In triangle $ABC$ , $\angle BAC=80^o$ . suppose $O$ is a point in triangle $ABC$ that $OBC=30^\circ$ and $OCB=10^\circ$ . Is it possible to find $\angle BOA$ ?
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$\begingroup$ What have you tried? Where do you get stuck? Hint: The triangle is isoceles, and $O$ lies on the perpendicular bisector of $AB$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 30 '15 at 17:22
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$\begingroup$ @GNUSupporter The triangle doesn't have to be isosceles. See the answers. $\endgroup$ – user236182 Dec 30 '15 at 17:35
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$\begingroup$ Sorry, I made a mistake. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 31 '15 at 2:57
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No, because this only fixes the $OBC$ triangle upto congruency.
In other words, for any $ABC$ with $BAC=80$ (and variable $ABC$ and $ACB$), you can find the point $O$.
Here's what I mean -
Triangle 1
Triangle 2
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1$\begingroup$ Though the point $O$ doesn't have to be inside the triangle. E.g., when $\angle ABC<30^{\circ}$. $\endgroup$ – user236182 Dec 30 '15 at 17:30
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No, because take such a triangle $ABC$ with $O$ fixed and consider the circumcircle of $ABC$. Then you can move the point $A$ along the circle without changing $\angle OBC$,$\angle OCB$ and $\angle BAC$, but surely $\angle BOA$ will change.
