# R S Aggarwal Solutions for Class 10 Maths Chapter 11 T-Ratios of Some Particular Angles

R S Aggarwal Class 10 Solutions for Chapter 11 T-Ratios of Some Particular Angles are available here. This chapter is composed of only one exercise which includes concepts like trigonometric ratios of different angles (45 degrees, 60 degrees and 30 degrees) and axioms for T-ratios. The solutions created are genuine and can be referred by the students to prepare for their Maths exam. The set of questions and answers helps students to master T-Ratios.

### Access Answers to Maths R S Aggarwal Chapter 11 T-Ratios of Some Particular Angles Exercise 11

Evaluate each of the following:

(Question 1 to Question 9)

Question 1:
Solution:

We know that,

sin 60Â° = âˆš3/2 = cos 30Â°

and sin 30Â° = 1/2 = cos 60Â°

Now,

sin 60Â° cos 30Â° + cos 60Â° sin 30Â° = (âˆš3/2)( âˆš3/2) + (1/2)(1/2)

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

= 4/4

= 1

Question 2:
Solution:

We know that,

cos 60Â° = 1/2 = sin 30Â°

and cos 30Â° = âˆš3/2 = sin 60Â°

Now,

cos 60Â° cos 30Â° â€” sin 60Â° sin 30Â° = (1/2) Ã— (âˆš3/2) – (âˆš3/2) Ã— (1/2)

= (âˆš3/4) – (âˆš3/4)

= 0

Question 3:
Solution:

We know that,

cos 45Â° = 1/âˆš2 = sin 45Â°

cos 30Â° = âˆš3/2 and

sin 30Â° = 1/2

Now,

cos 45Â° cos 30Â° + sin 45Â° sin 30Â° = (1/âˆš2)Ã—(âˆš3/2) + (1/âˆš2)(1/2)

= (âˆš3 / 2âˆš2) + (1 / 2âˆš2)

= (âˆš3 + 1) / (2âˆš2)

Question 4:
Solution:

We know that,

sin 30Â° = 1/2 , sin 60Â° = âˆš3/2 , sin 90Â° = 1

cos 30Â° = âˆš3/2 , cos 45Â° = 1/âˆš2 , cos 60Â° = 1/2

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

Now,

Question 5:

Solution:

We know that,

cos 30Â° = âˆš3/2 â‡¨ cos2 30Â° = 3/4

cos 60Â° = 1/2 â‡¨ cos2 60Â° = 1/4

sec 30Â° =(2/âˆš3) â‡¨ sec2 30Â° = 4/3

tan 45Â° = 1 â‡¨ tan2 45Â° = 1

sin 30Â° = 1/2 â‡¨ sin2 30Â° = 1/4

Now,

(5cos260Â°+4sec230Â°-tan245Â°)/(sin230Â°+cos230Â°)

= 67/12

Question 6:
Solution:

We know that,

sin 45Â° = 1/âˆš2 , cos 60Â° = 1/2

sin 30Â° = 1/2 and cos 90Â° = 0

Now,

2cos2Â 60Â° + 3sin2Â 45Â° âˆ’ 3sin2Â 30Â° + 2cos2Â 90Â°

Question 7:
Solution:

We know that,

Cot 30Â° = âˆš3, cos 30Â° = âˆš3/2

Sec 45Â° = âˆš2 , cosec 30Â° = 2

Now,

cot230Â° âˆ’ 2cos230Â° âˆ’ Â¾ sec245Â° + Â¼ cosec230Â°

= 1

Question 8:
Solution:

sin 30Â° = 1/2 â‡’ sin2 30Â° = 1/4

cos 45Â° = 1/âˆš2 = sin 45Â°

cot 45Â° = 1 â‡’ cot2 45Â° = 1

cos 60Â° = 1/2 â‡’ sec 60Â° = 2 â‡’ sec2 60Â° = 4

cos 30Â° = âˆš3/2 â‡’ sec 30Â° = 2/âˆš3 â‡’ sec2 30Â° = 4/3

cosec 45Â° = 1/sin 45Â° = âˆš2 â‡’ cosec2 45Â° = 2

(sin230Â° + 4cot245Â° âˆ’ sec260Â°)(cosec245Â° sec230Â°)

