The Ambiguous Case

You know that $\sin 30° = \tfrac{1}{2}$. Is there another angle between $0°$ and $180°$ with the same sine? If a triangle's angle has sine $\tfrac{1}{2}$, does knowing the sine tell you the angle?

Sine cannot distinguish an angle from its supplement. So when the sine rule returns an angle, there are two candidates, $\theta$ and $180° - \theta$, and only the angle sum can rule one out. If both survive, both triangles are real.

Lesson 5 established that $\sin(180° - \theta) = \sin\theta$. An angle and its supplement sit at the same height on the unit circle, so they share a sine exactly.

Every angle in a triangle lies strictly between $0°$ and $180°$, and that whole range is available. So if you know a triangle's angle has sine $0.6$, you know the angle is $36.9°$ or $143.1°$, and the sine value alone cannot choose between them.

Cosine has no such problem. It is positive for acute angles and negative for obtuse ones, so a cosine value names exactly one angle in the range. This is why the cosine rule never produces an ambiguous answer and the sine rule sometimes does.

The ambiguity only appears in one arrangement of information: two sides and an angle opposite one of them, often shortened to SSA.

Two sides and the included angle (SAS) fixes the triangle completely, and you would use the cosine rule anyway. Three sides (SSS) fixes it. Two angles and a side (AAS) fixes it, since the third angle follows from the angle sum.

SSA is the exception, and the reason is the compass picture. The angle and its adjacent side are locked in place. The opposite side is free to swing, and a swinging segment can meet a straight ray at two points.

Watch for the shape of the question rather than the words. If the sine rule is going to hand you an angle, ask about the supplement before you go any further.

Having found the acute candidate $\theta$ from the calculator, form the obtuse candidate $180° - \theta$ and add it to the angle you were given.

If the total is less than $180°$, there is room left for a third angle, so the second triangle exists and must be reported. If the total is $180°$ or more, no third angle fits and the obtuse candidate is impossible, so there is only one triangle.

That is the entire test, and it takes one line of arithmetic. The commonest mistake in this topic is not getting it wrong, it is never doing it.

Compare the two known sides. If the known angle is opposite the longer of them, there is exactly one triangle, always.

The reason is the relationship between sides and angles: in any triangle, the longer side faces the larger angle. Suppose the known angle $A$ is opposite the longer side $a$. Then $A$ is the larger of the two angles, so $B$ must be smaller than $A$, which rules out $B$ being obtuse whenever $A$ is acute.

If the known angle is opposite the shorter side, the second triangle may exist, and you must run the test. In the surveyor's figures, $40°$ sits opposite the $8$ m side while the other known side is $11$ m, which is the arrangement to be suspicious of.

The area rule $A = \tfrac{1}{2}ab\sin C$ contains a sine, so working backwards from a known area to the angle $C$ inherits the same double answer.

Here, though, the outcome is different: both answers are always valid. Two sides with any angle between $0°$ and $180°$ between them close into a triangle, so there is no angle sum left to violate. A given area, two given sides, and you have two genuinely different triangles, one thin and wide, one squat.

So the area rule is the cleaner illustration of the idea. The sine rule needs a test; the area rule simply gives you both.

Five items on the ambiguous case. Decide how many triangles there are before you write any angles down.

1. 1 Find all angles $\theta$ with $0° < \theta < 180°$ such that $\sin\theta = 0.6$.

   Show answer
   The calculator gives the acute one; its supplement has the same sine.$36.9°$ and $143.1°$

2. 2 A triangle has a known angle of $70°$, and the sine rule gives a second angle whose obtuse candidate is $115°$. Does the second triangle exist?

   Show answer
   Test the sum: $70° + 115° = 185°$, which already exceeds $180°$.No, one triangle only

3. 3 In triangle $ABC$, $a = 7$, $A = 30°$ and $b = 12$. How many triangles fit?

   Show answer
   $\sin B = \frac{12\sin 30°}{7} = 0.857\ldots$, so $B = 59.0°$ or $121.0°$, and $30° + 121.0° = 151.0° < 180°$.Two

4. 4 Two sides of $6$ cm and $10$ cm enclose an angle $\theta$ and the triangle has area $24$ cm$^2$. Find $\theta$.

   Show answer
   $30\sin\theta = 24$, so $\sin\theta = 0.8$.$53.1°$ or $126.9°$

5. 5 Why can the cosine rule never produce an ambiguous answer for an angle?

   Show answer
   Cosine is positive for acute angles and negative for obtuse ones, so the sign of the answer already decides which it is.Cosine distinguishes acute from obtuse; sine does not