Quantitative Aptitude Ques 547
Question: The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is 75 m and the angle of elevation of the top of the tower from the bottom of the building is $ 60{}^\circ $ ?
Options:
A) 25 m
B) 37.5 m
C) 50 m
D) 60 m
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ 2\theta =60{}^\circ $
$ \Rightarrow $ $ \theta =30{}^\circ $ In $ \Delta ABC, $ $ \tan 60{}^\circ =\frac{AB}{BC} $ $ \sqrt{3}=\frac{75}{BC} $ $ BC=\frac{75}{\sqrt{3}} $ . (i) In $ \Delta ADE, $ $ \tan 30{}^\circ =\frac{AD}{DE} $ $ \frac{1}{\sqrt{3}}=\frac{75-h}{DE} $ (ii) $ \because $ $ BC=DE $
$ \therefore $ $ \frac{75}{\sqrt{3}}=\sqrt{3}(75-h) $
$ \Rightarrow $ $ 75=225-3h $
$ \Rightarrow $ $ h=50,m $
BETA
