RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles

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RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles

Exercise 6

(Question 1 to Question 9)

Question 1:
Solution:

We know that,

sin 60° = √3/2 = cos 30°

and sin 30° = 1/2 = cos 60°

Now,

sin 60° cos 30° + cos 60° sin 30° = (√3/2)( √3/2) + (1/2)(1/2)

= (3/4) + (1/4)

= 4/4

= 1

Question 2:
Solution:

We know that,

cos 60° = 1/2 = sin 30°

and cos 30° = √3/2 = sin 60°

Now,

cos 60° cos 30° — sin 60° sin 30° = (1/2) × (√3/2) – (√3/2) × (1/2)

= (√3/4) – (√3/4)

= 0

Question 3:
Solution:

We know that,

cos 45° = 1/√2 = sin 45°

cos 30° = √3/2 and

sin 30° = 1/2

Now,

cos 45° cos 30° + sin 45° sin 30° = (1/√2)×(√3/2) + (1/√2)(1/2)

= (√3 / 2√2) + (1 / 2√2)

= (√3 + 1) / (2√2)

Question 4:
Solution:

We know that,

sin 30° = 1/2 , sin 60° = √3/2 , sin 90° = 1

cos 30° = √3/2 , cos 45° = 1/√2 , cos 60° = 1/2

sec 60° = 2, tan 45° = 1 and cot 45° = 1

Now,

rs aggarwal class 10 chapter 11 img 2

Question 5:

Solution:

We know that,

cos 30° = √3/2 ⇨ cos2 30° = 3/4

cos 60° = 1/2 ⇨ cos2 60° = 1/4

sec 30° =(2/√3) ⇨ sec2 30° = 4/3

tan 45° = 1 ⇨ tan2 45° = 1

sin 30° = 1/2 ⇨ sin2 30° = 1/4

Now,

(5cos260°+4sec230°-tan245°)/(sin230°+cos230°)

rs aggarwal class 10 chapter 11 img 3

= 67/12

Question 6:
Solution:

We know that,

sin 45° = 1/√2 , cos 60° = 1/2

sin 30° = 1/2 and cos 90° = 0

Now,

2cos2 60° + 3sin2 45° − 3sin2 30° + 2cos2 90°

rs aggarwal class 10 chapter 11 img 4

Question 7:
Solution:

We know that,

Cot 30° = √3, cos 30° = √3/2

Sec 45° = √2 , cosec 30° = 2

Now,

cot230° − 2cos230° − ¾ sec245° + ¼ cosec230°

rs aggarwal class 10 chapter 11 img 5

= 1

Question 8:
Solution:

sin 30° = 1/2 ⇒ sin2 30° = 1/4

cos 45° = 1/√2 = sin 45°

cot 45° = 1 ⇒ cot2 45° = 1

cos 60° = 1/2 ⇒ sec 60° = 2 ⇒ sec2 60° = 4

cos 30° = √3/2 ⇒ sec 30° = 2/√3 ⇒ sec2 30° = 4/3

cosec 45° = 1/sin 45° = √2 ⇒ cosec2 45° = 2

(sin230° + 4cot245° − sec260°)(cosec245° sec230°)

rs aggarwal class 10 chapter 11 img 6

=1/4 x 8/3

= 2/3

Question 9:
Solution:

4/cot230°+1/sin230° – 2cos245°-sin2

rs aggarwal class 10 chapter 11 img 7

= 4/3 + 4/1 – 1 – 0

= 26/6

= 13/3

Question 10:
Solution:

(i)

rs aggarwal class 10 chapter 11 img 9

(ii)

rs aggarwal class 10 chapter 11 img 10

Question 11:
Solution:

(i) L.H.S. = sin 60° cos 30° — cos 60° sin 30°

= (√3/2) × (√3/2) – (1/2)(1/2)

= (3/4) – (1/4)

= 2/4

= 1/2

R.H.S.:

sin 30° = 1/2

LHS = RHS

(ii)

L.H.S. = cos 60° cos 30° + sin 60° sin 30°

= (1/2) × (√3/2) + (√3/2)(1/2)

= (√3/4) + (√3/4)

= √3/2

R.H.S.

cos 30° = √3/2

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

(iii)

L.H.S. = 2 sin 30° cos 30°

= 2 × (1/2) × (√3/2)

= √3/2

R.H.S.

sin 60° = √3/2

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

(iv)L.H.S. = 2 sin 45° cos 45°

= 2 × (1/√2) × (1/√2)

= (2 × 1/2)

= 1

R.H.S. = sin 90° = 1

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

Question 12:
Solution:

A = 45° then 2 A = 90°

(i)Sin 2A = sin90°

RHS:

2 sin 45° cos 45° = 2 × (1/√2) × (1/√2)

= 1

LHS:

sin 90° = 1

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

(ii) cos 2A = cos90° = 0

rs aggarwal class 10 chapter 11 img 11

Question 13.
Solution:

A = 30 ⇒ 2A = 60

(i)

rs aggarwal class 10 chapter 11 img 13

(ii)

rs aggarwal class 10 chapter 11 img 14

(iii)

rs aggarwal class 10 chapter 11 img 15

Question 14:
Solution:

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

If A = 60° and B = 30°, then

To verify: sin 90° = sin 60° cos 30° + cos 60° sin 30°

RHS: sin 60° cos 30° + cos 60° sin 30°

= (√3/2) × (√3/2) + (1/2)(1/2)

= (3/4) + (1/4)

= 4/4

= 1

= sin 90°

= LHS

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

If A = 60° and B = 30°

Verify: cos (90°) = cos 60° cos 30° — sin60° sin 30°

R.H.S. = cos 60° cos 30° – sin 60° sin 30°

= (1/2) × (√3/2) – (√3/2)(1/2)

= (√3/4) – (√3/4)

= 0

= cos 90°

= L.H.S.

