question_answer

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 \[\text{tan}\,\beta \] is equal to :-     JEE Main Solved Paper-2017

A)  \[\frac{4}{9}\]                                  

B)  \[\frac{6}{7}\]

C)  \[\frac{1}{4}\]                                  

D)  \[\frac{2}{9}\]

Correct Answer: D

Solution :

 \[\frac{AB}{AP}=\frac{1}{2}\]                 Let          \[\angle APC=\alpha \]                 \[\tan \theta =\frac{AC}{AP}=\frac{1}{2}\frac{AB}{AP}=\frac{1}{4}\]                        \[\left( AC=\frac{1}{2}AB \right)\]                 Now \[\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\] \[\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1}{2}\left[ \begin{matrix}    \tan (\alpha +\beta )=\frac{AB}{AP}  \\    \tan (\alpha +\beta )=\frac{1}{2}  \\ \end{matrix} \right]\] on solving \[\tan \beta =\frac{2}{9}\]