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Question

In the given figure, O is the centre of a circle and AOB=130. Then, ACB=?

(a) 50

(b) 65

(c) 115

(d) 155

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Solution

ANSWER:
( c ) 115°
Join AD and BD.

Then chord AB subtends AOB at the centre and ADB at a point D of the remaining parts of a circle.∴ AOB = 2ADB
ADB=12AOB=12×130°=65°
In cyclic quadrilateral, we have:
ADB + ACB = 180°
⇒ 65 ° + ACB = 180°
ACB = (180° - 65°) = 115°


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