Question

O is the centre of the circle.Given that $\angle$OAB = 20${}^{\circ }$, $\angle$OCB = 55${}^{\circ }$. Find $\angle$BOC and $\angle$AOC in degree.
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Answer 1

OA = OB [radii of a circle]

$\angle$OAB = $\angle$OBA [Angles opposite to equal sides of a triangle]

Similarly,

OB = OC [radii of the same circle]

$\angle$OCB = $\angle$OBC [Angles opposite to equal sides of a triangle]

In $\Delta$OBC,

$\angle$OBC + $\angle$OCB + $\angle$BOC = 180${}^{\circ }$ [Angle sum property]

$\angle$BOC = 180${}^{\circ }$ - 110${}^{\circ }$ = 70${}^{\circ }$

In $\Delta$OAB,

$\angle$OAB + $\angle$OBA + $\angle$AOC + $\angle$BOC = 180${}^{\circ }$ [Angle sum property]

$\angle$AOC = 180${}^{\circ }$ - 20${}^{\circ }$ - 20${}^{\circ }$ - 70${}^{\circ }$ = 70${}^{\circ }$

So, $\angle$AOC = $\angle$BOC = 70${}^{\circ }$

Since, both have same value their difference is 0.