📚 The Sine Rule | 正弦定理
The sine rule is a fundamental identity in triangle trigonometry. It links each side of a triangle to the sine of its opposite angle, allowing you to find missing sides or angles in non-right-angled triangles. For Edexcel A-Level Mathematics, you must be able to use it confidently, understand the ambiguous case, and choose between the sine rule and the cosine rule.
正弦定理是三角学中的一个基本恒等式。它将三角形的每条边与对角的正弦联系起来，使你能在非直角三角形中求缺失的边或角。对于 Edexcel A-Level 数学，你必须能熟练运用它、理解歧义情形，并能在正弦定理与余弦定理之间做出正确选择。
This article restates the rule, shows exactly when it applies, works through both directions (finding a side and finding an angle), explains the ambiguous SSA case, proves the result, and finishes with exam technique and a quick-reference summary.
本文将重新陈述该定理，说明它究竟何时适用，完整演示两个方向（求边与求角），解释 SSA 歧义情形，给出定理证明，并以考试技巧和快速复习小结收尾。
1. What the Sine Rule Says | 正弦定理的内容
For any triangle ABC, label side a opposite angle A, side b opposite angle B, and side c opposite angle C. The sine rule states:
对于任意三角形 ABC，将边 a 标记为角 A 的对边，边 b 标记为角 B 的对边，边 c 标记为角 C 的对边。正弦定理指出：
a / sin A = b / sin B = c / sin C
The reciprocal form is equally valid and is the version you will use most often when the unknown is an angle:
其倒数形式同样成立，而且当未知量是角时，这往往是你最常用的形式：
sin A / a = sin B / b = sin C / c
Some versions also write the common ratio as 2R, where R is the radius of the circumcircle, but the core exam requirement is the equality of the three ratios.
有些版本还会把公共比值写成 2R，其中 R 是外接圆半径，但考试的核心要求是三个比值相等。
The rule holds for every triangle, acute or obtuse, and it works with degrees or radians as long as your calculator is in the matching mode.
该定理对任意三角形都成立，无论是锐角三角形还是钝角三角形；只要计算器处于对应的模式，用角度或弧度都可以。
The largest side always sits opposite the largest angle, because sin θ increases on the range 0° to 90° and then decreases symmetrically. This is a useful sanity check after any calculation.
最长的边总是对着最大的角，这是因为 sin θ 在 0° 到 90° 上递增，随后对称递减。任何计算之后，这都是一条有用的合理性检验。
2. When to Use the Sine Rule | 何时使用正弦定理
Use the sine rule when you know:
在以下情况下使用正弦定理：
- two angles and one side (AAS or ASA);
- two sides and a non-included angle (SSA), which is the ambiguous case;
- any three of the four quantities a, b, A, B, where the pairing matches side-with-opposite-angle.
- 两角及一边（AAS 或 ASA）；
- 两边及其中一边的对角（SSA），也就是歧义情形；
- a、b、A、B 四个量中的任意三个，前提是配对关系符合“边与其对角”。
In every case the rule needs one complete side-angle pair plus one other piece of information. Without a complete pair, the sine rule gives you nothing.
无论哪种情形，使用该定理都需要一个完整的“边—对角”配对，再加上另一条信息。没有完整的配对，正弦定理就无从下手。
You should not use the sine rule when you know two sides and the included angle (SAS), or all three sides (SSS). Those two situations belong to the cosine rule.
当你已知两边及其夹角（SAS）或已知三边（SSS）时，不应使用正弦定理。这两种情形属于余弦定理的范畴。
3. The Ambiguous Case (SSA) | 歧义情形（SSA）
equation sin θ = k can have two solutions in the range 0° to 180°, namely θ and 180° − θ.
这一节是多数学生最容易丢分的地方。当你已知两边及其中一边的对角、而要求的是另一个角时，方程 sin θ = k 在 0° 到 180° 范围内可能有两个解，即 θ 与 180° − θ。
So after computing a value for an angle, always ask: could the obtuse partner also be valid? Test it by adding it to the known angle and checking that the sum is less than 180°.
