# Law of Sines

How to solve the law of sines problems: formula, proof, examples, and their solutions.

## Formula

*a*/(sin *A*) = *b*/(sin *B*) = *c*/(sin *C*)*a*, *b*, *c*: Sides of a triangle*A*, *B*, *C*: Angles of a triangle

## Proof

The area of a triangle

can be written in three ways:

(by choosing the different angles)

(area) = (1/2)⋅*bc* sin *A*

= (1/2)⋅*ac* sin *B*

= (1/2)⋅*ab* sin *C*

Area of the triangle (using sine)

Divide each side by (1/2)⋅*abc*.

Then the dark gray factors will be cancelled.

So (sin *A*)/*a* = (sin *B*)/*b* = (sin *C*)/*c*.

Switch the numerators and the denominators.

Then *a*/(sin *A*) = *b*/(sin *B*) = *c*/(sin *C*).

## Example 1

The law of sines is used when

[1 side, 2 opposite angles] → [other opposite side]

or

[2 sides, 1 opposite angle] → [other opposite angle].

([Given] → [Find])

Sides: *x*, 12

Opposite angles: 45º, 60º*x*/(sin 45º) = 12/(sin 60º)

Draw a 45-45-90 triangle and a 30-60-90 triangle.

Sine: SOH.

So sin 45º = 1/√2.

And sin 60º = √3/2.

So *x* = 12⋅[(sin 45º)/(sin 60º)]

= 12⋅[(1/√2)/(√3/2)].

## Example 2

To use the law of sines,

find the measure of the brown angle.

(= the angle opposite to the known side: 4)

m∠(brown) = 45º.

Interior angles of a triangle

Sides: *x*, 4

Opposite angles: 30º, 45º*x*/(sin 30º) = 4/(sin 45º)

Draw a 30-60-90 triangle and a 45-45-90 triangle.

Sine: SOH.

So sin 30º = 1/2.

And sin 45º = 1/√2.

So *x* = 4⋅[(sin 30º)/(sin 45º)]

= 4⋅[(1/2)/(1/√2)].