question_answer

TF is a tower with F on the ground. The angle of elevation of T from A is \[x{}^\circ \]such that \[\tan x{}^\circ =\frac{2}{5}\] and AF = 200m. The angle of elevation of T from a nearer point B is \[y{}^\circ \]with BF = 80 m. The value of \[y{}^\circ \]is

A) \[75{}^\circ \]                          

B) \[45{}^\circ \]

C) \[60{}^\circ \]                          

D) \[30{}^\circ \]

Correct Answer: B

Solution :

Given, \[\tan x{}^\circ =\frac{2}{5}\]

\[\therefore \]      \[\frac{2}{5}=\frac{TF}{AF}\]\[\Rightarrow \]\[TF=\frac{2\times 200}{5}\]

\[\Rightarrow \] \[TF=80\,m\]

We have, \[BF=80\,m\]

\[\therefore \] \[\tan y{}^\circ =\frac{TF}{BF}\]\[\Rightarrow \]\[\tan y{}^\circ =\frac{80}{80}\]

\[\Rightarrow \]   \[\tan y{}^\circ =1=\tan 45{}^\circ \]

\[\therefore \]      \[y{}^\circ =45{}^\circ \]