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