# LAW OF SINES WORKSHEET

1. In a triangle ABC, if a = 2, b = 3 and sinA = 2/3, find B.

2. If the angles of a triangle are in the ratio 1 : 2 : 3, prove that the corresponding sides are in the ratio 1 : √3 : 2.

3. The angles of a triangle ABC are in arithmetic sequence If b/c = √3/2, find ∠A.

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

5. In ΔABC, ∠B = 45°∠C = 105° and a = 2, find b. Using Law of Sines,

a/sinA = b/sinB

Substitute.

2/(2/3) = 3/sinB

3 = 3/sinB

3sinB = 3

sinB = 1

B = 90°

From the ratio 1 : 2 : 3, the angles of a triangle are assumed to be x, 2x and 3x.

x + 2x + 3x = 180°

6x = 180°

x = 30°

So, the angles are 30°, 60° and 90°.

Let a, b and c be the corresponding side lengths of the angles 30°, 60° and 90° respectively.

Using Law of Sines,

a/sin30° = b/sin60° = c/sin90°

Let a/sin30° = b/sin60° = c/sin90° = k.

a/sin30° = k

a = ksin30°

a = k/2

b/sin60° = k

b = ksin60°

b = k√3/2

c/sin90° = k

c = ksin90°

c = k

a : b : c = k/2 : k√3/2 : k

= k : k√3 : 2k

= 1 : √3 : 2

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

∠B -  A = ∠C - ∠B

2∠B = ∠A + ∠C ----(1)

Add ∠B to both sides.

3∠B = ∠A + ∠B + ∠C

3∠B = 180°

∠B = 6

Using Law of Sines,

In (1), substitute ∠B = 60° and ∠C = 45° and solve for ∠A.

2∠B = ∠A + ∠C

2(60°) = ∠A + 45°

120° = ∠A + 45°

75° = ∠A

Law of Sines :

a/sinA = b/sinB = c/sinC

a/sin45° = b/sin60° = c/sin75°

Let a/sin45° = b/sin60° = c/sin75° = k

a/sin45° = k

a = ksin45°

a = k√2/2

b/sin60° = k

b = ksin60°

b = k√3/2

c/sin90° = k

c = ksin75°

Find the value of sin75°.

sin75° = sin(45° + 30°)

= sin(45° + 30°)

= sin45°cos30° + cos45°sin30°

In ΔABC,

A + B + C = 180°

A + 45° + 105° = 180°

A + 150° = 180°

A = 30°

By Law of Sines :

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