A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards - Maths -

Question

A man on the top of a vertical tower observes a
car moving at a uniform speed coming directly towards him If it
takes 12 minutes for the angle of depression to change from 30
degree to 45 degree, how soon after this, will the car reach the
tower? Give your answer to the nearest second

Lalit Mehra · Accepted Answer

Let AB be the vertical tower Suppose D and C be the positions of the car when the angle of depression from the top of the tower is 30Â° and 45Â° respectively Consider the uniform speed of car be v m/min Time taken for the angle of depression to change from 30Â° to 45Â° = 12 min âˆ EAD = âˆ ADB = 30Â° (Alternate angles) âˆ EAC = âˆ ACB = 45Â° (Alternate angles) Suppose AB = h m and BC = x m CD = Distance covered by car in $12 \min = v m/min × 12 \min = 12 v m$ In Î”ABC, $\tan 45 ° = \frac{AB}{BC} ∴ 1 = \frac{h}{x} \Rightarrow h = x (1)$ In Î”ADB, $\tan 30 ° = \frac{AB}{CD + BC} ∴ \frac{1}{\sqrt{3}} = \frac{h}{12 v + x} \Rightarrow \sqrt{3} h = 12 v + x \Rightarrow \sqrt{3} x = 12 v + x [U sing (1)] \Rightarrow (\sqrt{3} - 1) x = 12 v \Rightarrow \frac{x}{v} = \frac{12}{(\sqrt{3} - 1)} \min = \frac{12 (\sqrt{3} + 1)}{2} \min = 6 (\sqrt{3} + 1) \min = 983 \sec$ Time taken by car to reach the tower from $C = \frac{x}{v}$ âˆ´ Time taken by car to reach the tower is 983 sec

. A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards him. If it takes 12 minutes for the angle of depression to change from 30 degree to 45 degree, how soon after this, will the car reach the tower? Give your answer to the nearest second.