# Law of Sines

## Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

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

where $R$ is the circumradius of $\triangle ABC$.

## Proof 1

Construct the altitude from $B$.

From the definition of sine:

$\sin A = \dfrac h c$ and $\sin C = \dfrac h a$

Thus:

$h = c \sin A$

and:

$h = a \sin C$

This gives:

$c \sin A = a \sin C$

So:

$\dfrac a {\sin A} = \dfrac c {\sin C}$

Similarly, constructing the altitude from $A$ gives:

$\dfrac b {\sin B} = \dfrac c {\sin C}$

$\blacksquare$

## Proof 2

Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius.

Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$.

By the Inscribed Angle Theorem:

$\angle ACB = \dfrac {\angle AOB} 2$

From the definition of the circumcenter:

$AO = BO$

From the definition of altitude and the fact that all right angles are congruent:

$\angle AEO = \angle BEO$

Therefore from Pythagoras's Theorem:

$AE = BE$

and then from Triangle Side-Side-Side Equality:

$\angle AOE = \angle BOE$

Thus:

$\angle AOE = \dfrac {\angle AOB} 2$

and so:

$\angle ACB = \angle AOE$

Then by the definition of sine:

$\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$

and so:

$\dfrac c {\sin C} = 2 R$

The same argument holds for all three angles in the triangle, and so:

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

$\blacksquare$

## Proof 3

### Acute Case

Let $\triangle ABC$ be acute.

Construct the circumcircle of $\triangle ABC$.

Let its radius be $R$.

Construct its diameter $BX$ through $B$.

By Thales' Theorem, $\angle BAX$ is a right angle.

$\angle AXB = \angle ACB$

Then:

 $\ds \sin \angle AXB$ $=$ $\ds \dfrac {AB} {BX}$ Definition of Sine of Angle $\ds \leadsto \ \$ $\ds \sin \angle ACB$ $=$ $\ds \dfrac c {2 R}$ $\ds \leadsto \ \$ $\ds 2 R$ $=$ $\ds \dfrac c {\sin C}$

The same construction can be applied to each of the remaining vertices of $\triangle ABC$.

Hence the result.

$\Box$

Let $\triangle ABC$ be obtuse.

As for the acute case, construct the circumcircle of $\triangle ABC$.

Let its radius be $R$.

Construct its diameter $BX$ through $B$.

By Thales' Theorem, $\angle BAX$ is a right angle.

We note that $\Box ABXC$ is a cyclic quadrilateral.

$\angle BAC = 108 \degrees - A$

Hence using a similar argument to the acute case:

 $\ds a$ $=$ $\ds 2 R \map \sin {180 \degrees - A}$ $\ds$ $=$ $\ds 2 R \sin A$

and the result follows.

$\blacksquare$

## Also presented as

Some sources do not include the relation with the circumradius, but instead merely present:

$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C}$

## Also known as

This result is also known as the sine law, sine rule or rule of sines.

## Historical Note

The Law of Sines was documented by Nasir al-Din al-Tusi in his work On the Sector Figure, part of his five-volume Kitāb al-Shakl al-Qattā (Book on the Complete Quadrilateral).