question_answer

An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are \[60{}^\circ \] and \[45{}^\circ \] respectively. Then the height of the lower plane from the ground is

A) \[100\sqrt{3}m\]                             

B) \[\frac{100}{\sqrt{3}}m\]

C) \[50\,m\]                                            

D) \[150(\sqrt{3}+1)m\]

Correct Answer: A

Solution :

Given \[B{{P}_{2}}=300m\] In \[\Delta AB{{P}_{2}}\] \[\frac{B{{P}_{2}}}{AB}=\tan {{60}^{o}}=\sqrt{3}\] \[\therefore \]\[AB=\frac{300}{\sqrt{3}}=100\sqrt{3}m\] Also in\[\Delta ABP1\] \[\frac{{{P}_{1}}B}{AB}=\tan {{45}^{o}}=1\] \[\therefore \]\[PB=100\sqrt{3}m\]