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 min for the angle of elevation to change from $45^\circ$ to $60^\circ$. After how much more time will this car reach the base of the tower?

• A. 1.5 $(\sqrt{3} + 1)$
• B. 2.6 $(\sqrt{3} + \sqrt{2})$
• C. 7 $(\sqrt{3} - 1)$
• D. 8 $(\sqrt{3} - 2)$

Hint:

Use trigonometric ratios like $\tan$ for height and distance problems involving angles of elevation or depression.

Answer 1

The Correct Option is B

Let the height of the tower be $h$ and the distance between the car and the base of the tower at the initial position be $x_1$. From the tangent of the angle of elevation, we have: \[ \tan(45^\circ) = \frac{h}{x_1} \quad \Rightarrow \quad x_1 = h \] When the angle of elevation changes to $60^\circ$, the new distance is $x_2$: \[ \tan(60^\circ) = \frac{h}{x_2} \quad \Rightarrow \quad x_2 = \frac{h}{\sqrt{3}} \] The car travels from $x_1$ to $x_2$ in 10 minutes, and the distance travelled is: \[ x_1 - x_2 = h - \frac{h}{\sqrt{3}} = h\left(1 - \frac{1}{\sqrt{3}}\right) = h \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) \] The car will reach the base of the tower when the remaining distance is zero. Using this, we can calculate the remaining time to reach the base.