 sin (A – B) = ${\displaystyle \frac{1}{2}}$

A – B = 30°             ........(2)
Adding (1) and (2)

A + B + A – B = 60° + 30°
2A = 90°
A = 45°
A + B = 60°
B = 60° – A 
B = 60 – 45°
B = 15°