Question

If in triangle ABC, $a=(1+\sqrt{3})cm,b=2cm$ and $\angle C={60}^0$ , find the other two angles and third side.

Answer 1

$c^2=a^2+b^2-2abcosC=6\therefore c=\sqrt{6}$.

$\frac{b}{sinB}=\frac{c}{sinC}=\frac{\sqrt{6}}{sin{60}^0}=\frac{2\sqrt{6}}{\sqrt{3}}=2\sqrt{2}$.
$\therefore sinB=\frac{b}{2\sqrt{2}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$.

$\therefore B={45}^0({135}^0$ is ruled out.)

Hence$A={180}^0-(B+C)={75}^0$.