question_answer

A 1.5 m tall boy is standing at some distance from a 63/2 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. What is the distance he walked towards the building?

A)  \[\frac{20}{\sqrt{3}}m\]           

B)  \[30\sqrt{3}m\]

C)  \[20\sqrt{3}m\] 

D)  \[\frac{30}{\sqrt{3}}m\]

Correct Answer: C

Solution :

(c) In \[\Delta ABF,\,\,\]\[\tan 60{}^\circ =\frac{30}{x}\] \[\therefore \]\[x=\frac{30}{\tan 60{}^\circ }=\frac{30}{\sqrt{3}}\] In \[\Delta ABE,\]\[\tan 30{}^\circ =\frac{30}{x+y}\] \[\therefore \]      \[x+y=\frac{30}{\tan 30{}^\circ }\] \[\Rightarrow \]   \[x+y=30\sqrt{3}\] \[\Rightarrow \]   \[\frac{30}{\sqrt{3}}+y=30\sqrt{3}\] \[\Rightarrow \]   \[y=30\sqrt{3}-\frac{30}{\sqrt{3}}=\frac{60}{\sqrt{3}}=20\sqrt{3}\]