Law of Sines

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Theorem

For any triangle $\triangle ABC$:

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

where:

$a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively
$R$ is the circumradius of $\triangle ABC$.


Proof 1

Construct the altitude from $B$.

Law Of Sines 1.png

It can be seen from the definition of sine that:

$\sin A = \dfrac h c$ 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 $O$.

Law-of-sines.png

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:

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

$\blacksquare$


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.


Also see


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).


Sources