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In the given figure, `O` is the centre of a circle. If `angle OAB = 40^(@)` and C is a point on the circle then `angle ACB =` ?

A

`40^(@)`

B

`50^(@)`

C

`80^(@)`

D

`100^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`OA = OB implies /_ OBA = /_ OAB = 40^(@)`
`:. /_ AOB = 180^(@) - ( 40^(@) + 40^(@)) = 100^(@)`
`:. /_ ACB = (1)/(2) /_ AOB = (1)/(2) xx 100^(@) = 50^(@)`
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