A car is travelling on a straight road leading to a tower. From a point at a distance of 500 m from the tower, as seen by the driver, the angle of elevation of the top of the tower is 30°. 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°. Then the speed of the car is

BC = 500 m

Let CD be $$x$$ => BD = $$500 - x$$

From $$\triangle$$ABC

=> $$tan 30 = \frac{AB}{BC}$$

=> $$\frac{1}{\sqrt{3}} = \frac{AB}{500}$$

=> $$AB = \frac{500}{\sqrt{3}}$$ m

Now, from $$\triangle$$ABD

=> $$tan 60 = \frac{AB}{BD}$$

=> $$\sqrt{3} = \frac{\frac{500}{\sqrt{3}}}{500 - x}$$

=> $$3 (500 - x) = 500$$

=> $$3x = 1000$$

$$\therefore x = \frac{1000}{3}$$ metre = $$\frac{1}{3}$$ km

Also, speed of car = $$\frac{distance}{time}$$

= $$\frac{\frac{1}{3}}{\frac{10}{60 * 60}}$$ km/hr

= 120 km/hr