因此，算出某个角的数值后，总要问：它的钝角搭档是否也可能成立？检验方法是把它与已知角相加，看总和是否小于 180°。
Suppose the known angle is A and the known sides are a (opposite A) and b, with b > a. The number of triangles is decided as follows:
设已知角为 A，已知边为 a（A 的对边）与 b，且 b > a。三角形解的个数按以下规则判断：
| Condition | Number of triangles |
| a < b sin A | no triangle exists |
| a = b sin A | exactly one, and angle B = 90° |
| b sin A < a < b | two possible triangles |
| a ≥ b | exactly one triangle |
If A is already obtuse, then it is the largest angle in the triangle, so a must be the largest side and only one triangle is possible. The ambiguity disappears.
如果 A 本身已经是钝角，那么它就是三角形中最大的角，因此 a 必为最长边，此时只可能有一个三角形，歧义随之消失。
In the Edexcel specification the examiner usually signals the ambiguity with a phrase such as 'there are two possible values' or 'find both possible values'. If you see no such phrase, still check before discarding a solution.
在 Edexcel 考纲中，考官通常会用“有两个可能值”或“求出两个可能值”之类的说法来提示歧义。如果没有看到这类措辞，也应在舍弃一个解之前先做检验。
4. Worked Example: Finding a Missing Side | 例题精讲：求未知边
m. Find b. Because two angles and a non-included side are known, this is an AAS situation.
在三角形 ABC 中，角 A = 42°，角 B = 68°，a = 9.5 cm。求 b。由于已知两角及其中一角的对边，这属于 AAS 情形。
The unknown b is opposite the known angle B, and the known side a is opposite the known angle A. So the pair matches and we can write:
未知边 b 对着已知角 B，而已知边 a 对着已知角 A。配对关系成立，因此可写出：
b / sin B = a / sin A → b = a sin B / sin A
= 9.5 × 0.9272 / 0.6691 = 13.16 cm, which rounds to 13.2 cm to three significant figures.
代入数值：b = 9.5 × sin 68° / sin 42° = 9.5 × 0.9272 / 0.6691 = 13.16 cm，取三位有效数字得 13.2 cm。
se angle B = 68° is larger than angle A = 42°, and a longer side must face a larger angle.
注意 b 比 a 长。这是合理的，因为角 B = 68° 大于角 A = 42°，而更长的边必然对着更大的角。
C = 180° − 42° − 68° = 70° and then c = a sin C / sin A = 9.5 × 0.9397 / 0.6691 = 13.3 cm.
如果题目还要求角 C 和边 c，则用 C = 180° − 42° − 68° = 70°，再算 c = a sin C / sin A = 9.5 × 0.9397 / 0.6691 = 13.3 cm。
5. Worked Example: Finding a Missing Angle | 例题精讲：求未知角
ssible values of angle B. Here the unknown is an angle, so start from the reciprocal form.
在三角形 ABC 中，a = 7 cm，b = 9 cm，角 A = 40°。求角 B 的所有可能值。这里未知量是角，所以从倒数形式入手。
sin B / b = sin A / a, so sin B = b sin A / a = 9 × sin 40° / 7 = 9 × 0.6428 / 7 = 0.8264.
写出 sin B / b = sin A / a，于是 sin B = b sin A / a = 9 × sin 40° / 7 = 9 × 0.6428 / 7 = 0.8264。
incipal value B = 55.7° to one decimal place. The obtuse partner is 180° − 55.7° = 124.3°.
由此得到主值 B = 55.7°（保留一位小数）。钝角搭档为 180° − 55.7° = 124.3°。
still exists and both values are valid. This matches the condition a < b, since 7 < 9.
