🔍 Grade 10 · Lesson 8 of 12
Hinge Hanger: The Ambiguous Case of the Sine Rule
Give a triangle two sides and an angle that isn't between them, and it can hinge open into two different shapes — or none at all.
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The big idea: Most triangle information (like two angles and a side, or two sides and the included angle) pins down exactly one triangle. But two sides and a non-included angle (SSA) is different: the side opposite the given angle can swing like a hinge and land in more than one place, so the Law of Sines can produce zero, one, or two valid triangles.
Not every triangle is pinned down the same way
Give a triangle two angles and a side, or two sides and the angle between them, and exactly one triangle fits — you can build it and there's no other option. But give it two sides and an angle that is not between them (SSA) and something strange can happen: sometimes there's still only one triangle, sometimes there are two completely different ones, and sometimes there's none at all.
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The Law of Sines
In any triangle, each side is proportional to the sine of its opposite angle, and that ratio is the same for all three sides: a ÷ sin A = b ÷ sin B = c ÷ sin C.
\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \]
The Law of Sines: side over sine-of-opposite-angle is constant across a triangle.
🎮 Hinge Hanger LIVE
Explore the ambiguous case: one angle and two sides can hinge into two different triangles.
Picture side a on a hinge
Fix angle A and side b, then swing side a like a hinge from the far end of b. An arc of radius a sweeps around and may cross the triangle's base line at two points, just one point (tangent to it), or none at all. Each crossing point is a valid triangle, which is exactly why SSA can have more than one answer.
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Compare a to b·sin A and to b
Given angle A and sides a and b: if b sin A < a < b, there are two triangles. If a ≥ b, there is exactly one. If a < b sin A, there is none — the swinging side is too short to ever reach the base.
📝 Worked example: In a triangle, A = 30°, a = 5, b = 8. Find the possible value(s) of angle B.
- Check: b sin A = 8 × 0.5 = 4. Since 4 < 5 < 8, this is the two-triangle case.
- By the Law of Sines: sin B = (b sin A) ÷ a = 4 ÷ 5 = 0.8.
- B = sin−1(0.8) ≈ 53.1°, but sine is also positive in the second quadrant, so B could also be 180° − 53.1° = 126.9°.
- Check both against the triangle angle sum with A = 30°: 30° + 53.1° = 83.1° (valid) and 30° + 126.9° = 156.9° (also valid).
✓ Both work: B ≈ <strong>53.1°</strong> or B ≈ <strong>126.9°</strong> — two different triangles satisfy the given information.
📝 Worked example: In a triangle, A = 30°, a = 10, b = 8. How many triangles are possible?
- Here a = 10 ≥ b = 8, so this is the one-triangle case — side a is long enough to reach the base at only one point.
- sin B = (b sin A) ÷ a = (8 × 0.5) ÷ 10 = 0.4, giving B ≈ 23.6°.
- The other candidate, 180° − 23.6° = 156.4°, fails: 30° + 156.4° = 186.4° > 180°, so it is thrown out.
✓ Only <strong>one</strong> triangle exists, with B ≈ 23.6°.
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Always test the obtuse candidate
Whenever you solve sin B = (some value) in the SSA case, remember there are two angles between 0° and 180° with that sine: one acute and one obtuse (its supplement). Always check whether the obtuse one still keeps the angle sum under 180° before accepting or rejecting it.
Check your understanding
1. The ambiguous case (SSA) arises when you know:
SSA means side-side-angle, where the known angle is not sandwiched between the two known sides — this is the setup that can be ambiguous.
2. For A = 30°, a = 5, b = 8, computing b·sinA = 4 shows that:
Since 4 < 5 < 8 (that is, b·sinA < a < b), side a is long enough to cross the base twice, giving two valid triangles.
3. In that same triangle, one solution is B ≈ 53.1°. What is the other possible value of B?
The other candidate is the supplement: 180° − 53.1° = 126.9°, and it still keeps the angle sum under 180° with A = 30°.
4. For A = 30°, a = 10, b = 8 (a > b), how many triangles are possible?
When a ≥ b, the swinging side is long enough to meet the base at only one point, so the triangle is unique.
5. Why can side a 'hinge' into two positions when solving an SSA triangle?
Side a behaves like a hinged rod of fixed length; the arc it sweeps can intersect the base line twice, once, or not at all.
✅ Key takeaways
- The Law of Sines relates each side to the sine of its opposite angle: a/sinA = b/sinB = c/sinC.
- SSA (two sides, non-included angle) is ambiguous: it can produce 0, 1, or 2 valid triangles.
- Picture side a hinging from the far vertex — the number of times its arc crosses the base gives the count.
- Compare a to b·sinA and to b: two triangles if b·sinA < a < b, one if a ≥ b, none if a < b·sinA.
- Whenever you find one angle from sine, always test its supplement too before ruling it out.