In $\triangle \mathrm{ABC}, \mathrm{m} \angle \mathrm{B}=\frac{\pi}{3}$ and $\mathrm{m} \angle…

In $\triangle \mathrm{ABC}, \mathrm{m} \angle \mathrm{B}=\frac{\pi}{3}$ and $\mathrm{m} \angle \mathrm{C}=\frac{\pi}{4}$. Let point $\mathrm{D}$ divide $\mathrm{BC}$ internally in the ratio $1: 3$, then $\frac{\sin (\angle B A D)}{\sin (\angle C A D)}$ has the value
  1. $\frac{1}{3}$
  2. $\frac{1}{\sqrt{3}}$
  3. $\frac{1}{\sqrt{6}}$
  4. $\sqrt{\frac{2}{3}}$

Solution


In $\triangle \mathrm{ABD}$, $\begin{aligned} & \frac{\sin (\angle \mathrm{BAD})}{\mathrm{BD}}=\frac{\sin (\angle \mathrm{ABD})}{\mathrm{AD}} \\ & \Rightarrow \frac{\sin (\angle \mathrm{BAD})}{x}=\frac{\frac{\sqrt{3}}{2}}{\mathrm{AD}} \\ & \Rightarrow \mathrm{AD}=\frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})} \end{aligned}$ In $\triangle \mathrm{ADC}$, $\begin{aligned} & \frac{\sin (\angle \mathrm{CAD})}{\mathrm{DC}}=\frac{\sin (\angle \mathrm{ACD})}{\mathrm{AD}} \\ \Rightarrow & \frac{\sin (\angle \mathrm{CAD})}{3 x}=\frac{\frac{1}{\sqrt{2}}}{\mathrm{AD}} \\ \therefore \quad & \mathrm{AD}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \end{aligned}$ From (i) and (ii), we get $\begin{aligned} & \frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \\ \therefore & \frac{\sin (\angle \mathrm{BAD})}{\sin (\angle \mathrm{CAD})}=\frac{\sqrt{6}}{6}=\frac{1}{\sqrt{6}} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 2)

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