Law of Sines

For a random $\triangle ABC$ , $a$ , $b$ , and $c$ are the length of the sides of the triangle, and seperately, $\angle A$ , $\angle B$ , and $\angle C$ are the opposite angles, then the Law of Sines points out:

\frac{a}{\sin{\angle A}}=\frac{b}{\sin{\angle B}}=\frac{c}{\sin{\angle C}}= 2R

where $R$ is the length of the radius of the circumscribed circle of $\triangle ABC$ .

Derivation

$\odot O$ is the circumscribed circle of $\triangle ABC$ ; connect $OB$ and $OC$ ; draw a prependicular line from $O$ to side $BC$ and intersect with it at point $D$ .

Note that $\angle A$ is an inscribed angle, while $\angle BOC$ is the central angle subtending $\overset{\frown}{BC}$ , which means:

\angle A=\frac{1}{2}\angle BOC

\begin{align*} &\because OB=OC=R, \ OD\perp BC \\ &\therefore \angle BOD=\angle COD= \frac{1}{2}\angle BOC=\angle A\\ &\therefore \sin{\angle A}=\sin{\angle COD}=\frac{CD}{OC}=\frac{\frac{1}{2}a}{R}\\ &\implies \frac{a}{\sin {\angle A}}= 2R \end{align*}

Similarly, from other directions, we may also get:

\frac{b}{\sin {\angle B}}= 2R,\ \frac{c}{\sin {\angle C}}= 2R

Hence:

\frac{a}{\sin{\angle A}}=\frac{b}{\sin{\angle B}}=\frac{c}{\sin{\angle C}}= 2R \tag*{$\blacksquare$}