Question

Two sides of a triangle are $\sqrt{3}-1$ and $\sqrt{3}+1$ units and their included angle is ${60}^o$.Solve the triangle.

Answer 1

Let $A={60}^o\Rightarrow B+C={120}^o$

Also $\tan (B-C)/2=(b-c)/(b+c)\cot A/2$

$=(\sqrt{3}+1-\sqrt{3}+1)/(\sqrt{3}+1+\sqrt{3}-1)\cot {30}^o$

$=1$

$\Rightarrow (B-C)/2={45}^o$

$\Rightarrow B-C={90}^o$

$\Rightarrow B={105}^o,C={15}^o$

Also $a=b\sin /\sin B=(\sqrt{3}+1)/\sin {60}^o\cos {45}^o+\cos {60}^o\sin {45}^o$

$=\sqrt{2}(\sqrt{3}+1).\sqrt{3}/(\sqrt{3}+1)=\sqrt{6}$.

Alternatively:

$a^2=b^2+c^2-2bc\cos A$

$=(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-2(\sqrt{3}+1)(\sqrt{3}-1)\cos {60}^o$

$=8-4.1/2=6$