**Questions based on sine rule :**

Here we are going to see some example problems based on sine rule.

**Example 1 :**

In a triangle ABC, if a = 2, b = 3 and sin A = 2/3, find <B

**Solution :**

We have,

a/sin A = b/sin B = c/sin C

2/(2/3) = 3/sin B ==> 3 = 3/sin B ==> sin B = 1

Hence ∠B = 90 degree

**Example 2 :**

If any triangle the angles to be one another as 1 : 2 : 3, prove that the corresponding sides are 1 : √3 : 2

**Solution :**

Let the measures of the angles be x,2x and 3x. Then,

X + 2x + 3x = 180

6x = 180 ==> x = 30

So, the angles are 30˚, 60˚ and 90˚

Let a, b and c be the lengths of the sides opposite to these angles. Then,

a/sin 30˚ = b/sin 60˚ = c/sin 90˚

a : b : c = sin 30˚ : sin 60˚ : sin 90˚

a : b : c = 1/2 : √3/2 : 1 ==> 1 : √3 : 2

**Example 3 : **

The angles of a triangle ABC are in A.P and it is being that b:c = √3 : 2. Find ∠A

**Solution :**

It is given that the angles ∠A, ∠B and ∠C are in A.P

2∠B = ∠A + ∠C

3∠B = ∠A + ∠B + ∠C ==> 3∠B = 180 ==> ∠B = 60

Now,

b/sin B = c/sin C

b/c = sin B/sin C

√3/2 = sin 60/sin C

√3/2 = (√3/2)/sin C ==> sin C = 1/√2

∠A = 180 – (∠B + ∠C)

∠A = 180 - (60+45) ==> 75

**Example 4 : **

If in a triangle ABC, ∠A = 45˚, ∠B = 60˚ and ∠C = 75˚, find the ratio of its sides.

**Solution :**

To find the ratio of the sides, we need to use sin formula

a/sin A = b/sin B = c/sin C

a/sin45˚ = b/sin 60˚ = c/sin75˚

To find the value of sin 75˚, we need to use the compound angles formula

sin 75˚ = sin (45˚ + 30˚)

==> sin45˚cos30˚ + cos45˚sin30˚

==> (1/√2) (√3/2) + (1/√2) (1/2)

==> (√3/2√2) + (1/2√2) ==> (√3 + 1)/2√2

a/(√2/2) = b/(√3/2) = c/(√3+1/2√2)

a:b:c ==> √2 λ/2 : b = √3 λ/2 : (√3+1)/2√2

2a/√2 = λ a = √2 λ/2 |
2b/√3 = λ 2b = √3λ b = √3 λ/2 |
2√3c/(√3+1) = λ c = (√3+1) λ /2√2 |

Multiplying by 2, we get

√2 λ : √3 λ : (√3+1) λ/√2

Multiplying by √2, we get

2 λ : √6 λ : (√6 + √2) λ

Hence the sides are in the ratio 2 : √6 : √6 + √2

**Example 5 :**

If in any triangle ABC, ∠c = 105˚, ∠B = 45˚, a = 2 then find b.

**Solution :**

∠A + ∠B + ∠C = 180˚

∠A + 45˚ + 105˚ = 180˚

∠A = 30˚

a/sin A = b/sin B = c/sin C

2/sin 30˚ = b/sin 45˚ = c/sin 105

2/(1/2) = b/(1/√2) = c/sin 105

4 = √2b

b = 4/√2 ==> 4√2/2 ==> 2√2

Hence the value of b is 2√2

sin (B + C) = sin A sin (C + A) = sin B sin (A + B) = sin C |
cos (B + C) = -cos A, cos (C + A) = cos B, cos (A + B) = -cos C |

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