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, \[BF=80\,m\] |
\[\therefore \] \[\tan y{}^\circ =\frac{TF}{BF}\]\[\Rightarrow \]\[\tan y{}^\circ =\frac{80}{80}\] |
\[\Rightarrow \] \[\tan y{}^\circ =1=\tan 45{}^\circ \] |
\[\therefore \] \[y{}^\circ =45{}^\circ \] |
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