question_answer

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 metre) travelled by the car during this time is

A) \[100\sqrt{3}\]

B) \[\frac{200\sqrt{3}}{3}\]

C) \[\frac{100\sqrt{3}}{3}\]

D) \[200\sqrt{3}\]

Correct Answer: B

Solution :

[b] Suppose distance travelled by car\[=x\,m\] In \[\Delta \Alpha \Beta C,\]\[\tan 30{}^\circ =\frac{100}{x+l}\] \[\Rightarrow \]   \[\frac{1}{\sqrt{3}}=\frac{100}{x+l}\] \[\Rightarrow \]   \[x+l=100\sqrt{3}\,m\]                           (i) Now, in \[\Delta ABD,\]\[\tan 60{}^\circ =\frac{100}{l}\] \[\Rightarrow \]   \[\sqrt{3}=\frac{100}{l}\]\[\Rightarrow \]\[l=\frac{100}{\sqrt{3}}\,m\] On subtracting Eq. (ii) from Eq.(i), we get \[(x+l)-l=100\sqrt{3}-\frac{100}{\sqrt{3}}\] \[\Rightarrow \]   \[x=100\left( \sqrt{3=\frac{1}{\sqrt{3}}} \right)\] \[\Rightarrow \]   \[x=\frac{200}{\sqrt{3}}\,m\]\[\Rightarrow \]\[x=\frac{200\sqrt{3}}{3}\,m\]