The law of sines and the SSA ambiguous case

study guide · oblique triangles

The law of sines is the friendliest formula in the oblique-triangle toolkit — until the day your homework says "two sides and a non-included angle" and suddenly one problem has two correct answers and the next has none at all. This is the SSA ambiguous case, the most reliably misunderstood topic in a trigonometry course. It isn't actually hard; it's just never explained geometrically. Let's fix that.

The law itself

In any triangle — no right angle required — label the angles A, B, C and the sides opposite them a, b, c. Then the ratio of each side to the sine of its opposite angle is constant:

a / sin A = b / sin B = c / sin C

Given any "matched pair" (a side and its opposite angle) plus one more piece, you can solve for the rest. ASA and AAS configurations are safe: two angles pin down the third (angles sum to 180°), and every side follows uniquely. The trouble starts only with SSA: two sides and an angle not between them.

Why SSA is ambiguous: the swinging door

Picture building the triangle physically. Draw the known angle A at a vertex, and run the known side b along one of its rays to point C. From C, the second known side a has to swing down — like a door on a hinge — until its far end lands somewhere on the other ray of angle A. The question is: where can it land?

The answer depends on how long a is compared to the height h = b · sin A, the perpendicular distance from C down to the opposite ray:

• a < h: the door is too short to reach the ray at all. No triangle exists.
• a = h: the door just barely touches — one right triangle, with the right angle at the foot of the height.
• h < a < b: the door crosses the ray in two places, one on each side of the height's foot. Two different valid triangles share the same three given measurements.
• a ≥ b: only the far crossing produces a valid triangle. One triangle. (This also covers the case where A is obtuse and a > b.)

That's the whole phenomenon. Zero, one, or two triangles — decided entirely by comparing a to h = b · sin A and to b. No memorized flowchart needed if you can picture the swinging door.

The arcsin trap

Here is why calculators quietly hide the second triangle. Solving SSA by the law of sines gives you something like sin B = 0.82, and you press arcsin to get B ≈ 55.1°. But arcsin only ever returns angles up to 90° — while a triangle happily hosts obtuse angles. Since sin(180° − B) = sin B, the supplement B′ = 180° − 55.1° = 124.9° satisfies the same equation.

A worked two-triangle example

Suppose A = 35°, b = 10, and a = 7. First, the height test:

h = b · sin A = 10 × sin 35° ≈ 5.74

Since 5.74 < 7 < 10 (that is, h < a < b), we're in two-triangle territory. Apply the law of sines:

sin B / b = sin A / a → sin B = 10 × sin 35° / 7 ≈ 0.8194

Triangle 1: B ≈ arcsin(0.8194) ≈ 55.0°, so C ≈ 180° − 35° − 55.0° = 90.0°, and c = a · sin C / sin A ≈ 7 × 1 / 0.5736 ≈ 12.2.

Triangle 2: B′ ≈ 180° − 55.0° = 125.0°. Check: 35° + 125.0° = 160° < 180°, so it's valid. Then C′ ≈ 20.0° and c′ ≈ 7 × sin 20.0° / sin 35° ≈ 4.2.

Two genuinely different triangles — one nearly right-angled and lanky, one squat and obtuse — built from identical given data. An answer of "B ≈ 55°" alone would be half credit at best.

When to use the law of cosines instead

If the given data is SAS (two sides with the angle between them) or SSS (three sides), the law of sines can't start — you never have a matched side–angle pair. That's the law of cosines' territory: c² = a² + b² − 2ab·cos C. It's also unambiguous, because arccos, unlike arcsin, distinguishes acute from obtuse on its own: cos of an obtuse angle is negative, so the sign of the result tells you the angle's nature directly. A practical strategy for SSA fans: after finding one angle with the law of sines, find the next with the law of cosines to sidestep a second ambiguity entirely.

Checklist for any oblique triangle

1. Classify the given data: ASA / AAS / SAS / SSS / SSA.
2. SAS or SSS → law of cosines first.
3. ASA or AAS → third angle, then law of sines. One solution, guaranteed.
4. SSA → compute h = b · sin A, compare with a: zero, one, or two triangles. Solve each fully.
5. Sanity-check every result: largest side faces largest angle, angles sum to 180°.