📚 Ambiguous Case of the Sine Rule: Two-Solution Problem | 正弦定理的模糊情形：两解问题辨析
When we know two sides and a non-included angle of a triangle, the sine rule can sometimes produce two different triangles. This is known as the ambiguous case. Understanding exactly when two solutions appear is essential for IB Mathematics exams and for solving realistic geometry problems.
当我们已知三角形的两边和其中一边的对角（SSA）时，正弦定理有时会给出两个不同的三角形。这就是所谓的“模糊情形”。准确把握两解何时出现，是IB数学考试和实际几何问题的关键。
1. The Sine Rule and When to Use It | 正弦定理及其适用条件
For any triangle ABC, let side a be opposite angle A, side b be opposite angle B, and side c be opposite angle C. The sine rule states that the ratio of a side to the sine of its opposite angle is constant.
对于任意三角形 ABC，设边 a 对应角 A，边 b 对应角 B，边 c 对应角 C。正弦定理指出：一条边与其对角正弦的比值恒为常数。
a / sin A = b / sin B = c / sin C
The sine rule is commonly used when we know two angles and one side, or two sides and an angle that is not the included angle.
正弦定理通常用于已知两角一边，或已知两边及其中一边的对角的情形。
2. The Ambiguous Case: Why SSA Is Dangerous | 模糊情形：为何“SSA”具有危险性
When two sides and the included angle are known (SAS), the triangle is unique. But when two sides and a non-included angle are known (SSA), the given information may represent zero, one, or two different triangles.
当已知两边及其夹角（SAS）时，三角形唯一确定。但当已知两边和其中一边的对角（SSA）时，所给条件可能对应零个、一个或两个不同的三角形。
Consider a fixed angle A and fixed side b. The unknown side c is not fixed, so side a can be placed in different directions. Whether the circle of radius a centred at C meets the ray AB decides how many triangles exist.
考虑固定角 A 和固定边 b。未知边 c 并不固定，因此边 a 可以沿不同方向放置。以 C 为圆心、半径为 a 的圆与射线 AB 的交点个数，决定了三角形存在的个数。
3. The Height h and the First Comparison | 高 h 与首个比较
Draw side b from A to C. The perpendicular distance from C to the ray AB is the height formed by the given angle.
从 A 到 C 作边 b。点 C 到射线 AB 的垂线距离，就是由已知角形成的高。
h = b × sin A
This height is the shortest possible length for side a to create a triangle. If a is shorter than h, no triangle can be formed; if a equals h, exactly one right-angled triangle is formed.
这个高是边 a 能构成三角形的最短长度。若 a 比 h 还短，则无法构成三角形；若 a 恰好等于 h，则构成一个直角三角形。
4. The Four Possible Outcomes | 四种可能结果
Assume the given angle A is acute. The number of possible triangles can be summarised by comparing a with h and b.
假设已知角 A 为锐角。可能的三角形数量可通过比较 a 与 h、b 来概括。
| Condition | Number of Triangles | Comment |
|---|---|---|
| a < h | 0 | No triangle is possible. |
| a = h | 1 | B = 90°, a right triangle. |
| h < a < b | 2 | Two different triangles exist. |
| a ≥ b | 1 | Only one triangle exists; isosceles if a = b. |
In Chinese, the typical way to remember the two-solution condition is “h < a < b". This is the only situation that yields two triangles.
中文中常将两解条件记为“h < a < b”。这是唯一会产生两个三角形的情况。
If A is obtuse, no ambiguity occurs: we need a > b, and the resulting triangle is unique.
若 A 为钝角，则不存在模糊性：需要 a > b，并且所得三角形唯一。
5. Two-Solution Example: A = 30°, b = 10, a = 6 | 两解示例：A = 30°，b = 10，a = 6
First calculate the height.
首先计算高。
h = b × sin A = 10 × sin 30° = 10 × 0.5 = 5
Because 5 < 6 < 10, the condition h < a < b is satisfied, so two triangles are possible.
因为 5 < 6 < 10，满足 h < a < b，所以存在两个可能的三角形。
Use the sine rule to find sin B.
