Quantitative Aptitude Ques 2051
Question: The angle of elevation of the top of a tower from a point on the ground is $ 45{}^\circ . $ Moving 21m directly towards the base of the tower, the angle of elevation changes to $ 60{}^\circ . $ What is the height of the tower, to the nearest metre?
Options:
A) 48 m
B) 49 m
C) 50 m
D) 51 m
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta PBC, $ $ \tan 60{}^\circ =\frac{h}{x} $
$ \Rightarrow $ $ \frac{h}{x}=\sqrt{3} $
$ \Rightarrow $ $ x=\frac{h}{\sqrt{3}} $ In $ \Delta PAC, $ $ \tan 45{}^\circ =\frac{h}{21+x}=1 $
$ \Rightarrow $ $ h=21+x $
$ \Rightarrow $ $ h=21+\frac{h}{\sqrt{3}} $ [from Eq. (i)]
$ \Rightarrow $ $ h( 1-\frac{1}{\sqrt{3}} )=21 $
$ \Rightarrow $ $ h=\frac{21\sqrt{3}}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)} $
$ \Rightarrow $ $ h=\frac{21\sqrt{3},(\sqrt{3}+1)}{2}=49.68\approx 50m $
BETA
