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Question

In the given figure, O is the centre of a circle and AOB=140. Then ACB=?

(a) 70

(b) 80

(c) 110

(d) 40

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Solution

(c) 110°
Join AB.
Then chord AB subtends AOB at the centre and ADB at a point D of the remaining parts of a circle.
AOB = 2ADB
ADB=1/2AOB=1/2×140°=70°
In the cyclic quadrilateral, we have:
ADB + ACB = 180°
⇒ 70° + ACB = 180°
ACB = (180° - 70°) = 110°


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