Question

If $a=\sqrt{3},b=\frac{1}{2}(\sqrt{6}+\sqrt{2})$ and $c=\sqrt{2}$ then find $\angle A$ in degree

Answer 1

Given $a=\sqrt{3},b=\frac{1}{2}(\sqrt{6}+\sqrt{2})andc=\sqrt{2}$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{\left(\frac{1}{4}\right)(8+4\sqrt{3})+2-3}{\sqrt{12}+\sqrt{4}}=\frac{1+\sqrt{3}}{2(1+\sqrt{3})}=\frac{1}{2}=\cos {60}^{\circ }$

$\therefore \angle A=\frac{\pi }{3}={60}^0$