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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Maths book pdf download. Now you will get step by step solution to each question.

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

Question 1.
Solution:
On substituting the value of various T-ratios, we get
sin60° cos30° + cos60° sin30°

Question 2.
Solution:
On substituting the value of various T-ratios, we get
cos60° cos30° – sin60° sin30°

Question 3.
Solution:
On substituting the value of various Tratios, we get
cos45° cos30° + sin45° sin30°

Question 4.
Solution:
On substituting the value of various Tratios, we get

Question 5.
Solution:

Question 6.
Solution:
On substituting the value of various Tratios, we get

Question 7.
Solution:
On substituting the value of various Tratios, we get

Question 8.
Solution:
On substituting the value of various Tratios, we get

Question 9.
Solution:
On substituting the value of various Tratios, we get

Question 10.
Solution:
(i)

(ii)

Question 11.
Solution:
(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.
Solution:
A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1

(ii) cos2A = cos90° = 0

Question 13.
Solution:
A = 30 ⇒ 2A = 60

(i)

(ii)

(iii)

Question 14.
Solution:
(i)

(ii)

Question 15.
Solution:
(i)

(ii)

(iii)

Question 16.
Solution:

Hence, (A + B) = 45

Question 17.
Solution:
Putting A = 30° 2 A = 60°

Question 18.
Solution:
Putting A = 30° 2 A = 60°

Question 19.
Solution:
Putting A = 30° 2 A = 60°

Question 20.
Solution:
From right angled ∆ABC,

Question 21.
Solution:
From right angled ∆ABC,

Question 22.
Solution:
From right angled  ∆ABC,

(i)

(ii)
By Pythagoras theorem

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

Question 23.
Solution:
sin (A + B)= 1  sin (A + B) = sin90°

Adding (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1) we get
45° + B = 90° ⇒ B = 45°
Hence, A = 45° and B = 45°.

Question 24.
Solution:

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45 – 30° = 15°
Hence, A = 45°, B = 15°.

Question 25.
Solution:

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45° – 30°  = 15°
A = 45°, B = 15°

Question 26.
Solution:

Question 27.
Solution:

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