Question

In a triangle ABC, the median to the side BC is of length $\frac{1}{\sqrt{11-6\sqrt{3}}}$ & it divides the angle A into angles of ${30}^{\circ }$ & ${45}^{\circ }$Find the length of the side BC

Answer 1

Applying $m-n$ theorem we get $2\cot \theta =\cot 30-\cot {45}^{\circ }\Rightarrow \tan \theta =\sqrt{3}+1$ ...(i)
Now $\sin C=\sin (\frac{\pi }{4}+\theta )=\frac{1}{\sqrt{2}}(\sin \theta +\cos \theta )$
Using (i) we get $\sin C=\frac{\sqrt{3}+2}{\sqrt{2}\sqrt{5+2}\sqrt{3}}$
Now applying sine law in $\Delta$ADC we get $\frac{AD}{\sin C}=\frac{DC}{\sin {45}^{\circ }}$
$\therefore DC=\frac{1}{\sqrt{11-6\sqrt{3}}}\times \frac{\sqrt{2}\sqrt{5+2\sqrt{3}}}{\sqrt{3}+2}\cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{5+2\sqrt{3}}}{\sqrt{11-6\sqrt{3}}\sqrt{7+4\sqrt{3}}}=1$
Hence, $DC=BD=1$$\Rightarrow BC=2$