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Hey, I'm stuck on this question, I'm not even sure where to start, please could you give me some advice on how to go around this?
A chord AB divides a circle with centre O into 2 regions whose areas are in the ratio 2:1. If angle OAB=θradians, show that:
3θ-3sinθ=2π
Thank you in advance.
Draw a diagram. Draw a radius from O to A and from O to B. Let the radius be of length r.
The area of the larger part is equal to the area of the larger sector + the area of the triangle AOB.
The area of the smaller part is equal to the area of the smaller sector - the area of the triangle AOB.
Remember that the area of a sector is
and the area of a triangle is (where a and b = r here).Also note that the angle of the large sector = 2pi - theta.
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I get the larger region to equal:
[r²sinθ+r²(2π-θ)]/2
And the smaller region to equal:
[r²θ-r²sinθ]/2
Which means that:
[r²sinθ+2πr²-r²θ]/2 = r²θ-r²sinθ
=> r²sinθ+2πr²-r²θ = 2r²θ-2r²sinθ
=> 3r²sinθ+2πr²-3r²θ = 0
=> ÷r²] 3sinθ+2π-3θ = 0
=> 3θ-3sinθ = 2π
Yay! Thank you! ^^
No problem
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