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Math Help - Angles in a circle.

  1. #1
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    Angles in a circle.

    AB is a diameter of a circle with centre O and CD//BA. If <CAB=x, find
    i) <COB
    ii) <DOC
    iii) <DAC
    iv) <ADC
    Attached Thumbnails Attached Thumbnails Angles in a circle.-107nbd5.jpg  
    Last edited by mr fantastic; February 5th 2010 at 12:23 PM. Reason: Changed post title
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  2. #2
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    Hello snigdha
    Quote Originally Posted by snigdha View Post
    AB is a diameter of a circle with centre O and CD//BA. If <CAB=x, find
    i) <COB
    ii) <DOC
    iii) <DAC
    iv) <ADC
    (i) Use angle at centre = twice angle at circumference.

    (ii) \angle DCO = \angle COB (alternate angles). Then use the fact that \triangle DOC is isosceles ( OC = OD) to find its other angles.

    (iii) Use angle at centre again, as in (i).

    (iv) \angle DCA =\angle CAB (alternate angles). Then use the angle-sum of \triangle ADC.

    Can you complete these now?

    Grandad

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  3. #3
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    i sure can...thanks a lot..!!
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  4. #4
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    hello grandad..

    well....do u think we can work out i) and ii) as i've done below-??

    We know that AOB is a straight line...
    so <AOB=180
    <COB, <DOC & <AOD cut equal angles at the centre as triangles COB, DOC & AOD are equal isosceles triangles..
    => <COB=<DOC=<AOD=180/3
    =60.
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  5. #5
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    Hello snigdha
    Quote Originally Posted by snigdha View Post
    hello grandad..

    well....do u think we can work out i) and ii) as i've done below-??

    We know that AOB is a straight line...
    so <AOB=180
    <COB, <DOC & <AOD cut equal angles at the centre as triangles COB, DOC & AOD are equal isosceles triangles..
    => <COB=<DOC=<AOD=180/3
    =60.
    No, you need to use the theorem: the angle subtended at the centre of a circle (by an arc) is equal to twice the angle subtended at the circumference (by the same arc).

    In (i) we need \angle COB, which is the angle subtended at the centre, O, by the arc CB. So this is twice the angle subtended at the circumference by the same arc. Can you see that this is \angle CAB? So:
    \angle COB = 2\angle CAB = 2x^o
    For (ii), \angle DCO = \angle COB (alternate angles). Then, since OC = OD (radii), \triangle COD is isosceles.

    \Rightarrow \angle ODC = \angle DCO

    Then use angle sum of triangle to find \angle DOC.

    In (iii), you'll use the 'angle at the centre' theorem again, but this time you'll use the fact that \angle DOC = 2\angle DAC.

    And I've given you an outline for (iv).

    Have another go. Let us know if you need further help.

    Grandad
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  6. #6
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    i got it! ..thanks a tonne for clarifying...!
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