Question

In triangle $ABC,$ the median to the side $BC$ is of length $\frac{1}{2\sqrt{11-6\sqrt{3}}}$ and it divides the angle $A$ into angles ${30}^{\circ }$ and ${45}^{\circ }$. Length of the side $BC$ is

Answer 1

$(m+n)\cot \theta =m\cot \alpha -n\cot \beta \cdots \left(i\right)$
By applying Equation $\left(i\right)$ in $\Delta ABC$, we get
$(1+1)\cot \theta =1\cdot \cot {30}^{\circ }-1\cdot \cot {45}^{\circ }\Rightarrow 2\cot \theta =\sqrt{3}-1\Rightarrow \cot \theta =\frac{\sqrt{3}-1}{2}\Rightarrow \tan \theta =\frac{2}{\sqrt{3}-1}$
But $\theta =B+{30}^{\circ },\text{we get}\tan (B+{30}^{\circ })=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1$
$\Rightarrow \tan B=\frac{2+\sqrt{3}}{2\sqrt{3}+1}\cot B=3\sqrt{3}-4$
$\sin B=\frac{1}{2\sqrt{11-6\sqrt{3}}}\cdots \left(ii\right)$
Now in $\Delta ABD$, by applying the sine law, we get
$\frac{BD}{\sin {30}^{\circ }}=\frac{AD}{\sin B}\Rightarrow BD=\frac{AD}{\sin B}\times \sin {30}^{\circ }=1\times \frac{1}{2}=\frac{1}{2}$
$\therefore BC=2BD=1$ unit