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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'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,566

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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