A car is traveling on a straight road leading to a tower From a point at a distance of - Maths - Some Applications of Trigonometry

Question

A car is traveling on a straight road leading
to a tower From a point at a distance of 500m from the tower the
angle of elevation of the top of the tower as seen by the driver is
30o After driving towards the tower for 10secs The angle
of elevation of the top of the tower as seen by the driver is found
to be 60o Find the speed of the car

Ankush Jain · Accepted Answer

Let h be the height of the tower and x be the distance travelled by car in 10 seconds

In âˆ†ABD $\tan 30 ° = \frac{h}{500} \Rightarrow h = 500 \times \frac{1}{\sqrt{3}} = \frac{500 \sqrt{3}}{3} m (1)$ Now, In âˆ†ABC, we have $\tan 60 ° = \frac{AB}{BC} = \frac{h}{500 - x} \sqrt{3} = \frac{h}{500 - x} \Rightarrow h = (500 - x) \sqrt{3} \frac{500 \sqrt{3}}{3} = 500 \sqrt{3} - \sqrt{3} x [using (1)] \sqrt{3} x = 500 \sqrt{3} - \frac{500 \sqrt{3}}{3} \sqrt{3} x = \frac{1500 \sqrt{3} - 500 \sqrt{3}}{3} \sqrt{3} x = \frac{1000 \sqrt{3}}{3} \Rightarrow x = \frac{1000}{3} m Hence distance travelled by car in 10 sec = \frac{1000}{3} m ∴ speed of the car = \frac{Distance}{time} = \frac{1000}{\frac{3}{10}} m/s = \frac{100}{3} m/s$