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**Circles #4**

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AB is the diameter of a circle with center O.

C is a point on the circumference such that angle COB = θ

The area of the minor segment cut off by AC is equal to twice the area of the sector BOC.

Prove that

θ is in degrees

*Last edited by Agnishom (2014-02-02 00:50:06)*

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

hi Agnishom,

I must have time to spare today; I'm looking through the unanswered posts; and I found this one. Did you ever do it?

Use the following:

area of a segment = area of the sector less the triangular bit.

And then you simplify this and use sin(A) = 2 sin(a/2).cos(a/2)

Takes a few lines only.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

http://www.meritnation.com/ask-answer/question/q-ab-is-the-diameter-of-a-circle-center-o-c-is-a-point-on/areas-related-to-circles/1587959

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