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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles

**Exercise 6**

**Question 1.**

**Solution:**

On substituting the value of various T-ratios, we get

sin60° cos30° + cos60° sin30°

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**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°

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**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:**

Hope given RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles are helpful to complete your math homework.

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