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Question

In a circle with centre O, suppose A,P,B are three points on its circumference such that P is the mid-point of minor arc AB. Suppose when AOB=θ,
area(AOB)area(APB)=5+2,
If AOB is doubled to 2θ, then the ratio is area(AOB)area(APB) is

A
15
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B
52
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C
23+3
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D
512
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Solution

The correct option is A 15

Let the radius of the circle be 'r'
area(AOB)=12×r×rsinθ =12r2sinθ
Now,
area(APB)=area(OAP)+area(OBP)area(AOB)area(APB)=2(12r2sinθ2)12r2sinθarea(APB)=r2sinθ2(1cosθ2)

We know that area(AOB)area(APB)=12r2sinθr2sinθ2(1cosθ2)area(AOB)area(APB)=cosθ21cosθ25+2=cosθ21cosθ215+2=1cosθ2152=1cosθ21cosθ2=151=5+14

cosθ=2cos2θ21 =5+1+2581 =514
When the angle θ is doubled then,
area(AOB)area(APB)=cosθ1cosθ =5141514 =5155=15

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