Question

In the figure, $BC$ is the diameter, $PQ$ is the tangent, $O$ is the centre of circle and $\angle BAQ={65}^o$.
Find $\angle AOC$.

• A. ${25}^o$
• B. ${65}^o$
• C. ${50}^o$
• D. ${30}^o$

Answer 1

The correct option is C ${50}^o$


We know tangent is always perpendicular to the radius at the point of contact.
So, $\angle OAQ={90}^o$
$\angle OAB+\angle BAQ={90}^o$
$\angle OAB+{65}^o={90}^o$
$\angle OAB={90}^o-{65}^o$
$\angle OAB={25}^o$

In $△OAB$
$OA=OB$ (radius of circle)
so, $\angle OAB=\angle OBA$
$\angle OBA={25}^o$

Now, minor arc $AC$ subtending $\angle AOC$ at centre and $\angle ABC$ at point B on the circumference of circle.
So,
$\angle AOC=2\angle ABC\text{(angle made by arc at centre is double}\text{the angle made at any point on rest}\text{part of circumference of circle)}$

$\angle AOC=2\times {25}^o$
$\angle AOC={50}^o$

So, option $\left(c\right)$ correct.