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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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 $\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}$, and $\overrightarrow{OC}=\alpha\,\overrightarrow{OA}+\beta\,\overrightarrow{OB}$, then $\alpha+\sqrt{2}\,(\sqrt{3}-1)\,\beta$ is equal to:
Solution
Let radius $=1$ and take $O$ as originArc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$
Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$
$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$
Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$
$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$
$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$
Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$
$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$
So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$
$\boxed{2-\sqrt3}$
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