=1/4 x 8/3

= 2/3

Question 9:
Solution:

4/cot230Â°+1/sin230Â° â€“ 2cos245Â°-sin20Â°

= 4/3 + 4/1 â€“ 1 â€“ 0

= 26/6

= 13/3

Question 10:
Solution:

(i)

(ii)

Question 11:
Solution:

(i) L.H.S. = sin 60Â° cos 30Â° â€” cos 60Â° sin 30Â°

= (âˆš3/2) Ã— (âˆš3/2) â€“ (1/2)(1/2)

= (3/4) â€“ (1/4)

= 2/4

= 1/2

R.H.S.:

sin 30Â° = 1/2

LHS = RHS

(ii)

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

= (1/2) Ã— (âˆš3/2) + (âˆš3/2)(1/2)

= (âˆš3/4) + (âˆš3/4)

= âˆš3/2

R.H.S.

cos 30Â° = âˆš3/2

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

(iii)

L.H.S. = 2 sin 30Â° cos 30Â°

= 2 Ã— (1/2) Ã— (âˆš3/2)

= âˆš3/2

R.H.S.

sin 60Â° = âˆš3/2

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

(iv)L.H.S. = 2 sin 45Â° cos 45Â°

= 2 Ã— (1/âˆš2) Ã— (1/âˆš2)

= (2 Ã— 1/2)

= 1

R.H.S. = sin 90Â° = 1

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

Question 12:
Solution:

A = 45Â° then 2 A = 90Â°

(i)Sin 2A = sin90Â°

RHS:

2 sin 45Â° cos 45Â° = 2 Ã— (1/âˆš2) Ã— (1/âˆš2)

= 1

LHS:

sin 90Â° = 1

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

(ii) cos 2A = cos90Â° = 0

Question 13.
Solution:

A = 30 â‡’ 2A = 60

(i)

(ii)

(iii)

Question 14:
Solution:

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

If A = 60Â° and B = 30Â°, then

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

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

= (âˆš3/2) Ã— (âˆš3/2) + (1/2)(1/2)

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

= 4/4

= 1

= sin 90Â°

= LHS

(ii) cos (A + B) = cos A cos B â€” sin A sin B

If A = 60Â° and B = 30Â°

Verify: cos (90Â°) = cos 60Â° cos 30Â° â€” sin60Â° sin 30Â°

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

= (1/2) Ã— (âˆš3/2) – (âˆš3/2)(1/2)

= (âˆš3/4) – (âˆš3/4)

= 0

= cos 90Â°

= L.H.S.

Question 15:
Solution:

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

If A = 60Â° and B = 30Â°, then

LHS :

= sin(60Â°-30Â°)

= sin 30Â°

= 1/2

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

= (âˆš3/2) Ã— (âˆš3/2) – (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 = 60Â° and B = 30Â°, then

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

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

= (1/2) Ã— (âˆš3/2) + (âˆš3/2)(1/2)

= (âˆš3/4) + (âˆš3/4)

= âˆš3/2

= cos 30Â°

= L.H.S.

Verified.

Question 16:
Solution:

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

= 1

This implies, tan(A + B) = 1

= tan 450

Or A + B = 450 . Proved

Question 17:
Solution:

Put A = 30Â° â‡¨ 2A = 60Â°

The value of tan 60o is âˆš3.

Question 18:
Solution:

Put A = 30Â° then 2 A = 60Â°

$cos A = \sqrt{\frac{1 + cos2A}{2}}$

The value of cos 30Â° is âˆš3/2.

