Caption: The proof of the law of sines illustrated.
Proof:
We replace B and C by B' and C' as needed.
Note that h is a perpendicular dropped from vertex B.
c*sin(A) = h = a*sin(180°-C) = a*sin(C) ,
where we have used the trig identity sin(180°-C) = sin(C).
We now immediately see that sin(A)/a = sin(C)/c.
Thus, we arrive at the final result
sin(A) sin(B) sin(C) ------- = -------- = ------- which is the law of sines, QED. a b c
There are 2 possible solutions. This arises from the fact the law of sines gives the sine of an angle, not the angle itself.
Well both are allowed solutions if they satisfy the triangle-angles-sum-to-180° rule: i.e., A + C < 180° and A + C' = A + (180°-C) < 180°.
If both these inequalities hold, then both solutions lead to a triangle. Then B = 180°-A-C < 180° and B' = 180°-A-C' < 180° and then sin(B or B') is valid and then side b or b' can determined from the law of sines.
If the problem is specified has having only one solution, then more information is needed to determine which of the possible solutions is that one solution.
If A < 90° holds, there may be 2 solutions, but there may still be a unique solution. For example, say that A = 80°, C = 5°, and C' = (180° - 5°) = 175°. In this case, B = 180° - 80° - 5° = 95° for a solution, but B' = 180° - 80° - 175° = -75° does NOT give a solution.
However, if A < 90° is violated (i.e., A ≥ 90°), then one or both of A + C < 180° and A + (180°-C) < 180° is violated and there can only be a unique solution or NO solution (e.g., with A = 179° and C = 5°).
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File: Trigonometry file:
law_of_sines.html.