利用正弦定理求 sin B。
sin B = b × sin A / a = 10 × 0.5 / 6 = 5 / 6 ≈ 0.8333
The acute angle is B₁ = 56.44°. The obtuse candidate is B₂ = 180° − 56.44° = 123.56°. Both are valid because A + B₂ = 30° + 123.56° = 153.56° < 180°.
锐角解为 B₁ = 56.44°。钝角候选解为 B₂ = 180° − 56.44° = 123.56°。两者均有效，因为 A + B₂ = 30° + 123.56° = 153.56° < 180°。
For Triangle 1: C₁ = 180° − 30° − 56.44° = 93.56°.
对三角形 1：C₁ = 180° − 30° − 56.44° = 93.56°。
c₁ = a × sin C₁ / sin A = 6 × sin 93.56° / 0.5 ≈ 11.98
For Triangle 2: C₂ = 180° − 30° − 123.56° = 26.44°.
对三角形 2：C₂ = 180° − 30° − 123.56° = 26.44°。
c₂ = a × sin C₂ / sin A = 6 × sin 26.44° / 0.5 ≈ 5.34
6. One-Solution Example: A = 30°, b = 10, a = 12 | 一解示例：A = 30°，b = 10，a = 12
Here h = 5 and a = 12, so a > b. The two-solution condition is not satisfied.
这里 h = 5，而 a = 12，所以 a > b。两解条件不满足。
sin B = b × sin A / a = 10 × 0.5 / 12 = 5 / 12 ≈ 0.4167
The calculator gives the acute angle B₁ = 24.62°. The obtuse candidate is B₂ = 155.38°, but A + B₂ = 185.38° > 180°, so it cannot form a triangle.
计算器给出锐角 B₁ = 24.62°。钝角候选解为 B₂ = 155.38°，但 A + B₂ = 185.38° > 180°，因此不能构成三角形。
Thus the only possible angle is B = 24.62°, and C = 180° − 30° − 24.62° = 125.38°.
因此唯一可能的角度是 B = 24.62°，C = 180° − 30° − 24.62° = 125.38°。
7. No-Solution Example: A = 30°, b = 10, a = 4 | 无解示例：A = 30°，b = 10，a = 4
Now h = 5 and a = 4, so a < h. The side given is too short to reach the line AB.
此时 h = 5，而 a = 4，所以 a < h。所给边太短，无法到达 AB 所在直线。
sin B = b × sin A / a = 10 × 0.5 / 4 = 1.25
Because 1.25 > 1, the sine rule has no real solution. No triangle exists.
因为 1.25 > 1，正弦定理无实数解。因此不存在任何三角形。
8. The Algebraic Method: Test the Two Candidate Angles | 代数方法：检验两个候选角
Even without using the height h, we can decide the number of solutions using the sine rule.
即使不使用高 h，我们也可以用正弦定理判断解的个数。
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Step 1: Compute sin B = b sin A / a.
第一步：计算 sin B = b sin A / a。
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Step 2: If sin B > 1, no solution.
第二步：若 sin B > 1，则无解。
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Step 3: If sin B = 1, then B = 90°, giving one solution.
第三步：若 sin B = 1，则 B = 90°，得一解。
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Step 4: If sin B < 1, let B₁ = arcsin(sin B). The second candidate is B₂ = 180° − B₁.
第四步：若 sin B < 1，令 B₁ = arcsin(sin B)。第二个候选角为 B₂ = 180° − B₁。
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Step 5: For each candidate, compute C = 180° − A − B. If C > 0, that candidate is valid.
第五步：对每个候选角计算 C = 180° − A − B。若 C > 0，则该候选角有效。
This method is reliable and avoids relying on memorised inequalities.
这种方法非常可靠，也避免死记硬背不等式。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
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Mistake 1: Forgetting the obtuse angle when sin B is positive. Always consider B₂ = 180° − B₁.
错误 1：当 sin B 为正时忘记钝角解。始终要考虑 B₂ = 180° − B₁。
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Mistake 2: Assuming two solutions whenever 0 < sin B < 1. You must check whether A + B₂ < 180°.
错误 2：只要 0 < sin B < 1 就默认为两解。必须检查 A + B
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