Question

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

Answer 1

From $\cos C=\frac{a^2+b^2-c^2}{2ab}$, we have
$\frac{1}{2}=\frac{(1+\sqrt{3}{)}^2+4-c^2}{2(1+\sqrt{3})2}$
$2+2\sqrt{3}=1+3+2\sqrt{3}+4-c^2$
$c^2=6$
$\therefore c=\sqrt{6}$
Also, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\frac{\sin A}{1+\sqrt{3}}=\frac{\sin B}{2}=\frac{\sqrt{3}/2}{\sqrt{6}}$
$\therefore \sin B=\frac{1}{\sqrt{2}}$

$B={45}^{\circ }$ and $A={180}^{\circ }-({60}^{\circ }+{45}^{\circ })={75}^{\circ }$