In the given figure, O is the centre of a circle and ∠AOC = 130°. Then, ∠ABC = ?
(a) 50°
(b) 65°
(c) 115°
(d) 130°

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Solution

(c) 115°
Take a point D on the remaining part of the circumference.
Join AD and CD.

Then $∠ ADC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 130 °) = 65 °$
In cyclic quadrilateral ABCD, we have:
∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC + 65° = 180°
⇒ ∠ABC = (180° - 65°) = 115°

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