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A quadrilateral $ABCD$ is inscribed in a circle whose center is $O$. We know that $ \widehat{AOB}=90^\circ$, $ \widehat{BOC}= 60^\circ$, $ \widehat{COD}= 140^\circ$. Determine $\angle{A}$, $\angle{B}$, $\angle{C}$ and $\angle{D}$.

I really can't figure it out. Please help.

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    $\begingroup$ What does $\hat{A}$ mean? $\endgroup$
    – N. Owad
    Commented Sep 12, 2014 at 14:41
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    $\begingroup$ @N.Owad Probably angle $\widehat{BAD}$. $\endgroup$
    – Jean-Claude Arbaut
    Commented Sep 12, 2014 at 14:42
  • $\begingroup$ right, the angle described by AB and AC $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 14:44
  • $\begingroup$ and the same for the others $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 14:44

1 Answer 1

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The triangles from the center are isoceles, it is a trivial matter to compute their two other angles. Then add these pairwise.

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  • $\begingroup$ Hence one should be able to solve it...right? $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 14:52
  • $\begingroup$ Give me your results...i don't know their "sides" $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 14:53
  • $\begingroup$ Could you pleaseeee write all the steps? $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 14:55
  • $\begingroup$ Draw a figure, you'll see that the question is trivial. $\endgroup$
    – user65203
    Commented Sep 12, 2014 at 14:58
  • $\begingroup$ true...how to prove it? $\endgroup$
    – user175544
    Commented Sep 12, 2014 at 15:01

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