Question

In a triangle $ABC$, if $B=30°,C=45°$ and $a=\sqrt{3}+1$ , then find the area of triangle $ABC$.

Answer 1

$\frac{a}{\sin A}=\frac{b}{\sin B}\Rightarrow b=\frac{\sqrt{3}+1}{\sin 75}\left(\sin 30°\right)=\sqrt{2}$
$\therefore \Delta =\frac{1}{2}ab\sin C=\frac{1}{2}(\sqrt{3}+1)\left(\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{3}+1}{2}$