Question

O is the centre of the circle. $\angle OAB={20}^{\circ },\angle OCB={55}^{\circ }$. Find $\angle BOC$ and $\angle AOC$.

Answer 1

Given : $\angle OAB={20}^{\circ },\angle OCB=\angle {55}^{\circ }$
To find : $\angle BOC=?$ and $\angle AOC=?$
Join AC, OA = OB [radius of circle]
$\Rightarrow \angle OAB=\angle OBA={20}^{\circ }$
Now, As OC = OB [radii of circle]
$\therefore \angle OCB=\angle OBC$ [Angle opposite to equal sides are equal]
$\angle OCB=\angle OBA+\angle ABC$
$\angle ABC=\angle OCB–\angle OBA={55}^{\circ }–{20}^{\circ }={35}^{\circ }$
and $\angle AOC=2\angle ABC=2\left({35}^{\circ }\right)={70}^{\circ }$ [Angle subtended at centre is twice at arc]
Now, In $\Delta AOB$
$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$
$\angle AOC+\angle BOC+\angle OAB+\angle OBA={180}^{\circ }{70}^{\circ }+\angle BOC+{20}^{\circ }+{20}^{\circ }={180}^{\circ }$
$\Rightarrow \angle BOC={70}^{\circ }$