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

Step 1: Draw the diagram according to the question.

Let $AB$ be the tower. $D$ is the initial and$C$ is the final position of the car respectively.

As man is standing at the top of the tower so, Angles of depression are measured from $A$.

$BC$ is the distance from the foot of the tower to the car.

Step 2: Find relation between $BC$ and $CD.$

From

$$\begin{gathered} \Delta ABC, \\ \tan 60°=\frac{AB}{BC} \end{gathered}$$

$$\begin{gathered} \sqrt{3}=\frac{AB}{BC} \\ BC=\frac{AB}{\sqrt{3}} \\ AB=\sqrt{3}BC \end{gathered}$$

Again in $∆ABD$

$$\begin{gathered} \tan 30°=\frac{AB}{BD} \\ \frac{1}{\sqrt{3}}=\frac{AB}{BD} \\ AB=\frac{BD}{\sqrt{3}} \end{gathered}$$

Now, comparing Height $AB$, we get

$\sqrt{3}BC=\frac{BD}{\sqrt{3}}$ (Since LHS are same, so RHS are also same)

$$\begin{gathered} 3BC=BD \\ 3BC=BC+CD \\ 2BC=CDorBC=\frac{CD}{2} \end{gathered}$$

Here, distance of $BC$ is half of $CD.$

Step 3: Find the required time.

Hence, the time taken is also half.

Since, Time taken by car to travel distance $CD=6sec.$

So, Time taken by car to travel $BC=\frac{6}{2}=3sec$.