Q. Let the arc $AC$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $AC$, divides the arc $AC$ such that $\frac{\text{length of arc }AB}{\text{length of arc }BC}=\frac15$, and $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$, then $\alpha+\sqrt2(\sqrt3-1)\beta$ is equal to
A) $2\sqrt3$
B) $2-\sqrt3$
C) $5\sqrt3$
D) $2+\sqrt3$
Explanation
[Solution by Dr. Abubacker] $\angle AOB=15^\circ$, $\angle BOC=75^\circ$. Take $OA=(1,0)$, $OC=(0,1)$, $OB=(\cos15^\circ,\sin15^\circ)$. Then $\alpha+\beta\cos15^\circ=0$ and $\beta\sin15^\circ=1$, so $\beta=\frac1{\sin15^\circ}$, $\alpha=-\cot15^\circ=-(2+\sqrt3)$. With $\sin15^\circ=\frac{\sqrt2(\sqrt3-1)}4$: $\sqrt2(\sqrt3-1)\beta=4$. The value is $4-2-\sqrt3=2-\sqrt3$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.