JEE Main Solved Paper 2017 Question 13
Question: Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $ \angle BPC=\beta , $ then $ tan,\beta $ is equal to :- [JEE Main Solved Paper-2017]
Options:
A) $ \frac{4}{9} $
B) $ \frac{6}{7} $
C) $ \frac{1}{4} $
D) $ \frac{2}{9} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{AB}{AP}=\frac{1}{2} $
Let $ \angle APC=\alpha $
$ \tan \theta =\frac{AC}{AP}=\frac{1}{2}\frac{AB}{AP}=\frac{1}{4} $
$ ( AC=\frac{1}{2}AB ) $
Now $ \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } $
$ \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1}{2} \begin{bmatrix} \tan (\alpha +\beta )=\frac{AB}{AP} \\ \tan (\alpha +\beta )=\frac{1}{2} \\ \end{bmatrix} $ on solving $ \tan \beta =\frac{2}{9} $
