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

RS Aggarwal Class 10 Solutions

**Exercise 6**

**Question 1:**

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

sin60^{0 } cos30^{0 } + cos60^{0 } sin30^{0}

**Question 2:**

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

sin60^{0 }cos30^{0 }– cos60^{0 }sin30^{0}

**Question 3:**

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

cos45^{0} cos30^{0} + sin40^{0} sin30^{0}

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**Question 4:**

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

**Question 5:**

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

**Question 6:**

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

**Question 7:**

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

**Question 8:**

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

**Question 9:**

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

**Question 10:**

**Question 11:**

**Question 12:**

A = 45° ⇒ 2A = 90°

(i) sin 2A = sin90° = 1

(ii) cos 2A = cos90° = 0

**Question 13:**

A = 30° ⇒ 2A = 60°

**Question 14:**

Putting A = 30^{o} ⇒ 2A = 60^{o}

**Question 15:**

Putting A = 30^{o} ⇒ 2 A = 60^{o}

**Question 16:**

Putting A = 30^{o} ⇒ 2 A = 60^{o}

**Question 17:**

A = 45^{o} and B = 45^{o } ⇒ (A + B) = (45^{o} + 450^{o}) = 90^{o}

(A – B) = 45^{o} – 45^{o} = 0

(i) LHS = sin(A + B) = sin90^{o} = 1

RHS = sinA cosB + cosA sinB

= sin 45^{o} + cos45^{o} + cos45^{o} sin45^{o}

Hence, sin (A + B) = sin A cos B + cos A sin B

(ii) A – B = 45^{o} – 45^{o} = 0

LHS = sin (A – B) = sin 0^{o} = 0

RHS = sin A cos B – cos A sin B

Hence, sin(A – B) = sin A cos B – cos A sin B

(iii) LHS = cos(A + B) = cos90^{o} = 0

RHS = cos A cos B – sin A sin B

Hence, cos A cos B – sin A sin B

(iv) LHS = cos(A – B) = cos 0^{o}= 1

RHS = cos A cos B + sin A sin B

Hence, cos(A – B) = cos A cos B + sin A sin B

**Question 18:**

A = 60^{o} and B = 30^{o}, (A +B) = (60^{o} + 30^{o}) = 90^{o}

(A – B) = (60^{o} – 30^{o}) = 30^{o}

(i) LHS = sin(A + B) = sin90^{o} = 1

RHS = sin A cos B + cos A sin B

Hence, sin(A + B) = sin A cos B + cos A sin B

(ii) LHS = sin(A – B) = sin 30^{o} = 1/2

RHS = sin A cos B – cos A sin B

= sin 60^{o} cos30^{o} – cos60^{o} sin 30^{o}

Hence, sin(A – B) = sin A cos B – cos A sin B

(iii) LHS = cos(A + B) = cos90^{o} = 0

RHS = cos A cos B – sin A sin B

Hence, cos (A + B) = cos A cos B – sin A sin B

**Question 19:**

Hence, (A + B) = 45^{o}

**Question 20:**

From right angled ∆ABC,

**Question 21:**

From right angled ∆ABC,

**Question 22:**

From right angled ∆ABC,

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

**Question 23:**

sin (A + B) = 1 sin (A + B) = sin90^{o}

Adding (1) and (2), we get

2A = 90^{o} ⇒ A = 45^{o}

Putting A = 45^{o} in (1) we get

45^{o} + B = 90^{o} B = 45^{o}

Hence, A = 45^{o} and B = 45^{o}

**Question 24:**

Solving (1) and (2), we get

2A = 90^{o} ⇒ A = 45^{o}

Putting A = 45^{o} in (1), we get

45^{o} – B = 30^{o} B = 45^{o} – 30^{o} = 15^{o}

Hence, A = 45^{o}, B = 15^{o}

**Question 25:**

Solving (1) and (2), we get

2A = 90^{o} ⇒ A = 45^{o}

Putting A = 45^{o} in (1), we get

45^{o} – B = 30^{o} ⇒ B = 45^{o} – 30^{o} = 15^{o}

A = 45^{o}, B = 15^{o}

**Question 26:**

Let ABC be an equilateral triangle in which each side is 2a.

**Question 27:**

Let ABC is an equilateral triangle in which each side is 2a.

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