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(B) Let $AB$ be the tower of height $100 \ m$.Let $C$ be the initial position of the car and $D$ be the final position after some time.Let the distanc

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$\frac{200 \sqrt{3}}{3}$

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(B) Let $AB$ be the tower of height $100 \ m$.Let $C$ be the initial position of the car and $D$ be the final position after some time.Let the distance travelled by the car be $CD = x \ m$.In $\Delta ABD$,the angle of elevation is $60^{\circ}$.$	an 60^{\circ} = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{100}{BD} \Rightarrow BD = \frac{100}{\sqrt{3}} \ m$.In $\Delta ABC$,the angle of elevation is $30^{\circ}$.$	an 30^{\circ} = \frac{AB}{BC} = \frac{AB}{BD + CD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{\frac{100}{\sqrt{3}} + x}$.$\frac{100}{\sqrt{3}} + x = 100 \sqrt{3}$.$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} \ m$.Rationalizing the denominator,$x = \frac{200 \sqrt{3}}{3} \ m$.

$A$ man from the top of a $100 \ m$ high tower sees a car moving towards the tower at an angle of depression of $30^{\circ}$. After some time,the angle of depression becomes $60^{\circ}$. The distance (in $m$) travelled by the car during this time is

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