SATHEE: Quantitative Aptitude Ques 71

Quantitative Aptitude Ques 71

Question: 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

Options:

A) $ 75{}^\circ $

B) $ 45{}^\circ $

C) $ 60{}^\circ $

D) $ 30{}^\circ $

Show Answer

Answer:

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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