Question

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from ${30}^o$ to ${45}^o$ in $12minutes$, find the time taken by the car now to reach the tower.

Answer 1

Let the tower be represented by $CD$, the initial position of the car by $A$ and the second position of the car by $B$.

Let the distance travelled from $A$ to $B$ be $d_1$ in time $t_1$ and distance between $B$ to $C$ be $d_2$ in time $t_2$.

In $△DCB,\tan 45=\frac{CD}{BC}$.

$\therefore BC=CD=d_2$

In $△ACD,\tan 30=\frac{CD}{AC}$

$\therefore \frac{1}{\sqrt{3}}=\frac{d_2}{d_1+d_2}$

$\therefore d_1+d_2=\sqrt{3}{d}_2$

Since the speed of the car is uniform throughout its journey, we can say

$t_1+t_2=\sqrt{3}{t}_2$

$\therefore 12+t_2=\sqrt{3}{t}_2$

$\therefore t_2=16.39$ min.