Law of Sines

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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 $B$.

Law Of Sines 1.png

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 $O$.

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 Congruence:

$\angle AOE = \angle BOE$

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.

Law-of-sines-proof-3.png

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.

From Angles in Same Segment of Circle are Equal:

$\angle AXB = \angle ACB$

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.

Law-of-sines-proof-3-obtuse.png

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, $\angle BCX$ is a right angle.

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

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:

$\angle BXC = 180 \degrees - A$

Hence using a similar argument to the acute case:

\(\ds a\) \(=\) \(\ds 2 R \sin \angle BXC\)
\(\ds \) \(=\) \(\ds 2 R \map \sin {180 \degrees - A}\)
\(\ds \) \(=\) \(\ds 2 R \sin A\)

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

The Law of Sines 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

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