Question

In a $\Delta ABC,\angle c={60}^{\circ }$ & $\angle A={75}^{\circ }\cdot$ If D is a point on AC such that the area of the $\Delta BAD$ is $\sqrt{3}$ times the area of the $\Delta BCD,$ then the $\angle ABD=$

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${90}^{\circ }$
• D. none of these

Answer 1

Answer: (B)

${30}^{\circ }$

Area of $\Delta BAD=\sqrt{3}$ Area of $\Delta BCD$

$\Rightarrow \frac{1}{2}\times AD\times X=\sqrt{3}\times \frac{1}{2}\times DC\times X\Rightarrow \frac{AD}{DC}=\frac{\sqrt{3}}{1}$

Applying $m-n$ theorem

$(\sqrt{3}+1)\cot \theta =\cot {75}^{\circ }-\sqrt{3}\cot {60}^{\circ }$

$(\sqrt{3}+1)\cot \theta =\frac{\sqrt{3}-1}{\sqrt{3}+1}-1$

$\Rightarrow (\sqrt{3}+1)\cot \theta =\frac{-2}{\sqrt{3}+1}$

$\Rightarrow \cot \theta =\frac{-2}{{(\sqrt{3}+1)}^2}$

$=\frac{-2}{4+2\sqrt{3}}=\frac{-1}{2+\sqrt{3}}$

$=-(2-\sqrt{3})$

$\Rightarrow \cot \theta =\cot {105}^{\circ }$

$\therefore \theta ={105}^{\circ }$

$\therefore \angle ADB={75}^{\circ }$

$\therefore \angle ABD={30}^{\circ }$