Question

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angles of elevation of his eyes to the top of the building increase from ${30}^{\circ }$ to ${60}^{\circ }$ as he walks towards the building. Find the distance he walked towards the building.

• A. 32.91m
• B. 22.91m
• C. 35 m
• D. 30.15 m

Answer 1

The correct option is A

32.91m

Height of the building above his height is 30 - 1.5 = 28.5m

Hence BC = 28.5m

In $\Delta$BDC,

$tan{60}^{\circ }=\frac{BC}{BD}$

$\Rightarrow$ BD = $\frac{28.5}{\sqrt{3}}$ = 16.45 m

In $\Delta$ABC,

$tan{30}^{\circ }=\frac{BC}{AB}$

$\Rightarrow$ AB = BC$\sqrt{3}$ = 28.5$\sqrt{3}$

= 49.363m

Hence distance he travelled = AB - BD = 49.363 - 16.45 = 32.91m