Question

In the figure, AB and CD are diameters of a circle with centre O. If $\angle OBD={50}^{\circ }$, find $\angle AOC$.

Answer 1

Two diameters AB and CD intersect each other at O. AC, CB and BD are joined

$\angle DBA={50}^{\circ }$

$\angle DBAand\angle DCA$ are in the same segment

$\therefore \angle DBA=\angle DCA={50}^{\circ }$

In $\Delta OAC,OA=OC$ (Radii of the circle)

$\therefore \angle OAC=\angle OCA$$=\angle DCA={50}^{\circ }$

and $\angle OAC+\angle OCA+\angle AOC={180}^{\circ }$ (Sum of angles of a triangle)

$\Rightarrow {50}^{\circ }+{50}^{\circ }+\angle AOC={180}^{\circ }$

$\Rightarrow {100}^{\circ }+\angle AOC={180}^{\circ }$

$\Rightarrow \angle AOC={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Hence $\angle AOC={80}^{\circ }$