Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles download pdf.

## 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,

**Question 5:**

**Solution:**

We know that,

cos 30° = √3/2 ⇨ cos^{2} 30° = 3/4

cos 60° = 1/2 ⇨ cos^{2} 60° = 1/4

sec 30° =(2/√3) ⇨ sec^{2} 30° = 4/3

tan 45° = 1 ⇨ tan^{2} 45° = 1

sin 30° = 1/2 ⇨ sin^{2} 30° = 1/4

Now,

(5cos^{2}60°+4sec^{2}30°-tan^{2}45°)/(sin^{2}30°+cos^{2}30°)

= 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,**

**2cos ^{2} 60° + 3sin^{2} 45° − 3sin^{2} 30° + 2cos^{2} 90°**

**Question 7:****Solution:**

We know that,

**Cot 30° = √3, cos 30° = √3/2**

**Sec 45° = √2 , cosec 30° = 2**

Now,

**cot ^{2}30° − 2cos^{2}30° − ¾ sec^{2}45° + ¼ cosec^{2}30°**

= 1

**Question 8:****Solution:**

sin 30° = 1/2 ⇒ sin^{2} 30° = 1/4

cos 45° = 1/√2 = sin 45°

cot 45° = 1 ⇒ cot^{2} 45° = 1

cos 60° = 1/2 ⇒ sec 60° = 2 ⇒ sec^{2} 60° = 4

cos 30° = √3/2 ⇒ sec 30° = 2/√3 ⇒ sec^{2} 30° = 4/3

cosec 45° = 1/sin 45° = √2 ⇒ cosec^{2} 45° = 2

**(sin ^{2}30° + 4cot^{2}45° − sec^{2}60°)(cosec^{2}45° sec^{2}30°)**

=1/4 x 8/3

= 2/3

**Question 9:****Solution:**

**4/cot ^{2}30°+1/sin^{2}30° – 2cos^{2}45°-sin^{2}0°**

**= 4/3 + 4/1 – 1 – 0**

**= 26/6**

**= 13/3**

**Question 10:****Solution:**

**(i)**

**(ii)**

**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

**Question 13.****Solution:**

A = 30 ⇒ 2A = 60

**(i)**

**(ii)**

**(iii)**

**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.

**Verified.**

**Question 16:****Solution:**

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

= 1

This implies, tan(A + B) = 1

= tan 45^{0}

Or A + B = 45^{0} . Proved

**Question 17:****Solution:**

Put A = 30° ⇨ 2A = 60°

The value of tan 60^{o }is √3.

**Question 18:****Solution:**

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

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,**

Sin 30^{0} = 1/2

**Question 20:****Solution:**

Draw a right angled ∆ABC using given instructions:

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:

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,

- BC/AC = sin45

BC32√=12√

Or BC = 3

- 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 Î¸ and 3/x = cot Î¸

We know that: cose^{2} Î¸ – cot^{2} Î¸ = 1

Substituting the values, we get

(3x)^{2} – (3/x)^{2} = 1

9( x^{2} – 1/ x^{2}) = 1

( x^{2} – 1/ x^{2}) = 1/9

3 ( x^{2} – 1/ x^{2}) = 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**

If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share **alarity.in** to your friends.

**Best of Luck For Your Future!!**

## No comments: