Question

A sector $OABO$ of central angle $\theta$ is constructed in a circle with centre $O$ and radius $6.$ The radius of the circle that is circumscribed about the triangle $OAB,$ is

• A. $6\cos \frac{\theta }{2}$
• B. $6\sec \frac{\theta }{2}$
• C. $3(\cos \frac{\theta }{2}+2)$
• D. $3\sec \frac{\theta }{2}$

Answer 1

The correct option is D $3\sec \frac{\theta }{2}$

Let $OA=a,OB=b,AB=c$ and circumradius of the triangle $OAB$ be $R$


We know that $R=\frac{abc}{4\Delta }$
$\Delta =\frac{1}{2}ab\sin \theta =18\sin \theta$
In $△AOD,$ we have
$\sin \frac{\theta }{2}=\frac{c/2}{6}$
$\Rightarrow c=12\sin \frac{\theta }{2}$

Now, $R=\frac{6\cdot 6\cdot 12\sin \left(\frac{\theta }{2}\right)}{4\cdot 18\sin \theta }$
$\Rightarrow R=\frac{6\sin \left(\frac{\theta }{2}\right)}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$
$\therefore R=3\sec \frac{\theta }{2}$