Sine rule for an angle can be ambiguous: When you use the sine rule to find an ANGLE (given two sides and a non-included angle — the SSA case), there may be two triangles that fit, because sin θ = sin(180° − θ).
Finding a side is never ambiguous: Using the sine rule (or cosine rule) to find a side gives a single answer — only finding an angle can produce two.
Why two angles? sin θ is the up-coordinate (y). Two angles — one acute, one obtuse (180° − θ) — sit at the SAME height, so they have the same sine. The calculator gives only the acute one.
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Subtract from 180°: Your calculator gives the acute angle from sin⁻¹. The second possibility is its supplement, 180° − θ, because both have the same sine.
IB-style question — find both angles
In a triangle, sin B = 0.6 with B unknown. Find both possible values of B.
Step by step
- Acute solution.
- Supplementary solution.
Final answer
B ≈ 36.9° or B ≈ 143.1°.
Both answers — 36.9° and 143.1° — sit at the same height on the circle, so they share the same sine. That equal-height pair is exactly why the sine rule for an angle can give two solutions.
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Only sine does this: cos⁻¹ gives a single angle in 0°–180°, so a cosine-rule angle isn't ambiguous — only the sine rule is.
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Obtuse partner = 180° − acute: To get the obtuse possibility from an acute angle θ, compute 180° − θ. Whether it's valid depends on the rest of the triangle (next section).
IB-style question — the two triangles
A triangle has a = 9, A = 40°, and b = 12. Find both possible values of angle B.
Step by step
- Sine rule for sin B.
- Two angles with this sine.
Final answer
B ≈ 59.0° or B ≈ 121°.
Sine rule: 9 is opposite 40°, and 12 is opposite B → sin B = 12 sin 40° ÷ 9 ≈ 0.857. Two angles share that sine: B ≈ 59.0° (the acute triangle) or B ≈ 121° (a second, more open triangle).
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Both can be genuine triangles: Here both 59° and 121° leave a positive third angle (with A = 40°), so two triangles exist.
Check the angle sum: A second (obtuse) angle is only valid if, added to the known angle, it stays under 180° — leaving room for a positive third angle. If the known angle plus the obtuse partner ≥ 180°, reject it.
IB-style question — reject one
A triangle has A = 70° and the sine rule gives B ≈ 50° or 130°. Which are valid?
Step by step
- Check B = 130° with A = 70°.
- So the obtuse case is impossible.
Final answer
Only B ≈ 50° works (130° would overflow 180°).
Always test the obtuse option against the angle sum: with A = 70°, B = 50° leaves C = 60° (✓), but B = 130° would give 70° + 130° = 200° > 180° — impossible, so that second triangle is rejected.
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Always test the obtuse option: Don't assume two answers — add the obtuse partner to the given angle; keep it only if the total is below 180°.