### 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,

**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 θ and 3/x = cot θ

We know that: cose^{2} θ – cot^{2} θ = 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**

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