Mock Test 5 Ssc Cgl Tear II Paper 1 Question 55
Question: From the top of a building 60 m high the angles of depression of the top and bottom of a tower are observed to be $ 30{}^\circ $ and $ 60{}^\circ . $ Then, the height of the tower is
Options:
A) 10 m
B) 20 m
C) 30 m
D) 40 m
Show Answer
Answer:
Correct Answer: D
Solution:
- Let x be the height of the tower.
AB = 60m, BD = y m
In $ \Delta ABD, $
$ \frac{AB}{BD}=\tan 60{}^\circ $ or $ \frac{60}{y}=\sqrt{3} $
$ \Rightarrow $ $ y=\frac{60}{\sqrt{3}} $ ?(i) In $ \Delta AEC, $ $ \frac{AE}{EC}=\tan 30{}^\circ $
$ \Rightarrow $ $ \frac{AE}{BD}=\frac{1}{\sqrt{3}} $
$ \Rightarrow $ $ \frac{AE}{Y}=\frac{1}{\sqrt{3}} $
$ \Rightarrow $ $ \frac{AE}{\frac{60}{\sqrt{3}}}=\frac{1}{\sqrt{3}} $
$ AE=\frac{1}{\sqrt{3}}\times \frac{60}{\sqrt{3}}=20m $
$ \therefore $ $ x=AB-AE=60-20=40m $ So, height of tower = 40 m
BETA
