SATHEE: Quantitative Aptitude Ques 2051

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 $

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