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.

Question

A ladder 14 m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is:

Answer 1

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
The car's initial position is C, and it moves to position D in six seconds.

In right triangle ADB,

$\frac{AB}{ DB} = tan 60^{\degree}$

$\frac{AB}{ DB} = \sqrt3$

$DB = \frac{AB} {\sqrt3}$

In right triangle ABC,

$\frac{AB}{ BC}= tan 30^{\degree}$

$\frac{ AB }{ BD + DC} = \frac1{\sqrt3}$

$AB \sqrt3 = BD + DC$

$AB \sqrt3 = \frac{AB }{\sqrt3} + DC$

$ DC = AB \sqrt3 -\frac{ AB}{ \sqrt3} = AB (\sqrt3- \frac{1}{ \sqrt3})$

$DC= \frac{2AB}{ \sqrt3}$

The time taken for the car to travel distance DC (which is $\frac{2AB}{ \sqrt3}$) is 6 seconds.

The time taken for the car to travel distance DB (which is $\frac{AB}{ \sqrt3}$) is calculated as: $\frac{6}{\frac{ 2AB}{ \sqrt3}} \times \frac{AB}{ \sqrt3}$ = $\frac 6{2}$ = 3 seconds.

Solution and Explanation

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