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Maths - Circle Theorems Help - Prove angle x is equal to y-180 watch

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    Below is a picture. I want to know how you can prove that x or Angle CAB is equal to y-180 or angle CDB
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    Name:  2015-11-05 17.23.20.jpg
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Size:  365.2 KBThere is the picture. HELP! P.S: Clicking on it makes it bigger.
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    Just to be sure, was it stated that CD is a tangent to the circle?
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    (Original post by justinawe)
    Just to be sure, was it stated that CD is a tangent to the circle?
    Yes
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    Ok then, these are the things you should note:

    Angle CDB is 180-y degrees. as triangle BCD is an isosceles triangle, angle BCD will be the same.

    given that CD is a tangent to the circle, then what is angle ACD? using this and 180-y for angle BCD, you can find angle ACB.

    as triangle ABC is in a semi-circle, what does this make angle ABC?

    With this info you should be able to show that x = 180-y
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    Name:  2015-11-05 18.25.45.jpg
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Size:  401.3 KBI still can't work it out
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    (Original post by justinawe)
    Ok then, these are the things you should note:

    Angle CDB is 180-y degrees. as triangle BCD is an isosceles triangle, angle BCD will be the same.

    given that CD is a tangent to the circle, then what is angle ACD? using this and 180-y for angle BCD, you can find angle ACB.

    as triangle ABC is in a semi-circle, what does this make angle ABC?

    With this info you should be able to show that x = 180-y
    ...
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    (Original post by justinawe)
    Ok then, these are the things you should note:

    Angle CDB is 180-y degrees. as triangle BCD is an isosceles triangle, angle BCD will be the same.

    given that CD is a tangent to the circle, then what is angle ACD? using this and 180-y for angle BCD, you can find angle ACB.

    as triangle ABC is in a semi-circle, what does this make angle ABC?

    With this info you should be able to show that x = 180-y
    ...
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    Use the alternate segment theorem: the angle between a tangent and a chord at the point of contact is equal to the angle subtended by the chord.
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    Thanks
 
 
 
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