In $\triangle A B C, \angle B=60^{\circ}$ and $\angle A=75^{\circ}$. If a point $D$ divides $\mathrm{BC}$ in…

1. $\sqrt{2} ; \sqrt{3}$
2. $\sqrt{3}: 2$
3. $\sqrt{3}: \sqrt{2}$
4. $3: \sqrt{2}$

Solution

Given $\angle B=60^{\circ}, \angle A=75^{\circ}$ Let $\angle B A D=\theta$ and $\angle C A D=\phi$ Using sine rule in $\triangle A B D$,
$\frac{A D}{\sin 45^{\circ}}=\frac{C D}{\sin \phi}$ ...(ii) Also, given $\frac{B D}{C D}=\frac{2}{3}$ Now, Eqn. (i)/(ii), $\begin{aligned} & \Rightarrow \frac{\sin 45}{\sin 60}=\frac{B D}{C D} \times \frac{\sin \phi}{\sin \theta} \\ & \Rightarrow \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}=\frac{2}{3} \times \frac{\sin \phi}{\sin \theta}\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}}=\frac{2}{3} \times \frac{\sin \phi}{\sin \theta} \Rightarrow \frac{\sin \phi}{\sin \theta}=\frac{\sqrt{3}}{\sqrt{2}} \\ & \Rightarrow \frac{\sin \angle C A D}{\sin \angle B A D}=\frac{\sqrt{3}}{\sqrt{2}} \Rightarrow \frac{\sin \angle B A D}{\sin \angle C A D}=\frac{\sqrt{2}}{\sqrt{3}} .\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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