Question

In a triangle $ABC$, the median of side $BC$ is of length $\frac{1}{\sqrt{11-6\sqrt{3}}}$ and it divides $\angle A$ in to the angles of ${30}^0$ and ${45}^0$. What is the length of side $BC$?

Answer 1

To find: Length of BC

By $m-ncot$ theorem

$2\cot \theta =\cot 30-\cot 45$

$\cot \theta =\frac{\sqrt{3}-1}{2}$

$\sin R=\sin (180-45-\theta )$

$\sin R=\sin (45+\theta )$

$\sin R=\frac{1}{\sqrt{2}}\sin \theta +\cos \theta$

$\sin R=\frac{1}{\sqrt{2}}\frac{\sqrt{3}+1}{\sqrt{8-2\sqrt{3}}}$

Apply sine rule in $△ADC$

$\frac{AD}{\sin C}=\frac{\frac{BC}{2}}{\sin 45}$

$BC=\frac{2\sin 45\times AD}{\sin C}$

$BC=\frac{\sqrt{2}\times \sqrt{2}\times \sqrt{8-2\sqrt{3}}}{\sqrt{11-6\sqrt{3}}\times \sqrt{3}+1}$

$BC=\frac{(\sqrt{3}-1)\sqrt{8-2\sqrt{3}}}{\sqrt{11-6\sqrt{3}}}$

$BC=\frac{\sqrt{4-2\sqrt{3}}\sqrt{8-2\sqrt{3}}}{\sqrt{11-6\sqrt{3}}}$

$BC=\frac{\sqrt{44-24\sqrt{3}}}{\sqrt{11-6\sqrt{3}}}$

$BC=2$