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

Answer 1

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


$AB$ and $OD$ are the diameters of the circumscribed circle.
In rt. $△OBD,$
$\frac{x}{6}=\tan \frac{\theta }{2}$
So, $O{D}^2=6^2+6^2{\tan }^2\left(\frac{\theta }{2}\right)=6^2{\sec }^2\left(\frac{\theta }{2}\right)$
$\Rightarrow 4r^2=36{\sec }^2\left(\frac{\theta }{2}\right)$
$\therefore r=3\sec \left(\frac{\theta }{2}\right)$