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 $
BETA
