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Question
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
Options
130°
115°
65°
165°
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Solution
115°
We have the following information in the following figure. Take a point P on the circle in the given figure and join AP and CP.
Since, the angle subtended by a chord at the centre is twice that of subtended atany part of the circle.
So, `angleAPC = (angleAOC)/2 = 130/2 = 65°`
Since `squareAPBP` is a cyclic quadrilateral and we known that opposite angles are supplementary.
Therefore,
\[ \Rightarrow \angle ABC + 65° = 180° \]
\[ \Rightarrow \angle ABC = 180° - 65° \]
\[ \Rightarrow \angle ABC = 115°\]
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