Question

A 1.5 m tall boy is standing at some distance from a 30 m tall building; find the distance he walked towards the building if the angle of elevation of the top increases from \(30^\circ\) to \(60^\circ\).

Hint:

Always subtract observer height from object height in elevation problems to avoid common mistakes.

Answer 1

Concept: In height and distance problems:

• Use \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \).
• Effective height = building height − boy’s height.

Step 1: Find effective height.
Building height = 30 m, boy’s height = 1.5 m \[ \text{Effective height} = 30 - 1.5 = 28.5 \text{ m}. \]
Step 2: Let initial distance be \( x \).
When angle of elevation = \(30^\circ\): \[ \tan 30^\circ = \frac{28.5}{x}. \] \[ \frac{1}{\sqrt{3}} = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3}. \]
Step 3: Let new distance be \( y \).
When angle becomes \(60^\circ\): \[ \tan 60^\circ = \frac{28.5}{y}. \] \[ \sqrt{3} = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}}. \]
Step 4: Distance walked towards building.
\[ \text{Distance walked} = x - y = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}}. \] Take common factor: \[ = 28.5\left(\frac{3 - 1}{\sqrt{3}}\right) = 28.5 \cdot \frac{2}{\sqrt{3}} = \frac{57}{\sqrt{3}}. \] Rationalizing: \[ = \frac{57\sqrt{3}}{3} = 19\sqrt{3}. \] Conclusion:
The boy walked: \[ 19\sqrt{3} \text{ m} \approx 32.9 \text{ m}. \]