1. 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
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- By: anil on 05 May 2019 01.58 am
BC = 500 m Let CD be $$x$$ => BD = $$500 - x$$ From $$ riangle$$ABC => $$tan 30 = frac{AB}{BC}$$ => $$frac{1}{sqrt{3}} = frac{AB}{500}$$ => $$AB = frac{500}{sqrt{3}}$$ m Now, from $$ riangle$$ABD => $$tan 60 = frac{AB}{BD}$$ => $$sqrt{3} = frac{frac{500}{sqrt{3}}}{500 - x}$$ => $$3 (500 - x) = 500$$ => $$3x = 1000$$ $$ herefore x = frac{1000}{3}$$ 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
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