Question 19:
Solution:

Put A = 30Â° then 2 A = 60Â°

$sin A = \sqrt{\frac{1 – cos2A}{2}}$

Squaring both side, we get

$sin^{2} A = \frac{1 – cos2A}{2}$

And,

Sin 300 = 1/2

Question 20:
Solution:

Draw a right angled âˆ†ABC using given instructions:

Here sin 30 = BC/AC

Â½ = BC/20

Or BC = 10 cm

By Pythagoras theorem:

(AB)2 = (AC)2 â€“ (BC)2

=(20)2 â€“ (10)2

= 300

AB = 10 âˆš3 cm

Question 21:
Solution:

Draw a right angled âˆ†ABC using given instructions:

Here sin 30 = BC/AC

Â½ = 6/AC

Or AC = 12cm

By Pythagoras theorem:

(AB)2 = (AC)2 â€“ (BC)2

=(12)2 â€“ (6)2

= 108

AB = 6 âˆš3 cm

Question 22:
Solution:

From right angled âˆ†ABC,

1. BC/AC = sin45
$\frac{BC}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$

Or BC = 3

1. By Pythagoras theorem

(AB)2 = (AC)2 â€“ (BC)2

= (3âˆš2)2 – (3)2

= 18 – 9

= 9

AB = 3 cm

Question 23:
Solution:

sin (A + B)= 1 or sin (A + B) = sin90Â° [As sin90Â° = 1)

A + B = 90Â° â€¦(1)

Again, Cos(A-B) = 1

= cos 0Â°

A â€“ B = 0 â€¦(2)

Adding (1) and (2), we get

2A = 90Â° or A = 45Â°

Putting A = 45Â° in (1) we get

45Â° + B = 90Â° or B = 45Â°

Hence, A = 45Â° and B = 45Â°.

Question 24:
Solution:

sin (A – B)= 1/2

or sin (A – B) = sin 30Â°

A – B = 30Â° â€¦(1)

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

= cos 60Â°

A + B = 60Â° â€¦(2)

Solving (1) and (2), we get

2A = 90Â° or A = 45Â°

Putting A = 45Â° in (1), we get

45Â° â€“ B = 30Â° or B = 45 â€“ 30Â° = 15Â°

Therefore, A = 45Â°, B = 15Â°.

Question 25:

Solution:

tan (A – B)= 1/âˆš3

or tan(A – B) = tan 30Â°

A – B = 30Â° â€¦(1)

Again, tan(A+B) = âˆš3

= tan 60Â°

A + B = 60Â° â€¦(2)

Solving (1) and (2), we get

2A = 90Â° or A = 45Â°

Putting A = 45Â° in (1), we get

45Â° â€“ B = 30Â° or B = 45Â° â€“ 30Â° = 15Â°

Therefore, A = 45Â°, B = 15Â°

Question 26:

Solution:

Given: 3x = cosec Î¸ and 3/x = cot Î¸

We know that: cose2 Î¸ â€“ cot2 Î¸ = 1

Substituting the values, we get

(3x)2 â€“ (3/x)2 = 1

9( x2 â€“ 1/ x2) = 1

( x2 â€“ 1/ x2) = 1/9

3Â ( x2 â€“ 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 75Â°

Put A = 30Â° and B = 45Â°, then

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

= (1/2) Ã— (1/âˆš2) + (âˆš3/2) Ã— (1/âˆš2)

= (1/2âˆš2) + (âˆš3/2âˆš2)

= (1+âˆš3)/2âˆš2

(ii) Find cos 75Â°

Put A = 45Â° and B = 30Â°, then

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

= (1/âˆš2) Ã— (âˆš3/2) + (1/âˆš2) Ã— (1/2)

= (âˆš3 / 2âˆš2) + (1/2âˆš2)

= (1+âˆš3)/2âˆš2

## R S Aggarwal Solutions for Chapter 11 T-Ratios of Some Particular Angles Topics

In this chapter students will study important concepts on T-Ratios of Some Particular Angles as listed below:

• T-Ratios of Some Particular Angles introduction
• Trigonometric ratios of 45 degrees
• Trigonometric ratios of 60 degrees and 30 degrees
• Axioms for T-ratios of 0 degrees
• Axioms for T-ratios of 90 degrees

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 11 T-Ratios of Some Particular Angles

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