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Math Help - question on properties of circle 3

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    question on properties of circle 3

    Referring to the attached pic:
    (1) prove triangle BTQ ~ triangle QTA
    (2) find TQ
    (3) find the values of the radii
    Attached Thumbnails Attached Thumbnails question on properties of circle 3-ques-8.jpg  
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  2. #2
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    Hello ukorov
    Quote Originally Posted by ukorov View Post
    Referring to the attached pic:
    (1) prove triangle BTQ ~ triangle QTA
    (2) find TQ
    (3) find the values of the radii
    Produce QO_2 to meet the circle at R. Then \angle TBQ = \angle BRQ (alternate segment)

    \Rightarrow \angle TBQ = \tfrac12 \angle BO_2Q (angle at centre)

    \angle TAO_1 = \angle TBO_2 = 90^o (angle between tangent and radius)

    \Rightarrow AO_1 \parallel BO_2 (corresponding angles equal)

    \Rightarrow \angle AO_1T = \angle BO_2Q (corresponding angles)

    But \angle AQT =\tfrac12\angle AO_1T (angle at centre)

    \Rightarrow \angle AQT = \tfrac12\angle BO_2Q =\angle TBQ (proved)

    So in \triangle's BTQ, QTA:
    \angle AQT = \angle TBQ (proved)

    \angle ATQ = \angle QTB (same angle)
    \Rightarrow \triangle BTQ \sim \triangle QTA

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello ukorovProduce....
    alright that makes sense now.
    for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?
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  4. #4
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    Hello ukorov
    Quote Originally Posted by ukorov View Post
    alright that makes sense now.
    for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?
    Yes. Note that PA=PQ=PB (tangents from point to circle). So Q lies on a circle that has AB as a diameter. So \angle AQB = ?

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello ukorovYes. Note that PA=PQ=PB (tangents from point to circle). So Q lies on a circle that has AB as a diameter. So \angle AQB = ?

    Grandad
    i saw that but i used angle AQT = a and angle BQR = (180 - a)/2 instead. with adj. angle on straight line, angle AQB = 90
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