The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is $50$ m high, then find the height of the building.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Let the height of the building be $h$ m and the horizontal distance between the tower and building be $d$ m.
From the foot of the building (angle of elevation of tower = 60°):
$$\tan 60° = \frac{50}{d} \implies \sqrt{3} = \frac{50}{d} \implies d = \frac{50}{\sqrt{3}}$$
From the foot of the tower (angle of elevation of building = 30°):
$$\tan 30° = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{h}{d} \implies h = \frac{d}{\sqrt{3}}$$
Substituting the value of $d$:
$$h = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3} = 16\frac{2}{3} \text{ m}$$
The height of the building is $\dfrac{50}{3}$ m $\approx 16.67$ m.
Source: Exercise 9.1, Q.9, Chapter 9
---
Explanation
- Set up two right triangles sharing the same base $d$ (horizontal distance between foot of tower and foot of building).
- Use $\tan 60°$ to find $d$ from the tower's height (given = 50 m), then use $\tan 30°$ to find the building's height.
- The answer $\frac{50}{3}$ m is exact; examiners accept this or its decimal equivalent.
- Draw a neat labelled diagram — it fetches 1 mark in most CBSE marking schemes.