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Question:

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?

Explanation:

Let x be the distance from the later position of the car and the tower (i.e. when the angle of elevation was 60°).

Since the triangle formed (i.e. ∆ABD) is a 30°-60°-90° triangle, we have,

height of the tower, h = x$\sqrt{3}$

Now, since the triangle formed by the initial position of the car (i.e. ∆ABC) is an isosceles triangle, AB = BC

i.e. BC = $x\sqrt{3}$

∴ DC = x$\sqrt{3}$ - x = x($\sqrt{3}$ - 1)

Time taken to travel distance DC is 10 minutes, thus,

Speed $s=\frac{x(\sqrt{3}-1)}{10}$

Time taken to travel distance x = $\frac{x}{{\displaystyle \frac{x(\sqrt{3}-1)}{10}}}=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

Hence, option (a).