I’m trying to solve triangles with the sine rule when I know two sides and a non-included angle (SSA). I understand that sin⁻¹ gives an acute angle but there might also be an obtuse option, and sometimes there are two valid triangles or none. I’m not sure how to decide the case from the given numbers alone.
For example, if I fix angle A and I’m given sides a and b, my calculator returns an acute angle for B, but the diagram sometimes suggests an obtuse B or even two mirror possibilities. Other times the triangle shouldn’t exist at all. I keep mixing up which side comparison tells me the outcome.
Is there a simple, reliable decision rule to tell-before committing to a full solution-whether SSA gives 0, 1 (acute), 1 (obtuse), or 2 triangles? Do I compare one of the sides to a “height” relative to the known angle, or am I thinking about the wrong quantity?
Analogy: it feels like a hinged ruler with fixed lengths reaching a point-sometimes there are two mirror positions, sometimes just one, and sometimes it can’t reach at all. I’d appreciate a clear way to check which situation I’m in.
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
Here’s the quick, reliable rule for SSA (given A, a, b with a opposite A, b opposite B): compute the “height” h = b·sin(A). Then compare a to h and b. If a < h, no triangle exists (the side a is too short to “reach” the ray at angle A). If a = h, there’s exactly one right triangle at B. If a > h, you have at least one triangle; now check A and the a vs b comparison: if A is obtuse or right, there’s exactly one triangle; if A is acute, then you get two triangles precisely when h < a < b, and otherwise (a ≥ b) just one triangle (acute). This matches the sine-rule view: sin B = (b sin A)/a must be ≤ 1 to exist, giving one acute B from arcsin; the “mirror” obtuse B = 180° − arcsin(...) is valid only when A is small enough to leave room, which boils down to a < b. Analogy time: fix a ray making angle A and mark off a segment of length b on it; now swing a compass of radius a from the far end of that segment-if the circle misses the base line you get 0, if it just kisses you get 1 (right), if it cuts twice you get 2, and if it cuts once beyond the “turning point” you’re in the a ≥ b one-triangle case.
Quick rule-of-thumb: treat b as a fixed pole and h = b sin A as the “reach” a must clear-if A is acute then a < h ⇒ 0, a = h ⇒ 1 (right at B), h < a < b ⇒ 2, a ≥ b ⇒ 1; if A is obtuse, a triangle exists only when a > b. Example: A = 40°, b = 10 gives h ≈ 6.43, so a = 5 → 0, a = 6.43 → 1 right, a = 8 → 2, a = 12 → 1; see Khan Academy’s ambiguous case guide: https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-laws-of-sines/a/ambiguous-case.
Yes-use the “height” h = b sin A with A known and sides a (opposite A) and b: if A is acute then a < h → 0, a = h → 1 right (B = 90°), h < a < b → 2 (one B acute, one B obtuse), and a ≥ b → 1 (B acute); if A is obtuse then a ≤ b → 0 and a > b → 1 (B acute). I finally memorized this by always sketching the height first-“compare a to h, then to b”-and this explainer helped: https://www.mathsisfun.com/algebra/trig-ambiguous-case.html.