1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Circle question (triangles too)
  1. Circles and triangles (Replies: 2)

  2. Circles and Triangles (Replies: 1)