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How can I find the radius of this circle?

  1. Nov 13, 2005 #1
    A, B, C, D are points on a circle of centre O and TA is the tangent to the circle at A.

    TA = 7.1 cm
    AD = 4.5 cm

    Basically if we were to draw this circle we would first draw (Clockwise) the point D on the circle then A then C and then B.
    Point T being outside of the circle slightly on the right of it.

    Now angle BDA = 36 degrees and DAT = 28 degrees

    From this info I need to find the radius of the circle and say if OT bisects AD?

    I have demonstrated (using the circle property) that angle BOA = 72 degrees and BCA = 144 degrees and also OAD = 62 degrees.

    But how can I find the radius and prove the bisection? Please any hints or answers to this treaky one :-)
  2. jcsd
  3. Nov 13, 2005 #2
    Divide AD in two with a point E. Now from the triangle OAE you get OA=AE/cos(62).

    Draw the bisector of AD from O to the line AT, call the point it makes on line AT point F. The projection of AF onto AD is AE, from this we get AF=AE/cos(28)=2,55 cm, which is not equal to TA and thus OT does not bisect AD (OF does).

    edit: seems deceptively simple, maybe i missed something???

    - Kamataat
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3
    seems right thanks
    Last edited: Nov 13, 2005
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