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    Can someone help me how to find angle BDC and OCD
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    (Original post by Chocolate...s)
    Can someone help me how to find angle BDC and OCD
    Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

    Where did you find this Q?
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    (Original post by Chocolate...s)
    Can someone help me how to find angle BDC and OCD
    ACO is 90 as tangent and radius are perpendicular, from there you can easily work out AOC and then BDC follows as angle at centre is 2 x angle at circumference.
    The key here is just to work out each angle and make and put it on the diagram - you should be able to work out all angles.
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    (Original post by RDKGames)
    Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

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    ahem... the angle at the centre is double the angle at the edge

    :console:
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    (Original post by B_9710)
    ACO is 90 as tangent and radius are perpendicular, from there you can easily work out AOC and then BDC follows as angle at centre is 2 x angle at circumference.
    The key here is just to work out each angle and make and put it on the diagram - you should be able to work out all angles.
    I wouldn’t have gone the way you went, I saw that ABDC is a parallelogram, meaning that opposite angles are equal BCD is 32 and therefore OCD would be ((360 -(32*2))-90)
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    (Original post by RDKGames)
    Unless I'm being thick or forgotten something, I don't see how those angles can be obtained from the given information.

    Where did you find this Q?
    https://corbettmaths.files.wordpress...e-theorems.pdf
    Question 20
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    (Original post by the bear)
    ahem... the angle at the centre is double the angle at the edge

    :console:
    Ah, I know the rule but I was struggling to apply it but after 'untangling' the quadrilateral BOCD it's a smack in the face
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    (Original post by RDKGames)
    Ah, I know the rule but I was struggling to apply it but after 'untangling' the quadrilateral BOCD it's a smack in the face
    https://shechive.files.wordpress.com...100&strip=info
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    Luckily I didn't have to use my trusty rotating tool there.
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    (Original post by Daniel100499)
    I wouldn’t have gone the way you went, I saw that ABDC is a parallelogram, meaning that opposite angles are equal BCD is 32 and therefore OCD would be ((360 -(32*2))-90)
    The diagram is deceptive - ABDC is not a parallelogram. Although one pair of sides are parallel, the other pair are not.
 
 
 
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