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.

Answer 1

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Answer 2

The speed of the car is 100/3 m/s

Let the tower be denoted as line AB where B is the foot of the tower

Initial position of car be point C and after 10 seconds car moves to point D.

Given that Distance BC = 500 m .

Angle ACB = 30 and Angle ADB = 60

In triangle ABC,

using Tan ACB = Tan 30 = AB/BC = AB/500

1/√3 =  AB/ 500

AB = 500/√3 m

In triangle ADB,

Tan ADB = Tan 60 = AB/BD

√3 = (500/√3) / BD.

BD = 500/3 m

Distance covered by car = 500 -  500/3 m = 1000/3 m

Speed of car = (1000/3)/ 10 = 100/3 m/s