SATHEE: Quantitative Aptitude Ques 2192

Quantitative Aptitude Ques 2192

Question: AB is a vertical pole. The end A is on the ground, C is the middle point of AB, P is a point on the level ground. The portion BC subtends an angle $ \alpha $ at P. If $ AP=n\cdot AB, $ then $ \tan \alpha $ is equal to

Options:

A) $ \frac{n}{2n^{2}+1} $

B) $ \frac{n}{n^{2}-1} $

C) $ \frac{n}{n^{2}+1} $

D) $ \frac{n^{2}-1}{n^{2}+1} $

Show Answer

Answer:

Correct Answer: A

Solution:

Let $ \angle CPA=\beta , $ $ \tan \beta =\frac{AC}{AP}=\frac{\frac{1}{2}AB}{nAB}=\frac{1}{2n} $ and $ \tan (\alpha +\beta )=\frac{AB}{AP}=\frac{AB}{nAB}=\frac{1}{n} $

$ \Rightarrow $ $ \frac{\tan \alpha +\tan \beta }{1-\tan \alpha tan\beta }=\frac{1}{n} $

$ \Rightarrow $ $ n,(\tan \alpha +\tan \beta )=1-\tan \alpha \tan \beta $

$ \Rightarrow $ $ n,tan\alpha +\tan \alpha \tan \beta =1-n\tan \beta $

$ \Rightarrow $ $ \tan \alpha =\frac{1-n\tan \beta }{n+\tan \beta }=\frac{1-\frac{1}{2}}{n+\frac{1}{2n}}=\frac{n}{2n^{2}+1} $

Quantitative Aptitude Ques 2191

Quantitative Aptitude Ques 2193

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