In the figure, $AB$ and $CD$ are diameiers of a circle with centre $O$. If $\angle OBD = 50^o$, find $\angle AOC$.
"
Given:
In the figure, $AB$ and $CD$ are diameters of a circle with centre $O$.
$\angle OBD = 50^o$.
To do:
We have to find $\angle AOC$.
Solution:
$\angle DBA = 50^o$
$\angle DBA$ and $\angle DCA$ are in the same segment.
Therefore,
$\angle DBA = \angle DCA = 50^o$
In $\triangle OAC$,
$OA = OC$ (Radii of the circle)
$\angle OAC = \angle OCA = \angle DCA = 50^o$
$\angle OAC + \angle OCA + \angle AOC = 180^o$ (Sum of angles of a triangle)
$50^o + 50^o + \angle AOC = 180^o$
$100^o + \angle AOC = 180^o$
$\angle AOC = 180^o - 100^o = 80^o$
Hence $\angle AOC = 80^o$.
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