Trigonometric Equations Question 63
Question: AB is a vertical tower. The point A is on the ground and C is the middle point of AB. The part CB subtend an angle $ \alpha $ at a point P on the ground. If $ AP=n,AB, $ then the correct relation is
[MNR 1989; IIT 1980]
Options:
A) $ n=(n^{2}+1)\tan \alpha $
B) $ n=(2n^{2}-1)\tan \alpha $
C) $ n^{2}=(2n^{2}+1)\tan \alpha $
D) $ n=(2n^{2}+1)\tan \alpha $
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Answer:
Correct Answer: D
Solution:
- $ \tan \alpha =\frac{-\frac{AC}{AP}+\frac{AB}{AP}}{1+\frac{AC}{AP}.\frac{AB}{AP}} $ $ {AP=n(AB) $
$ \Rightarrow AP=2n(AC)} $ $ \tan \alpha =\frac{-\frac{1}{2n}+\frac{1}{n}}{1+\frac{1}{2n^{2}}} $
$ \Rightarrow $ $ \frac{n}{(2n^{2}+1)}=\tan \alpha $
$ \Rightarrow $ $ n=(2n^{2}+1)\tan \alpha $ .
BETA
