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Circle question (triangles too)

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations

    I think the following might be relevant:

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

    Area of sector = [tex]\frac{1}{2}r^2\theta[/tex]

    3. 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?
    Last edited: Mar 15, 2007
  2. jcsd
  3. Mar 15, 2007 #2
    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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