Q15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30º, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60º. Find the time taken by the car to reach the foot of the tower.

Let AB be the tower of height h. Let C be the initial position of the car and let after 6 seconds, the car be at D. It is given that the angles of depression at C and D are 30º and 60º respectively. Let the speed of the car be v m/s. Then ,

CD = Distance travelled by the car in 6 seconds

=> \( CD \ = \ 6v \) metres [∵ Distance = Speed × Time]

Let the car takes t seconds to reach the tower AB from C. Then,

\( DA \ = \ vt \) metres

In \( ∆ \ ABC \), we have

\( \frac{AB}{AC} \ = \ tan60º \)

=> \( \frac{h}{vt} \ = \ \sqrt{3} \)

\( => \ h \ = \sqrt{3} vt \) (1)

In \( ∆ \ ABD \) , we have

\( \frac{AB}{AD} \ = \ tan30º \)

=> \( \frac{h}{vt + 6v} \ = \frac{1}{\sqrt{3}} \)

\( => \ h\sqrt{3} \ = vt + 6v \) (2)

Substituting the value of h from (1) in (2), we get

\( \sqrt{3} \ × \ \sqrt{3}vt \ = \ vt + 6t \)

=> \( 3vt \ = \ vt \ + \ 6v \)

\( => \ 3t \ - \ t \ = \ 6 \)

=> \( t \ = \ 3 \)

∴ the car will reach the tower from C in 3 seconds.