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# Class 10 RD SHARMA Solutions Maths Chapter 10 - Trigonometric Ratios

## Trigonometric Ratios Exercise Ex. 10.1

### Solution 2

In ABC by applying Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7)2
= 576 + 49
= 625
AC =  = 25 cm

Given:

## Trigonometric Ratios Exercise Ex. 10.2

### Solution 30 (ii)

Given: tan(A + B) = 1 and

Therefore,

A + B = 45o … (i)

A - B = 30o … (ii)

Adding the two equations, we get

## Trigonometric Ratios Exercise Ex. 10.3

### Solution 6 (iii)

Given: A = 90o

For a triangle ABC, A + B + C = 90o

### Solution 9 (xi)

Using the identities

### Solution 11 (i)

For a triangle ABC, A + B + C = 90o

### Solution 18

Given: tan 2A = cot(A - 18o)

As tan x = cot(90o - x), we have

cot(90o - 2A) = cot(A - 18o)

90o - 2A = A - 18o

3A = 108o

Therefore, A = 36o.

## Trigonometric Ratios Exercise 10.56

### Solution 4

So, the correct option is (a).

## Trigonometric Ratios Exercise 10.57

### Solution 10

So, the correct option is (d).

### Solution 11

So, the correct option is (c).

### Solution 15

So, the correct option is (b).

### Solution 16

So, the correct option is (a).

### Solution 17

So, the correct option is (b).

### Solution 18

So, the correct option is (d).

## Trigonometric Ratios Exercise 10.58

### Solution 20

So, the correct option is (b).

### Solution 27

So, the correct option is (d).

### Solution 28

So, the correct option is (a).

## Trigonometric Ratios Exercise 10.59

### Solution 33

So, the correct option is (c).