Law of Sines/Proof 3

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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

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.

$\blacksquare$


Sources

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