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Thread: Angles - Circle Properties

  1. #1
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    Angles - Circle Properties



    In the diagram, TP and TQ are tangents to the circle at the point P and Q respectively. TQY is a straight line. Given that ∠RPQ = 26, ∠SQY = 42 and ∠RXQ = 101, calculate the following angles.
    (a) ∠QRS
    (b) ∠RQP
    (c) ∠RQS
    (d) ∠PTQ
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  2. #2
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    Hello Punch
    Quote Originally Posted by Punch View Post

    In the diagram, TP and TQ are tangents to the circle at the point P and Q respectively. TQY is a straight line. Given that ∠RPQ = 26, ∠SQY = 42 and ∠RXQ = 101, calculate the following angles.
    (a) ∠QRS
    (b) ∠RQP
    (c) ∠RQS
    (d) ∠PTQ
    Do you know the Alternate Segment Theorem? This gives:

    (a) $\displaystyle \angle QRS = \angle SQY = 42^o$

    (b) $\displaystyle \angle RQP$ can then be calculated from the angles of $\displaystyle \triangle RQX$.

    (c) Using the alternate segment theorem again, $\displaystyle \angle RQT = \angle RPQ = 26^o$. So you can now find $\displaystyle \angle RQS$ from the straight line $\displaystyle TQY$.

    (d) $\displaystyle \angle TQP = \angle TQR + \angle RQX=\angle TPQ$ (since $\displaystyle \triangle TPQ$ is isosceles). So you can now work out $\displaystyle \angle PTQ$.

    Can you fill in the gaps?

    Grandad
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  3. #3
    MHF Contributor Amer's Avatar
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    as I can see you find most angles but you can find all of them by finding QPS

    there is a theorem said that the angle between the tangent of the cirlce and the chord equal to the angle on that chord so

    QPS=SQY = 42

    I think now you can find all

    you can use this to find RQT and RPT

    RQT = QSR
    RPT = RSP
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  4. #4
    Senior Member Shanks's Avatar
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    Yeah, Amer is right!
    When I was studying geometry in the plane, I saw the circle to be the most Unbelievable math Object.
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