Quantitative Aptitude Ques 1634

Question: The height of a tower is h and the angle of elevation of the top of the tower is $ \alpha . $ On moving a distance h/2 towards the tower, the angle of elevation becomes $ \beta . $ What is the value of $ (\cot \alpha -\cot \beta ) $ ?

Options:

A) $ \frac{1}{2} $

B) $ \frac{2}{3} $

C) 1

D) 2

Show Answer

Answer:

Correct Answer: A

Solution:

  • [a] $ \tan \beta =\frac{h}{CD} $
    $ \Rightarrow $ $ CD=\frac{h}{\tan \beta }=h\cot \beta $ and $ \tan \alpha =\frac{h}{\frac{h}{2}+CD} $

$ \Rightarrow $ $ \frac{h}{2}+CD=\frac{h}{\tan \alpha }=h\cot \alpha $

$ \Rightarrow $ $ \frac{h}{2}+h\cot \beta =h\cot \alpha $

$ \Rightarrow $ $ h\cot \alpha -h\cos \beta =\frac{h}{2} $
$ \Rightarrow $ $ (\cot \alpha -\cot \beta )=\frac{h}{2} $

$ \therefore $ $ \cot \alpha -cot\beta =\frac{1}{2} $

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