Question

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

Answer 1

Using tan $\frac{(A-B)}{2}=\frac{(a-c)}{a+b}cot\frac{C}{2},$ we get

$tan\frac{(A-B)}{2}=\frac{2}{2\sqrt{3}}cot{30}^{\circ }=(\frac{1}{\sqrt{3}}\times \sqrt{3})=1$

$\Rightarrow \frac{A-B}{2}={45}^{\circ }\Rightarrow A-B={90}^{\circ }$

Also, A + B + C = ${180}^{\circ }\Rightarrow A+B={120}^{\circ }$

On solving (i) and (ii), we get A = ${105}^{\circ }andB={15}^{\circ }$.

Also, $cosC=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow cos{60}^{\circ }=\frac{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-c^2}{2(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{8-c^2}{4}$

$\therefore \frac{8-c^2}{4}=\frac{1}{2}\Rightarrow 8-c^2=2\Rightarrow c^2=6\Rightarrow c=\sqrt{6}.$

Hence, $c=\sqrt{6}cm,\angle A={105}^{\circ }and\angle B={15}^{\circ }$