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

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{3}{4}\right)$

$\theta ={\tan }^{-1}\sqrt{2}$
$C{Q}_1=2$
$\tan \varphi =\frac{1}{2}$
Required angle $=\theta -\varphi ={\tan }^{-1}2-{\tan }^{-1}\frac{1}{2}$
$={\tan }^{-1}\left(\frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}}\right)={\tan }^{-1}\frac{3}{4}$