Question

If $R$ is the radius of circumcircle of $\Delta ABC$, prove the following formula :
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Answer 1

Describe a circle with centre O and passing through the vertices of the $\Delta$ ABC. Clearly $\Delta$ BOD = $\Delta$ COD, we know that the angle subtended at the centre is double the angle subtended on the circumference, i.e.,
$\angle$ BOC = 2A
$\therefore \angle$ BOD = $\angle$ COD = A.
Now if OD is perp. to side BC, then
$BD=DC=\frac{a}{2}and\frac{BD}{OB}=\sin A$
or $\frac{a}{2R}=\sin A;\therefore \frac{a}{\sin A}=2R.$
Hence by sine formula,
$\frac{a}{\sin A}=\frac{b}{sinB}=\frac{c}{\sin C}=2R.$