Triangle Congruence: Side-Side-Angle Explained

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ibc
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from high school mathematics I remember that there is a side-side-angle triangle congruence statement, which says it only proves the congruence if the angle oppose the larger of the 2 sides.

I don't understand why is that, and if in both triangles I have 2 equal sides, and an equal angle opposing 1 side, shouldn't I immediately get the angle opposing the other side, from the sins law?
 
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ibc said:
from high school mathematics I remember that there is a side-side-angle triangle congruence statement, which says it only proves the congruence if the angle oppose the larger of the 2 sides.

I don't understand why is that, and if in both triangles I have 2 equal sides, and an equal angle opposing 1 side, shouldn't I immediately get the angle opposing the other side, from the sins law?
You are completely correct. The sine law will give you the sine of the angle. But then you have a choice between two angles having that same same sine, one acute and the other obtuse. If you have some way of knowing that the angle MUST be either acute or obtuse. Now, if the given angle is opposite the longer of the two given sides, then that angle must be larger than the angle you are looking for. Since there is at most only one obtuse angle in a triangle, the angle you are looking for must be acute. But if the two sides are of the same length, the situation you are referring to, the opposite angles must be the same and, again because there is at most one obtuse angle in a triangle, both must be acute.

The theorem you are remembering is that if you are given two sides and an angle not between them, then the triangles are congurent if and if the side the angle is opposite is longer than or equal to the other side.