A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Step-by-Step Solution
Key Concept: Use the definition of tangent in right‑angled triangles: \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\). Set up two equations for the two positions of the boy, solve for the two horizontal distances, and subtract to obtain the distance walked.
1. Draw a figure (conceptual):
- Let the foot of the building be point \(B\).
- Let the boy’s eyes be at height \(1.5\,\text{m}\) above the ground.
- Let the top of the building be point \(T\) (height \(30\,\text{m}\)).
- When the boy is at the initial position \(P\), the horizontal distance \(BP = x\) and \(\angle TPB = 30^{\circ}\).
- After walking towards the building to point \(Q\), the new horizontal distance \(BQ = y\) and \(\angle TQB = 60^{\circ}\).
2. Apply the tangent definition for the two positions.
- For position \(P\):
$$\tan 30^{\circ} = \frac{\text{height difference}}{\text{horizontal distance}} = \frac{30-1.5}{x} = \frac{28.5}{x}$$
Hence
$$x = \frac{28.5}{\tan 30^{\circ}}.$$
- For position \(Q\):
$$\tan 60^{\circ} = \frac{28.5}{y}$$
Hence
$$y = \frac{28.5}{\tan 60^{\circ}}.$$
3. Use the known values \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\tan 60^{\circ}=\sqrt{3}\).
- $$x = \frac{28.5}{\frac{1}{\sqrt{3}}}=28.5\sqrt{3}$$
- $$y = \frac{28.5}{\sqrt{3}}=\frac{28.5}{\sqrt{3}}.$$
4. Find the distance walked (the reduction in horizontal distance):
$$\text{Distance walked}=x-y = 28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}
=28.5\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)
=28.5\left(\frac{3-1}{\sqrt{3}}\right)
=\frac{57}{\sqrt{3}} = \frac{57\sqrt{3}}{3}=19\sqrt{3}\ \text{metres}.$$
5. Numerical value (optional):
$$19\sqrt{3}\approx 19\times1.732 \approx 32.9\ \text{m}.$$
Thus the boy walked approximately 33 metres towards the building.
Correct Answer: 19√3 m (≈ 33 m)