The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Step-by-Step Solution
Key Concept: Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \frac{\text{opposite side}}{\text{adjacent side}}\). Set up two equations for the unknown distance between the tower and the building and the unknown height of the building, then solve simultaneously.
1. Let \(d\) be the horizontal distance between the foot of the tower and the foot of the building.\
2. Let \(h\) be the height of the building (required). The tower height is given as \(50\) m.\
3. From the foot of the tower, the angle of elevation to the top of the building is \(30^{\circ}\). Using \(\tan30^{\circ}=\frac{h}{d}\):\
$$\tan30^{\circ}=\frac{h}{d}\quad\Rightarrow\quad \frac{1}{\sqrt{3}}=\frac{h}{d}\quad\Rightarrow\quad h = \frac{d}{\sqrt{3}}.\tag{1}$$\
4. From the foot of the building, the angle of elevation to the top of the tower is \(60^{\circ}\). Using \(\tan60^{\circ}=\frac{50}{d}\):\
$$\tan60^{\circ}=\frac{50}{d}\quad\Rightarrow\quad \sqrt{3}=\frac{50}{d}\quad\Rightarrow\quad d = \frac{50}{\sqrt{3}}.\tag{2}$$\
5. Substitute the value of \(d\) from (2) into equation (1):\
$$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3}\ \text{metres}.$$
6. Numerically, \(\frac{50}{3} \approx 16.67\) m.
Thus the height of the building is \(\displaystyle \frac{50}{3}\) metres (≈ 16.7 m).
Correct Answer: Height of the building = \(\frac{50}{3}\) m \(\approx 16.7\) m