Question

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 30°, 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 60°. Find the time taken

by the car to reach the foot of the tower from this point.​

Answer 1

Question :

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 30°, 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 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer :

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ∆ ADB,

AB/DB = tan 60°

$\frac{AB}{DB} = \sqrt{3}$

$DB = \frac{AB}{\sqrt{3}}$

In ∆ ABC,

AB/BC = tan 30°

$\frac{AB}{BD + DC} = \frac{1}{\sqrt{3}}$

$AB\sqrt{3} = BD+ DC$

=> $AB\sqrt{3} = \frac{AB}{\sqrt{3}} + DC$

=> $DC = AB\sqrt{3} - \frac{AB}{\sqrt{3}}$

=> $DC = AB{\sqrt{3} - \frac{1}{\sqrt{3}}}$

=> $DC = \frac{2AB}{\sqrt{3}}$

Time taken by the car to travel distance ,

DC $({i.e-}\frac{2AB}{\sqrt{3}})$ = 6 second

Time taken by the car to travel distance ,

DC $({i.e-}\frac{2AB}{\sqrt{3}})$ = $\bf\frac{6}{\frac{2AB}{\sqrt{3}}}$× $\bf\frac{AB}{\sqrt{3}}$

=>$\large\frac{6}{2}$

=> 3 seconds

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Answer 2

Answer:

Answer refer in the above attachment!!