检验钝角选项：A + B = 40° + 124.3° = 164.3° < 180°，所以三角形仍然存在，两个值都成立。这与 a < b（7 < 9）的条件一致。
/ sin 40° = 10.8 cm. If B = 124.3° then C = 15.7° and c = 7 sin 15.7° / sin 40° = 2.9 cm.
补全两个三角形：若 B = 55.7°，则 C = 84.3°，c = 7 sin 84.3° / sin 40° = 10.8 cm；若 B = 124.3°，则 C = 15.7°，c = 7 sin 15.7° / sin 40° = 2.9 cm。
s explicitly: (B = 55.7°, C = 84.3°, c = 10.8 cm) and (B = 124.3°, C = 15.7°, c = 2.9 cm).
因此完整答案是明确给出两个三角形：(B = 55.7°，C = 84.3°，c = 10.8 cm) 与 (B = 124.3°，C = 15.7°，c = 2.9 cm)。
6. The Circumradius and the Area Formula | 外接圆半径与面积公式
The common ratio is not just a computational convenience. It equals the diameter of the circumcircle of the triangle, that is, twice the radius R of the unique circle passing through A, B and C.
那个公共比值并不只是计算上的便利。它等于三角形外接圆的直径，也就是经过 A、B、C 三点的唯一圆的半径 R 的两倍。
a / sin A = b / sin B = c / sin C = 2R
ch is often the fastest route to an angle when the circumradius is given instead of a second side.
整理得 sin A = a / (2R)，当题目给出的是外接圆半径而非第二条边时，这往往是求角最快的途径。
The sine rule also feeds straight into the standard area formula for a triangle, in which you need two sides and the included angle:
正弦定理还直接关联到三角形面积的标准公式，该公式需要两边及其夹角：
Area = ½ a b sin C = ½ b c sin A = ½ c a sin B
rmula with the 2R form produces another result worth remembering, where Area = abc / (4R).
将面积公式与 2R 形式结合，可得到另一个值得记住的结果：Area = abc / (4R)。
7. Where the Sine Rule Comes From | 正弦定理的由来与证明
You do not need the proof in the exam, but knowing it makes the rule far harder to misremember. Drop a perpendicular of length h from vertex C to side c.
考试中并不需要写出证明，但掌握它会让这个定理难以被记错。从顶点 C 向边 c 作一条长度为 h 的垂线。
so h = b sin A. In the right-angled triangle on the right, sin B = h / a, so h = a sin B.
在左侧的直角三角形中，sin A = h / b，故 h = b sin A。在右侧的直角三角形中，sin B = h / a，故 h = a sin B。
sin B. Repeating the construction with a different perpendicular produces the third ratio.
令 h 的两个表达式相等，得 b sin A = a sin B；两边同除以 sin A sin B，得 a / sin A = b / sin B。换一条垂线重复同样的作法，即可得到第三个比值。
An alternative proof places the triangle in its circumcircle and uses the fact that the angle subtended by a chord at the circumference is constant, which is where the 2R factor comes from.
另一种证明把三角形放进它的外接圆中，利用“同弧所对的圆周角相等”这一事实，2R 这个因子正源于此处。
8. Sine Rule or Cosine Rule? | 用正弦还是用余弦定理？
Choosing the right rule is the single most common decision point in a trigonometry question. Use the information you are given, not what you are asked to find.
选对定理是三角题目中最常见的决策点。判断依据应该是题目给了什么信息，而不是题目要求什么。
| Given | Use |
| two angles + one side (AAS / ASA) | sine rule |
| two sides + non-included angle (SSA) | sine rule, watch the ambiguous case |
| two sides + included angle (SAS) | cosine rule (then sine rule for a second angle) |
| three sides (SSS) | cosine rule |
| right-angled triangle | SOHCAHTOA or Pythagoras first |
A frequent efficient strategy is mixed: use the cosine rule to find a third side from SAS, then switch to the sine rule to find the smaller remaining angle, because the smaller angle can never be obtuse and the ambiguity vanishes.
