Question

# O is the centre of the circle. AB is a minor arc of the circle. The angle subtended by AB at centre ∠AOB = 110∘, then angle subtended by the arc at any point on the remaining part of the circle i.e. ∠APB is (where P is any point on the circle)?

A

50

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B
45
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C
55
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D
65
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Solution

## The correct option is C 55∘ As P is any point on the circle we can take that as shown in the figure. Now we can easily see that in ΔPOB, OP=OB since both are radii of the circle. So, ∠OPB = ∠OBP As exterior angle is equal to sum of interior opposite angles, we can write ∠COB = ∠OPB + ∠OBP = 2 ∠OPB So, ∠COB = 2 ∠CPB Similarly, in ΔPOA, OP=OA since both are radii of the circle. So, ∠OPA = ∠OAP As the exterior angle is equal to the sum of interior opposite angles, we can write ∠COA = ∠OPA + ∠OAP = 2 ∠OPA So ∠COA = 2 ∠CPA Hence, ∠COA+∠COB = 2(∠CPA + ∠CPB) Or, ∠ AOB = 2∠APB Substituting we can get ∠APB = 55∘.

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