Question 15:
Solution:

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

If A = 60° and B = 30°, then

LHS :

= sin(60°-30°)

= sin 30°

= 1/2

R.H.S. = sin 60° cos 30° – cos 60° sin 30°

= (√3/2) × (√3/2) – (1/2) (1/2)

= (3/4) – (1/4)

= 2/4

= 1/2

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

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

If A = 60° and B = 30°, then

Verify: cos (30°) = cos 60° cos 30° + sin 60° sin 30°

R.H.S. = cos 60° cos 30° + sin 60° sin 30°

= (1/2) × (√3/2) + (√3/2)(1/2)

= (√3/4) + (√3/4)

= √3/2

= cos 30°

= L.H.S.

rs aggarwal class 10 chapter 11 img 16

Verified.

Question 16:
Solution:

rs aggarwal class 10 chapter 11 img 17

=( 5/6) /( 5/6)

= 1

This implies, tan(A + B) = 1

= tan 450

Or A + B = 450 . Proved

Question 17:
Solution:

Put A = 30° ⇨ 2A = 60°

rs aggarwal class 10 chapter 11 img 18

The value of tan 60is √3.

Question 18:
Solution:

Put A = 30° then 2 A = 60°cosA=1+cos2A2−−−−−−√

rs aggarwal class 10 chapter 11 img 19

The value of cos 30° is √3/2.

Question 19:
Solution:

Put A = 30° then 2 A = 60°

sinA=1–cos2A2−−−−−−√

Squaring both side, we get

sin2A=1–cos2A2

And,

rs aggarwal class 10 chapter 11 img 20

Sin 300 = 1/2

Question 20:
Solution:

Draw a right angled ∆ABC using given instructions:

rs aggarwal class 10 chapter 11 img 21

Here sin 30 = BC/AC

½ = BC/20

Or BC = 10 cm

By Pythagoras theorem:

(AB)2 = (AC)2 – (BC)2

=(20)2 – (10)2

= 300

AB = 10 √3 cm

Question 21:
Solution:

Draw a right angled ∆ABC using given instructions:

rs aggarwal class 10 chapter 11 img 22

Here sin 30 = BC/AC

½ = 6/AC

Or AC = 12cm

By Pythagoras theorem:

(AB)2 = (AC)2 – (BC)2

=(12)2 – (6)2

= 108

AB = 6 √3 cm

Question 22:
Solution:

From right angled ∆ABC,

rs aggarwal class 10 chapter 11 img 23
  1. BC/AC = sin45

BC32√=12√

Or BC = 3

  1. By Pythagoras theorem

(AB)2 = (AC)2 – (BC)2

= (3√2)2 – (3)2

= 18 – 9

= 9

AB = 3 cm

Question 23:
Solution:

sin (A + B)= 1 or sin (A + B) = sin90° [As sin90° = 1)

A + B = 90° …(1)

Again, Cos(A-B) = 1

= cos 0°

A – B = 0 …(2)

Adding (1) and (2), we get

2A = 90° or A = 45°

Putting A = 45° in (1) we get

45° + B = 90° or B = 45°

Hence, A = 45° and B = 45°.

Question 24:
Solution:

sin (A – B)= 1/2

or sin (A – B) = sin 30°

A – B = 30° …(1)

Again, Cos(A+B) = 1/2

= cos 60°

A + B = 60° …(2)

Solving (1) and (2), we get

2A = 90° or A = 45°

Putting A = 45° in (1), we get

45° – B = 30° or B = 45 – 30° = 15°

Therefore, A = 45°, B = 15°.

Question 25:

Solution:

tan (A – B)= 1/√3

or tan(A – B) = tan 30°

A – B = 30° …(1)

Again, tan(A+B) = √3

= tan 60°

A + B = 60° …(2)

Solving (1) and (2), we get

2A = 90° or A = 45°

Putting A = 45° in (1), we get

45° – B = 30° or B = 45° – 30° = 15°

Therefore, A = 45°, B = 15°

Question 26:

Solution:

Given: 3x = cosec θ and 3/x = cot θ

We know that: cose2 θ – cot2 θ = 1

Substituting the values, we get

(3x)2 – (3/x)2 = 1

9( x2 – 1/ x2) = 1

( x2 – 1/ x2) = 1/9

3 ( x2 – 1/ x2) = 1/3

Question 27:
Solution:

Given: sin (A + B) = sin A cos B + cos A sin B and

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

(i) To find: sin 75°

Put A = 30° and B = 45°, then

sin 75° = sin 30° cos 45° + cos 30° sin 45°

= (1/2) × (1/√2) + (√3/2) × (1/√2)

= (1/2√2) + (√3/2√2)

= (1+√3)/2√2

(ii) Find cos 75°

Put A = 45° and B = 30°, then

cos 15° = cos 45° cos 30° + sin 45° sin 30°

= (1/√2) × (√3/2) + (1/√2) × (1/2)

= (√3 / 2√2) + (1/2√2)

= (1+√3)/2√2

Complete RS Aggarwal Solutions Class 10

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