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Circle question (triangles too)

  • Thread starter Couperin
  • Start date
  • #1
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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 = [tex]\theta[/tex] radians.

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

Homework 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]

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:

Answers and Replies

  • #2
57
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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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