Question

In the given figure, O is the centre of the circle, $\angle OAB={30}^{\circ }$ and $\angle OCB={55}^{\circ }$. Find the measures of $\angle BOC$ and $\angle AOC$.

Answer 1

$\Delta AOB$ is an isosceles triangle
AO = OB = radii of circle
So, $\angle OAB=\angle OBA={30}^{\circ }$
$\angle AOB={180}^{\circ }-({30}^{\circ }+{30}^{\circ })$
$={120}^{\circ }$ (sum of all angles of $\Delta ={180}^{\circ }$)
$\Delta BOC,OC=OB=$ radius of circle
$\angle OCB=\angle OBC={55}^{\circ }$
$\angle COB={180}^{\circ }-({55}^{\circ }+{55}^{\circ })={70}^{\circ }$ (Property of $\Delta$)
$\angle AOC+\angle COB=\angle AOB$
$\angle AOC={120}^{\circ }-{70}^{\circ }={50}^{\circ }$