# Class 9 SELINA Solutions Maths Chapter 25 - Complementary Angles

## Complementary Angles Exercise Ex. 25

### Solution 1(a)

Correct option: (iii)
45^{o}

sin A = cos A

Now,

### Solution 1(b)

Correct option:
(iv) A + B = 90^{o}

If sin A = cos B (A ≠ B), then A and B are complementary angles.

⇒ A + B = 90^{o}

OR

### Solution 1(c)

Correct option: (i) 1

If A + B = 90^{o},
then

cos A = sin B

cot A = tan B

### Solution 1(d)

Correct option:
(ii) 2^{}

### Solution 1(e)

Correct option:
(iv) ^{}

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 2(iv)

### Solution 2(v)

### Solution 2(vi)

### Solution 2(vii)

### Solution 2(viii)

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 3(v)

### Solution 4(i)

L.H.S.

= tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°

= cot 80° cot 75 ° tan 75° tan 80°

= (cot 80° tan 80°)(cot 75° tan 75°)

= (1)(1)

= 1

= R.H.S.

### Solution 4(ii)

### Solution 5

### Solution 6

(i) We know that for a triangle ABC

A + B + C = 180°

(ii) We know that for a triangle ABC

A + B + C = 180°

B + C* *= 180° - A

### Solution 7

(i)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

### Solution 8

### Solution 9(i)

### Solution 9(ii)

## Complementary Angles Exercise Test Yourself

### Solution 1

### Solution 2

### Solution 3

### Solution 4

Now,

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

So,

### Solution 11

### Solution 12

### Solution 13

*Back answer is A = 14^{o}.
This is possible if the question is as follows:

### Solution 14