Question

In a triangle ABC, the median to the side BC is of length $\frac{1}{\sqrt{(}(11-6\sqrt{3})}$ and it divides angle A into angles of ${30}^0$ and ${45}^0$. Prove that sides BC is of length 2 units.

Answer 1

By m-n theorem, $cot\theta =\frac{\sqrt{3}-1}{2}$
or $tan\theta =\frac{2}{\sqrt{3}-1}$
$\therefore sin\theta =\frac{2}{\sqrt{(8-2\sqrt{3}}},cos\theta =\frac{\sqrt{3}-1}{\sqrt{(8-2\sqrt{3})}}$
Now by sine rule on $△ADC$ in which
$AD=\frac{1}{\sqrt{(11-6\sqrt{3})}}$ and $DC=\frac{a}{2}$, we have
$\frac{a/2}{sin{45}^0}=\frac{AD}{sin\left[\pi \right(\theta +{45}^0\left)\right]}=\frac{AD}{\frac{1}{\sqrt{2}}(sin\theta +cos\theta )}$
$\frac{a}{2}=\frac{AD}{sin\theta +cos\theta }=\frac{\sqrt{8-2\sqrt{3}}}{\sqrt{3}+1}.\frac{1}{\sqrt{(11-6\sqrt{3})}}$
$=\frac{\sqrt{8-2\sqrt{3}}}{\sqrt{(4+2\sqrt{3})}\sqrt{(11-6\sqrt{3})}}=$$\frac{\sqrt{8-2\sqrt{3}}}{\sqrt{(8-2\sqrt{3})}}=1.$
$\therefore a=2\because (\sqrt{3}+1{)}^2=4+2\sqrt{3}$