Geometry • Topic 13

Law of Sines

The Law of Sines relates the side lengths of a triangle to the sines of their opposite angles. It is particularly useful for finding circumradii.

Statement

Theorem (Law of Sines)

In any triangle

A BC

with side lengths

a, b, c

and circumradius

R

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R

From the Law of Sines, we derive a powerful area formula:

A re a = \frac{1}{2} ab sin C = \frac{ab c}{4 R}

When given two sides and a non-included angle (e.g.,

a, b, ∠ A

), the Law of Sines can yield two possible values for

∠ B

(since

sin B = sin (18 0^{\circ} - B)

This corresponds to the geometric ambiguity where two different triangles can be formed.

Exercise (Problem 1)

△ A BC

a = 10

∠ A = 3 0^{\circ}

, and

∠ B = 4 5^{\circ}

. Find side

b

and the circumradius

R

Exercise (Problem 2)

Prove that

\frac{a - b}{a + b} = \frac{t a n ( \frac{A - B}{2} )}{t a n ( \frac{A + B}{2} )}

(Law of Tangents) using the Law of Sines.

Exercise (Problem 3)

In a cyclic quadrilateral

A BC D

, prove that

\frac{A C}{B D} = \frac{A B \cdot D A + BC \cdot C D}{A B \cdot BC + C D \cdot D A}