Does SSA congruence criterion work if the non-included angle is obtuse? I was trying to make some triangles when $2$ sides and the non-included angle was given. When the angle given was below $90$ degrees, there were two possible triangles which could be made. But, when it went above $90$ degrees, only one possibility came. I was using a compass to determine the third side's possibilities. I checked the internet if there were questions like this, but there were none. Also, no site said specifically that it works for $90+$ angles.
  Does this mean that the SSA congruence can work for angles greater than $90$ degrees? If so, I was taught very wrong in school.
 A: Here is a trigonometric proof. Say $\triangle ABC$ has an angle at $A$ with a measure of $x$, and we know side $AC$ has length $b$ and side $BC$ has length  $a$. We now want to find the side length of $AB$, or $c$.  Using the law of cosines, $$b^2 + c^2 -2bc(\cos(x)) = a^2,$$ and rearranging gives us a quadratic in $c$: $$c^2 - 2b(\cos(x))(c) + (b^2- a^2) = 0.$$ Using the quadratic formula, we get c = 2acos(x) +/- sqrt(4a^2cos^2(x)-4(b^2-a^2))/2. Taking the 4s out of the square root and dividing, we get c = acos(x) +/- sqrt(a^2cos^2(x) - (b^2-a^2))
We know that if x > 90, or triangle ABC is obtuse, then cos (x) must be negative. Therefore, when triangle ABC is obtuse, we have that c is a negative number plus or minus a positive number. (Square roots are always positive. ) This means that only the positive sign works, since the negative sign means a negative minus a positive, or a negative, and side lengths can't be negative, so there's only one solution. You're right! It does work. Following this logic, we know that there is only one solution when sqrt(a^2cos^2(x) - (b^2-a^2)) > acos(x), since then the negative sign is invalid. Doing a bunch of algebra yields b < a, which is the general formula for when SSA works.
A: There is the following theorem.

Let $AB=PQ$, $BC=QR$, $\measuredangle A=\measuredangle P$ and $BC>AB$.
Thus, $\Delta ABC\cong\Delta PQR.$

If $\measuredangle A\geq 90^{\circ}$ we can use this theorem immediately, but it's true also for $\measuredangle A<90^{\circ}.$
A: If another triangle has parts congruent to angle A and sides AB and BC of our triangle, and angle A is not acute, then the two triangles are congruent. If angle A is acute, and BC > or = to AB, then they are congruent. Also congruent if either angle B or C is a right angle. Otherwise (i.e., angle A is acute, ABC is not a right triangle, and BC < AB) there are two triangles, the smaller of which has an obtuse angle at C which is supplementary to its cousin of the larger triangle.  For this latter case, there is only congruence if we also know both original triangles (ours and the other one) are obtuse or both acute.
