Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

We have $\angle A={30}^{\circ }+{45}^{\circ }={75}^{\circ }$
So, $\angle C={180}^{\circ }-({75}^{\circ }+B)$
$\Rightarrow \angle C={105}^{\circ }-B$
In $\Delta ABD,$
$\frac{BD}{\sin {30}^{\circ }}=\frac{AD}{\sin B}\Rightarrow BD=\frac{AD}{2\sin B}\cdots \left(i\right)$
In $\Delta ADC,$
$\frac{CD}{\sin {45}^{\circ }}=\frac{AD}{\sin C}\Rightarrow CD=\frac{AD}{\sqrt{2}\sin ({105}^{\circ }-B)}\cdots \left(ii\right)$
Now, $BD=CD$
$\Rightarrow \frac{AD}{2\sin B}=\frac{AD}{\sqrt{2}\sin ({105}^{\circ }-B)}$
$\Rightarrow \sqrt{2}\sin B=\sin ({105}^{\circ }-B)$
$\Rightarrow \sqrt{2}\sin B=\sin {105}^{\circ }.\cos B-\cos {105}^{\circ }.\sin B$
$\Rightarrow \sqrt{2}\sin B=\frac{\sqrt{3}+1}{2\sqrt{2}}\cos B+\frac{\sqrt{3}-1}{2\sqrt{2}}\sin B$
$\Rightarrow 4\sin B=(\sqrt{3}+1)\cos B+(\sqrt{3}-1)\sin B$
$\Rightarrow \cot B=3\sqrt{3}-4\Rightarrow \sin B=\frac{1}{2\sqrt{11-6\sqrt{3}}}$
Hence $BC=2BD=\frac{AD}{\sin B}=\frac{2\sqrt{11-6\sqrt{3}}}{\sqrt{11-6\sqrt{3}}}=2$