I’m trying to use the sine rule in a triangle and I keep tripping over the “two angles” thing. My calculator is being a little chaos gremlin and I’m not sure if it’s me or it.
Setup: Triangle ABC with A = 38°, a = 9 (opposite A), and b = 12 (opposite B). Using the sine rule, sin B / 12 = sin 38° / 9, so sin B ≈ (12/9)·sin 38° ≈ 0.8209. Then B ≈ arcsin(0.8209) ≈ 55.1°. But also, 180° − 55.1° = 124.9° seems to work because sin θ = sin(180° − θ). My brain: cue the “two triangles?” dance.
I tried finishing it both ways:
– If I take B = 55.1°, then C = 180° − 38° − 55.1° = 86.9°, and c comes out large.
– If I take B = 124.9°, then C = 17.1°, and c comes out much smaller.
Both sets appear to satisfy the sine rule. Am I actually allowed to have two different triangles here, or is one of these supposed to be ruled out?
Direct question: how do I decide between the acute and the obtuse angle when using the sine rule in this kind of SSA situation? Is there a simple check before committing (like the “longer side faces the larger angle” idea), and how do I use that without just assuming the answer?
Follow-up: is there a good trick to avoid my calculator silently picking the acute angle and making me forget the other option? Or is there a quick “height” test with the given numbers I should be doing first?
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3 Responses
Ah yes-the infamous SSA “ambiguous case”! Your calculator isn’t trolling you; arcsin always returns the acute angle, but sin θ = sin(180° − θ) means a second (obtuse) option can also fit. The clean way to decide is the height test: drop the altitude from A onto side b; its length is h = b·sin A. Then: if a < h, no triangle; if a = h, exactly one right triangle (B = 90°); if h < a < b, two triangles (one with B acute, one with B obtuse); if a ≥ b, exactly one triangle (B must be ≤ A). For your data A = 38°, a = 9, b = 12, we get h = 12·sin 38° ≈ 7.39, and since 7.39 < 9 < 12, this is the two-triangle case. That’s why sin B ≈ 0.8209 gives both B ≈ 55.1° and B ≈ 124.9°, and both lead to valid C (86.9° or 17.1°) and consistent side-angle ordering (b > a so B > A). Quick check habit: compare a to h and b before pressing equals-this tells you whether to expect 0, 1, or 2 solutions. Also use “longer side ↔ larger angle”: if a ≥ b, then B ≤ A, which instantly rules out the obtuse option. Simple worked example: take A = 30°, a = 5, b = 4. Here h = 4·sin 30° = 2 and a ≥ b, so there’s exactly one solution with B ≤ 30°. Law of sines gives sin B = (4·sin 30°)/5 = 0.4, so B = 23.6° (the supplement 156.4° would make A + B > 180°, so it’s impossible). In short: do the height test first, and if you do use arcsin, always consider both B and 180° − B and then filter them with angle-sum and side-length logic.
You’ve hit the classic SSA ambiguous case-use the quick height test h = b sin A: if a ≤ h there’s 0/1 (right) triangle, if a ≥ b there’s exactly one (with B acute), and if h < a < b there are two. Here h = 12 sin 38° ≈ 7.39 and since 7.39 < 9 < 12, both B ≈ 55.1° and B ≈ 124.9° are valid; does sketching with b as the base and “swinging” the side of length a help you see the two positions?
Yep, you really have two triangles here: with A=38°, a=9, b=12 the “height” is h = b·sin A ≈ 7.39, and since h < a < b you get both B ≈ 55.1° and 124.9°. Quick rule: compute h = b·sin A (picture a door hitting a wall-too short can’t reach, just right touches once, mid-length hits twice, long enough hits once), and whenever you’re in that two-hit zone, take B = arcsin(...) and also its supplement 180° − B.