  Question

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

A
60  B
30  C
90  D
none of these  Solution

The correct option is D $$\displaystyle 30^{\circ}$$Area of $$\displaystyle \Delta BAD=\sqrt{3}$$ Area of $$\displaystyle \Delta BCD$$ $$\displaystyle \Rightarrow \frac{1}{2}\times AD\times X=\sqrt{3}\times \frac{1}{2}\times DC\times X\Rightarrow \frac{AD}{DC}=\frac{\sqrt{3}}{1}$$ Applying $$m - n$$ theorem $$\displaystyle \left ( \sqrt{3}+1 \right )\cot \theta =\cot 75^{\circ}-\sqrt{3}\cot 60^{\circ}$$ $$\displaystyle \left ( \sqrt{3}+1 \right )\cot \theta =\frac{\sqrt{3}-1}{\sqrt{3}+1}-1$$ $$\Rightarrow\displaystyle \left ( \sqrt{3}+1 \right )\cot \theta =\frac{-2}{\sqrt{3}+1}$$ $$\Rightarrow\displaystyle \cot \theta =\frac{-2}{\left ( \sqrt{3}+1 \right )^{2}}$$$$=\dfrac{-2}{4+2\sqrt{3}}=\dfrac{-1}{2+\sqrt{3}}$$$$=-\left ( 2-\sqrt{3} \right )$$ $$\Rightarrow\displaystyle \cot \theta =\cot 105^{\circ}$$$$\therefore \theta =105^{\circ}$$$$\therefore \angle ADB=75^{\circ}$$$$\therefore \angle ABD=30^{\circ}$$ Physics

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