question_answer

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is \[30{}^\circ \]. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is \[60{}^\circ \]. Then the time taken (in minutes) by him, form B to reach the pillar, is : [JEE Main Solved Paper-2016 ]

A) 5                                             

B) 6

C) 10                                          

D) 20

Correct Answer: A

Solution :

\[\Delta QPA:\frac{h}{x+y}=\tan {{30}^{o}}\Rightarrow \sqrt{3}h=x+y\]       ?(1) \[\Delta QPB:\frac{h}{y}=\tan {{60}^{o}}\Rightarrow h=\sqrt{3}y\]                           ?(2) By (i) and (ii) : \[3y=x+y\Rightarrow y=\frac{x}{y}\] \[\therefore \]speed is uniform Distance x in 10 mins \[\Rightarrow \]Distance\[\frac{x}{2}\]in 5 mins