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If we are given two sides say $a$, $b$ and an angle $X$. How can we determine whether this angle $X$ is opposite to $a$ or $b$ (i.e. $A$, $B$) or the third included angle $C$ ?

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Generally, you can't. It is quite possible that the given angle measure and side lengths allow the angle to be the included angle C (determining 1 triangle that way), or either A or B (possibly determining 0, 1, or 2 triangles in each of these two configurations).

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$X$ can always be the included angle $C$. Just draw an angle $X$, mark off lengths $a$ and $b$ along the sides, and connect the endpoints. Your are following the side-angle-side method of proving triangles congruent. If $X \ge \pi/2$, it must be opposite the longest side of the triangle, so can only be opposite the longer of $a$ and $b$. If $X \lt \pi/2$, and it is opposite $a$, you can use the law of sines to find $B$, which may have one or two legal values, then use $\pi-A-B$ to find $C$. Then you can try making $X$ opposite $b$.

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In the last cases, with the Law of Sines, I think it's also possible that there would be no legal values for the angle. – Isaac Mar 12 '11 at 20:00
I believe you are correct. If $b/a$ is too large and you try to make $X$ be $A$ it could make the required $\sin B \gt 1$ – Ross Millikan Mar 12 '11 at 21:11

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