Law of Sines

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Theorem

For any triangle $\triangle ABC$:

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

where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.


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 $O$.

Sine.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:

$\sin C = \sin \left({\angle AOE}\right) = \dfrac {c / 2} R$

and so:

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


Because the same argument holds for all three angles in the triangle:

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


Note that this proof also yields a useful extension of the law of sines:

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

$\blacksquare$


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).


Sources