Question

A car is travelling on a straight road leading to a tower. From a point at a distance of $500m$ from the tower as seen by the driver is ${30}^{\circ }$. After driving towards the tower for $10$ seconds, the angle of elevation of the top of the tower as seen by the driver is found to be ${60}^{\circ }$. Find the speed of the car.

Answer 1

R.E.F image

Acc. to Rest.

Let $OC=$ Tower

$AC=500m$

$AB=xm$ in 10 seconds

$BC=(500-x)m$

$\angle OAC={30}^{\circ }$

$\angle OBC={60}^{\circ }$

To find $\to$ Speed of car = ?

Method $\to$ Step 1 $\to$ Find Height of tower (OC) = ?

Step 2 $\to$ Then Find x = ?

Step 3 $\to$ Then speed of car $=\frac{Distance}{Time}=\frac{xm}{10sec}$

Sol $\to$ Step $\to$ In $\Delta OAC$

$Tan{30}^{\circ }=\frac{OC}{AC}=\frac{OC}{500}=\frac{1}{\sqrt{3}}$

$\therefore OC=\frac{500}{\sqrt{3}}m$ $(\because Tan{30}^{\circ }=\frac{1}{\sqrt{3}})$

Step 2 $\to$ Finding $x=?$

In $\Delta OBC,Tan{60}^{\circ }=\frac{OC}{BC}=\frac{500/\sqrt{3}}{500-x}$

$\sqrt{3}=\frac{500}{\sqrt{3}(500-x)}$

$\Rightarrow 3(500-x)=500$

$\Rightarrow 1500-500=3x=1000$

$\Rightarrow x=333.3m$

Step 3 $\to \therefore$ speed $=\frac{Distance\left(x\right)}{Time\left(10sec\right)}=\frac{333.3}{10}m/sec$

$=33.33m/sec$