A 10 inch long pencil AB with mid-point C and a small eraser P are placed on the horizonta…

Question

A 10 inch long pencil $AB$ with mid-point $C$ and a small eraser $P$ are placed on the horizontal top of a table such that $PC=\sqrt5$ inches and $\angle PCB=	an^{-1}(2)$. 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

Accepted Answer

Let $	heta=\angle PCB=	an^{-1}(2)$, initially giving perpendicular distance $PM=PC\sin	heta=\sqrt5\cdot\dfrac{2}{\sqrt5}=2$ inches.

After rotating by an angle $\alpha$ about $C$, we need the new perpendicular distance $PC\sin\alpha=1\Rightarrow\sqrt5\sin\alpha=1\Rightarrow\sin\alpha=\dfrac{1}{\sqrt5}\Rightarrow	an\alpha=\dfrac12$.

Rotated angle $=	heta-\alpha=	an^{-1}2-	an^{-1}\dfrac12=	an^{-1}\left(\dfrac{2-\frac12}{1+2\cdot\frac12}ight)=	an^{-1}\dfrac34$.

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