Question

In triangle $ABC,$ $\angle B=\frac{\pi }{3}$ and $\angle C=\frac{\pi }{4}.$ Let $D$ divide $BC$ internally in the ratio $1:3$ and the angles $\angle BAD=\theta ,$ $\angle CAD=\varphi .$ The value of $\frac{{\sin }^2\varphi }{{\sin }^2\theta }=$

Answer 1

Using sine rule in $\Delta ABD,$
$\frac{\sin \theta }{BD}=\frac{\sin {60}^{\circ }}{AD}$
$\Rightarrow \frac{AD}{BD}=\frac{\sin {60}^{\circ }}{\sin \theta }\cdots \left(i\right)$

Using sine rule in $\Delta ACD,$
$\frac{\sin \varphi }{DC}=\frac{\sin {45}^{\circ }}{AD}$
$\Rightarrow \frac{DC}{AD}=\frac{\sin \varphi }{\sin {45}^{\circ }}\cdots \left(ii\right)$

Multiplying equation $\left(i\right),\left(ii\right),$ we get
$\frac{DC}{BD}=\frac{\sin {60}^{\circ }}{\sin {45}^{\circ }}\frac{\sin \varphi }{\sin \theta }$
Hence, $\frac{{\sin }^2\varphi }{{\sin }^2\theta }=6$