Question

A $10$ inches long pencil $AB$ with mid point $C$ and a small eraser $P$ are placed on the horizontal top of a table such that $\quad P C=\sqrt{5} \quad$ inches and $\angle PCB =\tan ^{-1}(2) .$ The acute angle through which the pencil must be rotated about $C$ so that the perpendicular distance between eraser and pencil becomes exactly $1 $ inch is:

• A. $\tan ^{-1}\left(\frac{3}{4}\right)$
• B. $\tan ^{-1}(1)$
• C. $\tan ^{-1}\left(\frac{4}{3}\right)$
• D. $\tan ^{-1}\left(\frac{1}{2}\right)$

Answer 1

Answer: (A)

From figure.
$\sin \beta=\frac{1}{\sqrt{5}} $
$\therefore \tan \beta=\frac{1}{2} $
$\tan (\alpha+\beta)=2 $
$\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=2 $
$\tan \alpha+\frac{1}{2} $
$1-\tan \alpha\left(\frac{1}{2}\right) $
$\tan \alpha=\frac{3}{4} $
$\alpha=\tan ^{1}\left(\frac{3}{4}\right)$