Question

Solve the triangle in which $a=(\sqrt{3}+1),b=(\sqrt{3}-1)$ and $\angle C={60}^{\circ }$

Answer 1

We know that,
In a triangle,
$cosC=\frac{a^2+b^2-c^2}{2ab}$
=> $cos{60}^o=\frac{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-c^2}{2(\sqrt{3}+1)(\sqrt{3}-1)}$
=> $\frac{1}{2}=\frac{4+2\sqrt{3}+4-2\sqrt{3}-c^2}{4}$
=> $c^2=6$
=> Hence, $c=\sqrt{6}$
From sine rule,
$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$
=> $\frac{sinA}{\sqrt{3}+1}=\frac{sinB}{\sqrt{3}+1}=\frac{sin{60}^o}{\sqrt{6}}$
=> $sinA=\frac{\sqrt{3}+1}{2\sqrt{2}}$ & $sinB=\frac{\sqrt{3}-1}{2\sqrt{2}}$
=> Hence, $\angle A={75}^o$ & $\angle B={15}^o$