Question
A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. The distance (in metres) travelled by the car during this time is
A.
$$100\sqrt 3 $$
B.
$$200\frac{{\sqrt 3 }}{3}$$
C.
$$100\frac{{\sqrt 3 }}{3}$$
D.
$$200\sqrt 3 $$
Answer :
$$200\frac{{\sqrt 3 }}{3}$$
Solution :
$$\eqalign{ & {\text{In }}\Delta CBD,\tan {60^ \circ } = \frac{{100}}{3} \cr & \Rightarrow \,\,x = \frac{{100}}{{\sqrt 3 }} \cr & {\text{In }}\Delta ACB,\tan {30^ \circ } = \frac{{100}}{{x + d}} \cr & \Rightarrow \,\,x + d = 100\sqrt 3 \cr & \Rightarrow \,\,d = 100\sqrt 3 - \frac{{100}}{{\sqrt 3 }} \cr & = \frac{{200}}{{\sqrt 3 }} \cr & = \frac{{200\sqrt 3 }}{3}m \cr} $$
$$\eqalign{ & {\text{In }}\Delta CBD,\tan {60^ \circ } = \frac{{100}}{3} \cr & \Rightarrow \,\,x = \frac{{100}}{{\sqrt 3 }} \cr & {\text{In }}\Delta ACB,\tan {30^ \circ } = \frac{{100}}{{x + d}} \cr & \Rightarrow \,\,x + d = 100\sqrt 3 \cr & \Rightarrow \,\,d = 100\sqrt 3 - \frac{{100}}{{\sqrt 3 }} \cr & = \frac{{200}}{{\sqrt 3 }} \cr & = \frac{{200\sqrt 3 }}{3}m \cr} $$