Law of Sines

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.

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

a, b, c: Sides of a triangle
A, B, C: Angles of a triangle

Proof

Law of Sines: Proof of the Formula

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

Find the value of x. Sides: x, 12. Opposite angles: 45, 60 degrees.

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)].

Complex fraction

Rationalizing a denominator

Example 2

Find the value of x. Sides: 4, x. Angles adjacent to side 4: 30, 105 degrees.

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)].

Complex fraction