# R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 6 TRatios of Some Particular Angles

## Chapter 6 – T-Ratios of Some Particular Angles Exercise Ex. 6

Question 1

Evaluate the following:

sin 60o cos 30o + cos 60o sin 30o

Solution 1

On substituting the value of various T-ratios, we get

sin60o cos30o + cos60o sin30o

Question 2

Evaluate the following:

cos 60o cos 30o – sin 60o sin 30o

Solution 2

On substituting the value of various T-ratios, we get

cos60o cos30o – sin60o sin30o

Question 3

Evaluate the following:

cos 45o cos 30o + sin 45o sin 30o

Solution 3

On substituting the value of various Tratios, we get

cos45o cos30o + sin45o sin30o

Question 4

Evaluate the following:

Solution 4

On substituting the value of various Tratios, we get

Question 5

Solution 5

Question 6

Evaluate the following :

Solution 6

On substituting the value of various Tratios, we get

Question 7

Evaluate the following:

Solution 7

On substituting the value of various Tratios, we get

Question 8

Evaluate the following:

Solution 8

On substituting the value of various Tratios, we get

Question 9

Evaluate the following:

Solution 9

On substituting the value of various Tratios, we get

Question 10

Show that:

Solution 10

(i)

(ii)

L.H.S = R.H.S.

Question 11

Verify each of the following:

(i)sin 60° cos30° – cos60° sin30° = sin30°

(ii)cos60° cos30° + sin60° sin30° = cos30°

(iii)2sin30° cos30° = sin60°

(iv)2sin45° cos45° = sin90°

Solution 11

(i)

R.H.S. = L.H.S.

Hence, sin60° cos30° – cos60° sin30° = sin30°

(ii)L.H.S. = cos60° cos30° + sin60° sin30°

(iii)

R.H.S. = L.H.S.

Hence,2sin30° cos30° = sin60°

(iv)

R.H.S. = sin90° = 1

R.H.S. = L.H.S.

Hence, 2 sin 45° cos45° = sin90°

Question 12

If A = 45°, verify that

(i) sin 2A = 2 sin A cosA

(ii)  cos 2 A =

Solution 12

A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1

(ii) cos2A = cos90° = 0

Question 13

If A = 30o, verify that:

Solution 13

A = 302A = 60

(i)

(ii)

(iii)

Question 14

If
A = 60° and B = 30°, verify that:

sin
(A + B) = sin A cos B + cos
A sin B

Solution 14

Question 15

If
A = 60° and B = 30°, verify that:

cos (A + B) = cos A cos B – sin A sin B

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

If A and B are acute angles such that , show that A + B = 45.

Solution 19

Hence, (A + B) = 45

Question 20

Using the formula, find the value of tan60, it being given that .

Solution 20

Putting A = 30o 2 A = 60o

Question 21

Using the formula, find the value of cos 30, it being given that .

Solution 21

Putting A = 30o 2 A = 60o

Question 22

(v)Using the formula, find the value of sin 30, it being given that .

Solution 22

Putting A = 30o 2 A = 60o

Question 23

In the adjoining figure, ?ABC is a right-angled triangle in which B = 90o, A = 30o and AC = 20cm. Find (i) BC, (ii) AB.

Solution 23

From right angled ABC,

Question 24

In the adjoining figure, ABC is right-angled at B and A = 30o. If BC = 6 cm, find (i) AB, (ii) AC.

Solution 24

From right angled ABC,

Question 25

In the adjoining figure, ABC is right-angled at B and A = 45o. If AC = cm, find (i) BC, (ii)AB.

Solution 25

From right angled  ABC,

(i)

(ii)     By Pythagoras theorem

Hence, (i) BC = 3cm and (ii) AB = 3cm

Question 26

If sin(A + B) = 1 and cos(A – B) = 1, and A > B, then find A and B.

Solution 26

sin (A + B)= 1  sin (A + B) = sin90

Adding (1) and (2), we get

2A = 90oA = 45o

Putting A = 45o in (1) we get

45o + B = 90oB = 45o

Hence, A = 45o and B = 45o

Question 27

If and A > B, find A and B.

Solution 27

Solving (1) and (2), we get

2A = 90oA = 45o

Putting A = 45o in (1), we get

45o – B = 30oB = 45 – 30o = 15o

Hence, A = 45o, B = 15o

Question 28

If and A > B, then find A and B.

Solution 28

Solving (1) and (2), we get

2A = 90oA = 45o

Putting A = 45o in (1), we get

45o – B = 30o B = 45o – 30o  = 15o

A = 45o, B = 15o

Question 29

Solution 29

Question 30

If
sin (A + B) = sin A cos B + cos
A sin B and cos (A – B) = cos
A cos B + sin A sin B, find the values of (i) sin 75° and (ii) cos 15°.

Solution 30

error: Content is protected !!