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    Hey everyone,
    My maths teacher asked me to work this out and she said it is really, i mean really, hard. Any one have an idea.

    This is what i got so far...
    <TSU = as exterior angle of pentagon = 72, that's it, lol

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    (Original post by Carthaginian)
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    (Original post by NothingButWaleed)
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    If you were to join R and T to make the triangle RST, it would be isosceles as PQRST is a regular pentagon. Therefore, RS = ST.

    Due to it being a regular pentagon angle <RST = 108.
    Therefore, <STR = <SRT, as isosceles so, (180-108)/2 = 72/2 = 36.
    As a result of this, we can work out that <QRT = <QRS - <SRT = 108 - 36 = 72.

    Therefore, due to alternate segment theorem:
    <QRT = <SUT = 72.
    Therefore, <TSU = <SUT. Thus, ST = UT, so TSU is an isosceles triangle!
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    (Original post by TheMightyBadger)
    ...
    Please don't post full solutions - it's against the rules of this forum.

    It would have been great if you gave some hints and let the OP try it themselves.
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    (Original post by notnek)
    Please don't post full solutions - it's against the rules of this forum.

    It would have been great if you gave some hints and let the OP try it themselves.
    Sorry, its my first time answering. I joined a few days ago.
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    (Original post by TheMightyBadger)
    Sorry, its my first time answering. I joined a few days ago.
    No problem. I hope you continue to help in this forum
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    (Original post by NothingButWaleed)
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    You can also reach a solution whereby you use the circle theorem that the angle at the centre of a circle is twice that of an angle at the circumference at a stage of your proof.
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    (Original post by TheMightyBadger)
    You can also reach a solution whereby you use the circle theorem that the angle at the centre of a circle is twice that of an angle at the circumference at a stage of your proof.
    And how would that be done. I know the theorem but how would i know the angle at the centre as i cannot split <108 from its center?
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    (Original post by NothingButWaleed)
    And how would that be done. I know the theorem but how would i know the angle at the centre as i cannot split <108 from its center?
    You are given that QR and PT are tangents to the circle. You can join them up somewhere and make a shape. The angle is not 108.
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    (Original post by TheMightyBadger)
    You are given that QR and PT are tangents to the circle. You can join them up somewhere and make a shape. The angle is not 108.
    I dont understand?
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    (Original post by NothingButWaleed)
    I dont understand?
    Well a tangent meets the centre of the circle at 90 degrees.
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    (Original post by TheMightyBadger)
    Well a tangent meets the centre of the circle at 90 degrees.
    Yea i know that..
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    (Original post by NothingButWaleed)
    Yea i know that..
    Okay so if you join the tangents to the centre of the circle, you will make the irregular pentagon QROTP. Two of the interior angles are right angles and the other two are regular interior angles of a pentagon. Then you work out <ROT.
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    Really helpful thank you.
 
 
 
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