Circle question (triangles too)

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The discussion focuses on finding the length of chord AB in a circle with center O and radius 6 cm, given angle AOB as θ radians. The solution involves using the cosine rule to relate the sides of triangle AOB to the angle θ. By treating triangle AOB appropriately, the length AB is derived as √{72(1 - cos θ)}. The initial confusion about the use of Pythagoras and the relevance of circle chord theory is addressed, leading to a successful application of the cosine rule. Ultimately, the problem is resolved by rearranging and factorizing the equation correctly.
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Homework Statement



O is the centre of the circle of radius 6cm, and A and B are two points on the circumference such that angle AOB = \theta radians.

Show that the length AB is equal to \sqrt {72 (1 - cos \theta)}

Homework Equations



I think the following might be relevant:

Cosine rule: \theta = cos^-1\frac{b^2 + c^2 - a^2}{2bc}

Area of sector = \frac{1}{2}r^2\theta

The Attempt at a Solution



I don't really know where to start here. I think pythagoras is involved somewhere. The radius 6 must therefore be squared, multplied by 2 = 72. But I don't want to work backwards from the given soultion.

Am I missing some theory about chords of circles?
 
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No matter, I solved it.

Treat AOB as a triangle. Apply cosine rule to find AB, rearrange and factorise under the squareroot.
 
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