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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
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B
30
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C
90
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D
none of these
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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 
$$\displaystyle \left ( \sqrt{3}+1 \right )\cot \theta =\cot 75^{\circ}-\sqrt{3}\cot 60^{\circ}$$ 

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

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

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

$$=-\left ( 2-\sqrt{3} \right )$$ 
$$\Rightarrow\displaystyle \cot \theta =\cot 105^{\circ}$$

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

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

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

366823_254412_ans_83b7bb2302b2465389285a349d7fa8e3.png

Physics

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