Question

A 1.6 m tall boy is standing at some distance from a 40 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

Answer 1

Answer: 44.34 m

Step-by-step explanation:

• In the given figure let AB be the building and CD be the boy.
• AB=40m,CD=1.6m and AF=38.4m
• In ΔACF, tan 30°=$\frac{AF}{FC}$

                      ⇒$\frac{1}{\sqrt{3} }=\frac{38.4}{BG+GD}$              [∵FC=BD]

                      ⇒BG+GD=(38.4×1.732)

                      ⇒BG+GD=66.51 m

• Now in ΔAFE, tan 60°=$\frac{AF}{FE}$

                              ⇒ $\sqrt{3}=\frac{38.4}{BG}$

                              ⇒BG=$\frac{38.4}{\sqrt{3} }$

                              ∴ BG=22.17 m

• BG+GD=66.51

    ⇒GD=(66.51-22.17)m

    ∴ GD=44.34 m

• Hence,the boy walked 44.34 m towards the building.