Quantitative Aptitude Ques 1318

Question: If the angles of elevation of a tower from two distant points a and $ b(a>b) $ from its foot and in the same straight line and on the same side of it, are $ 30{}^\circ $ and $ 60{}^\circ , $ then the height of the tower is

Options:

A) $ \sqrt{\frac{a}{b}} $

B) $ \sqrt{a+b} $

C) $ \sqrt{ab} $

D) $ \sqrt{a-b} $

Show Answer

Answer:

Correct Answer: C

Solution:

  • In the above figure, there are two elevation angles $ 60{}^\circ $ and $ 30{}^\circ $ of a tower. Let the height of the tower he h and distance points are a and $ b(a>b). $ Now, in $ \Delta ACB $ $ \tan 60{}^\circ =\frac{AB}{BC}=\frac{h}{b} $
    $ \Rightarrow $ $ \sqrt{3}=\frac{h}{b} $ $ h=b\sqrt{3} $ … (i) Now, again in $ \Delta ADB $ $ \tan 30{}^\circ =\frac{AB}{BD}=\frac{h}{a} $
    $ \Rightarrow $ $ \frac{1}{\sqrt{3}}=\frac{h}{a} $

$ \Rightarrow $ $ h=\frac{a}{\sqrt{3}} $ … (ii) On multiplying Eqs. (i) and (ii), we get $ h^{2}=(b\sqrt{3})\times ( \frac{a}{\sqrt{3}} ) $
$ \Rightarrow $ $ h^{2}=ab $

$ \therefore $ $ h=\sqrt{ab} $