The Ambiguous Case of the Sine Law (SSA)

The root cause is that the sine ratio gives the same value for an angle and its supplement: $\sin \theta =\sin (180°-\theta )$. On the unit circle, the points for $\theta$ and $180°-\theta$ are mirror images across the $y$-axis, so they share the same height $y=\sin \theta$. For example, $\sin 50°=\sin 130°\approx 0.7660$. When the sine law produces $\sin B=0.7660$, the calculator reports $B=50°$, but $B=130°$ satisfies the equation just as well. In ASA or AAS you never solve for an angle with the sine law, and in SAS or SSS the cosine law is used, whose inverse has no such double answer. Only SSA forces you to find an angle from its sine, so only SSA can be ambiguous.

Suppose you're given angle $A$ and the two sides $a$ (opposite $A$) and $b$ (adjacent to $A$). Imagine swinging side $a$ like a compass hinged at the far end of $b$ — depending on how long $a$ is, it might not reach the base at all, might just barely touch it (one triangle), might cross it twice (two triangles), or might reach past it in only one way (one triangle). The height of that swing, $h=b\sin A$, is the dividing line: it is the shortest possible distance from the hinge down to the base line, measured along the perpendicular.

Each row has a geometric reason. If $a<h$, even the shortest path to the base (the perpendicular) is longer than $a$. If $a=h$, side $a$ IS the perpendicular, so the triangle has a right angle. If $h<a<b$, the circle of radius $a$ crosses the base twice, once on each side of the foot of the perpendicular, and both crossings lie on the correct side of vertex $A$. If $a\ge b$, the second crossing lands at or behind vertex $A$, where it no longer makes a triangle with the given angle, so only one triangle survives.

This table only matters when angle $A$ is ACUTE. If the given angle is $90°$ or obtuse, there's a simpler rule: exactly one triangle exists if $a>b$, and none if $a\le b$. The reason: a triangle can hold at most one angle of $90°$ or more, so that angle must be the largest, and the largest angle always faces the longest side.

The Ambiguous Case for Sine Law - Nerdstudy

When two triangles exist, the sine law first gives you $\sin B$. Its inverse sine gives ONE acute angle $B_1$ — but $B_2=180°-B_1$ is also a valid solution as long as $A+B_2$ still comes out under $180°$. Solve the rest of each triangle separately: each has its own third angle $C=180°-A-B$ and its own third side from the sine law. For $A=40°$, $a=8$, $b=10$ the two triangles look very different, as the comparison shows.

Law of Sines (SSA) Ambiguous Case 2 Triangles Possible

Two triangles are not a mathematical curiosity; they often describe two real moments. A ship leaves port on a straight course that makes a $32°$ angle with the line to a lighthouse $11$ km away. When is the ship exactly $7$ km from the lighthouse? This is the last worked example in disguise: the ship is $7$ km away TWICE, once after about $5.45$ km of sailing (approaching) and again after about $13.20$ km (moving away). Sometimes context rules one triangle out, and then you say why. On a test with no context, report BOTH triangles. Follow the same routine every time so no case slips past you:

1. Confirm the information really is SSA (the angle is not between the two sides).
2. If the given angle is $90°$ or more: one triangle if $a>b$, otherwise none.
3. If it is acute: compute $h=b\sin A$ and compare $a$ with $h$ and $b$.
4. Two triangles: find $B_1$ with ${\sin }^{-1}$, set $B_2=180°-B_1$, then solve each triangle completely.
5. State all answers clearly, labelled triangle 1 and triangle 2.