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Thread: 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; Feb 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) $\displaystyle \angle DCO = \angle COB$ (alternate angles). Then use the fact that $\displaystyle \triangle DOC$ is isosceles ($\displaystyle OC = OD$) to find its other angles.

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

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

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

    $\displaystyle \Rightarrow \angle ODC = \angle DCO$

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

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

    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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