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    could a genius please work this out.......pretty please!

    ps. please excuse the terrible diagram ....:rolleyes:

    it should say:

    ABCD is a cyclic quadrilateral
    PAQ is a tangent to the circle at A
    BC = CD
    AD is parralel to BC
    angle BAQ = 32 degrees

    find the size of angle BAD

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    anyone?
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    (Original post by Saf!)
    could a genius please work this out.......pretty please!

    ps. please excuse the terrible diagram ....:rolleyes:

    it should say:

    ABCD is a cyclic quadrilateral
    PAQ is a tangent to the circle at A
    BC = CD
    AD is parralel to BC
    angle BAQ = 32 degrees

    find the size of angle BAD
    do you know what the answer is.....i got 64 degrees, but i could easily be wrong, since i've forgotten all these circle theoroms
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    I get 116 degrees. Is that right?
    Allternate segment theorem to get <ADB and <ABD , and because it's an isoceles, 2x + y = 180
    2 times 32 - 64 180 -64
    <BAD = 116 degrees?
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    i did tht paper in exam conditions in class and i got 64 degrees which was wrong but i dont know the right answer! anybody know the right answer??
    xxx
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    hey PPl

    I just tried that Q and a mate of mine did aswell. We both got BAD = 64
    Are you sure thats wrong ? Could your teacher of marked it badly ?

    Chris
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    this is how i did it...
    angle BDA = 32 , because of angle in opposite segment rule
    angle DBC = 32 , because of alternate ('Z') angles
    angle BDC = 32 , because triangle BCD is isoceles
    angle BCD = 116 , because angles in a triangle add up to 180
    therefore angle BAD = 64 , because opposite angles in a cyclic quad. add up to 180.
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    the answer is 64.
 
 
 
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