Question

A 1.8 m tall man is standing at some distance from a 31.8 m tall building. The angle of elevation from his eyes to the top of the building increases from ${30}^{\circ }$ to ${60}^{\circ }$ as he walks towards the building. Find the distance he walked towards the building

• A. 30
• B. 20√3
• C. 20
• D. 30√3

Answer 1

The correct option is B

20√3

The height of the building above the man's line of sight = (31.8 - 1.8) m = 30 m

We know BC = 30 m

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

BD = 10$\sqrt{3}$

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

AB= 30$\sqrt{3}$ m

AD = $30\sqrt{3}$ - $10\sqrt{3}$ = $20\sqrt{3}$ m