Question

A man on the top of vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes $12$ minutes for the angle of depression to change from ${30}^o$ to ${45}^o$, how long will the car take to reach the observation tower from this point?

Answer 1

Let $A$ be the position of the man and $AB$ be the tower.

Now consider the $△ABD,$

$\tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BD}$

$\Rightarrow \frac{BD}{\sqrt{3}}=AB$

$\Rightarrow \frac{BC+CD}{\sqrt{3}}=AB$ .......$\left(1\right)$

Consider the $△ABC$

$\tan {45}^{\circ }=\frac{AB}{BC}$

$\Rightarrow 1=\frac{AB}{BC}$

$\Rightarrow AB=BC$ ........$\left(2\right)$

Substitute the value of $BC$ from eqn$\left(2\right)$ in eqn$\left(1\right)$, we have

$\frac{AB+CD}{\sqrt{3}}=AB$

$\Rightarrow \sqrt{3}AB-AB=CD$

$\Rightarrow (\sqrt{3}-1)AB=CD$ .......$\left(3\right)$

Let $v$ m\min be the speed of the car.

Since the car takes $12$min to cover the distance $CD$, we have

$CD=12v$ ......$\left(4\right)$ [Distance=speed$\times$time]

Substitute the value of $CD$ from eqn$\left(4\right)$ in equation $\left(3\right)$, we have

$AB(\sqrt{3}-1)=12v$

$\Rightarrow \frac{AB(\sqrt{3}-1)}{12}=v$ .........$\left(5\right)$

We need to find the time taken by the car to cover the distance $BC$

Time$=\frac{BC}{v}$ since time=distance\speed

$\Rightarrow$Time$=\frac{12BC}{AB(\sqrt{3}-1)}$ from eqn$\left(5\right)$

$\Rightarrow$Time$=\frac{12AB}{AB(\sqrt{3}-1)}$ from eqn$\left(2\right)$

$\Rightarrow$Time$=\frac{12}{(\sqrt{3}-1)}\approx 16.4$minutes