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Math Help - question on properties of circle 7

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    question on properties of circle 7

    This is yet to be solved.
    AB and DC are tangents to the circle.
    angle APC = 125^o and angle PDC = 30^o
    Find angle PAB
    Attached Thumbnails Attached Thumbnails question on properties of circle 7-ques-15.jpg  
    Last edited by ukorov; November 8th 2009 at 07:46 AM.
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    MHF Contributor red_dog's Avatar
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    AP intersects the curcle at E and DP intersects the circle at F.

    \frac{m(arcCF)+m(arcBE)}{2}=55

    \frac{m(arcEC)+m(arcBF)}{2}=125

    \frac{m(arcEC)-m(arcCF)}{2}=30

    From the second equality substract the sum of the other two.
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    not quite understand.
    ...what does m stand for?
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    MHF Contributor red_dog's Avatar
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    Is the measure of the arc.
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    angle bpa + cpa = 180 (we know because all lines are 180 degrees)
    cpa = 125
    bpa = 55
    pba + pab + bpa = 180 (we know because all real triangles are 180 degrees)
    bpa = 55
    pba = 90 (we know because its a tangent of a radius)
    90 + pab + 55 = 180
    pab = 35
    ? i think i did it right
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    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by orange gold View Post
    angle bpa + cpa = 180 (we know because all lines are 180 degrees)
    cpa = 125
    bpa = 55
    pba + pab + bpa = 180 (we know because all real triangles are 180 degrees)
    bpa = 55
    pba = 90 (we know because its a tangent of a radius)
    90 + pab + 55 = 180
    pab = 35
    ? i think i did it right
    No, BC is not the diameter in the circle, so AB is not perpendicular to BP.
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    Be careful all of you, something is not given here:
    BPC is not diameter and P is not radius.
    BPC is straight line but APD may be NOT.

    yes AB and DC are tangents to the circle. However, angles B and C would be 90 degrees BUT ONLY IF BP and CP are radius....

    The correct answer is 40^o but i cannot quite understand Red Dog's method...
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    Quote Originally Posted by ukorov View Post
    This is yet to be solved.
    AB and DC are tangents to the circle.
    angle APC = 125^o and angle PDC = 30^o
    Find angle PAB
    In triangle PCD the angles are 30 at D, 55 at P, and therefore 95 at C. Now extend the tangent DC to a point E say, on the far side of C from D. Then angle BCE is 85. But the angles between a chord and the two tangents at the ends of the chord are equal. Therefore angle ABC is also 85. Then in triangle ABP the angles are 85 at B, 55 at P, and hence 40 at A.
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    Quote Originally Posted by Opalg View Post
    In triangle PCD the angles are 30 at D, 55 at P, and therefore 95 at C. Now extend the tangent DC to a point E say, on the far side of C from D. Then angle BCE is 85. But the angles between a chord and the two tangents at the ends of the chord are equal. Therefore angle ABC is also 85. Then in triangle ABP the angles are 85 at B, 55 at P, and hence 40 at A.
    argh...that make sense.....
    but what is the formal name of the theorem for the fact "the angles between a chord and the two tangents at the ends of the chord are equal" ?
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    Quote Originally Posted by ukorov View Post
    what is the formal name of the theorem for the fact "the angles between a chord and the two tangents at the ends of the chord are equal" ?
    I don't know that it is a theorem with a formal name. But if you draw the two radii from the centre of the circle to the ends of the chord, and also the line from the centre of the circle to the midpoint of the chord, then you get two congruent triangles. That, together with the fact that the radii are perpendicular to the tangents, gives you the result.
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