Mathematics — Standard (041) — CBSE Class 10 board question

Diagram: Let AB = h (height of tower), C and D be the initial and later positions of the car, with D closer to the foot B.

Setting up equations:

In △ABD (angle of depression = 60°):
$$\tan 60° = \frac{h}{BD} \Rightarrow \sqrt{3} = \frac{h}{BD} \Rightarrow BD = \frac{h}{\sqrt{3}}$$

In △ABC (angle of depression = 30°):
$$\tan 30° = \frac{h}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BC} \Rightarrow BC = h\sqrt{3}$$

Distance covered in 10 seconds:
$$CD = BC - BD = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$

Time to travel remaining distance BD:

Speed $= \dfrac{CD}{10} = \dfrac{2h}{10\sqrt{3}}$

$$\text{Time} = \frac{BD}{\text{speed}} = \frac{\dfrac{h}{\sqrt{3}}}{\dfrac{2h}{10\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{10\sqrt{3}}{2h} = 5 \text{ seconds}$$

The car will reach the foot of the tower in 5 seconds.

Source: Chapter 9 (Some Applications of Trigonometry), Exercise 9.1

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• Note: The textbook (Q.15) uses 6 seconds, giving answer 3 seconds. This question uses 10 seconds, giving 5 seconds. The method is identical — only the numbers differ.
• Key step: angles of depression from the top equal angles of elevation from the base (alternate angles), so use tan directly in the right triangles.
• The ratio CD : BD = 2 : 1, so time for BD = half the time for CD. Examiners award marks for correct tan equations, correct CD, correct speed ratio, and final answer.