一种高效且常见的混合策略是：先用余弦定理从 SAS 求出第三边，再改用正弦定理求较小的那个未知角，因为较小的角不可能是钝角，歧义自然消失。
² + c² − a²) / (2bc), also tells you the type of triangle: if cos A is negative the triangle is obtuse at A.
余弦定理变形后，cos A = (b² + c² − a²) / (2bc) 还能判断三角形的类型：若 cos A 为负，则三角形在 A 处为钝角。
9. Bearings and Real-World Problems | 方位角与实际应用
Bearings questions are the classic applied setting for the sine rule. A bearing is measured clockwise from due north and written with three digits, so 047° and 210° are typical answers.
方位角问题是正弦定理最典型的应用题背景。方位角从正北方向顺时针度量，写成三位数，因此 047° 和 210° 是典型的答案形式。
Always sketch the situation before calculating: mark north lines at each point, draw the given bearings, and label every side and angle you know or can deduce.
计算之前一定要先画图：在每个点处画出正北方向线，画出已知方位角，并标注出所有已知或可推导的边和角。
Two north lines are parallel, so alternate angles between them are equal, and a 'back bearing' differs from the forward bearing by 180°.
两条正北线互相平行，因此它们之间的内错角相等，而“反方位角”与正方位角相差 180°。
Once the sketch is reduced to a single triangle with a known side-angle pair, the sine rule applies exactly as in the pure examples above, with lengths usually in metres or kilometres.
一旦图形被简化为一个已知“边—角”配对的三角形，正弦定理的用法就与上面的纯数学例题完全一致，长度单位通常为米或千米。
10. Exam Technique and Common Mistakes | 考试技巧与常见错误
Most lost marks in sine rule questions come from a small set of avoidable errors rather than from not knowing the rule.
正弦定理题目的失分大多来自一小撮本可避免的错误，而不是因为不知道定理本身。
- Leaving the calculator in radian mode when the question is in degrees, or the reverse;
- Writing a / sin A = b / sin B but substituting a side under the wrong angle;
- Rounding intermediate values too early, which distorts the final answer;
- Ignoring the obtuse solution in an SSA problem;
- Quoting an angle whose sum with the known angle exceeds 180°, which is impossible;
- Reaching for the sine rule when the data is actually SAS or SSS.
- 题目用角度制，计算器却留在弧度模式，或反之；
- 写出 a / sin A = b / sin B 后，却把边代到了错误的角下面；
- 过早对中间结果取近似，导致最终答案失真；
- 在 SSA 问题中忽略了钝角解；
- 给出与已知角之和超过 180° 的角，这不可能成立；
- 数据其实属于 SAS 或 SSS，却仍去用正弦定理。
Keep full accuracy on your calculator display and round only at the final line, normally to three significant figures for lengths and to one decimal place for angles in degrees.
在计算器显示中保留完整精度，只在最后答案处取近似，长度通常保留三位有效数字，角度通常保留一位小数。
Write the general formula first, then substitute. Method marks are awarded for a correctly stated relationship even when an arithmetic slip follows.
先写出通用公式，再代入数值。即使随后出现算术失误，只要关系式书写正确，仍可获得方法分。
11. Quick Reference Summary | 快速复习小结
sin A / a = sin B / b = sin C / c, and the common ratio equals 2R for the circumradius R.
正弦定理表述为 a / sin A = b / sin B = c / sin C，等价于 sin A / a = sin B / b = sin C / c，且公共比值等于 2R，其中 R 为外接圆半径。
Use it for AAS, ASA and SSA data, always pairing each side with its opposite angle, and use the cosine rule for SAS and SSS.
它适用于 AAS、ASA 与 SSA 数据，务必把每条边与其对角配对；而 SAS 与 SSS 则应使用余弦定理。
When the unknown is an angle and two sides with a non-included angle are given, always test the supplement 180° − θ, and state both valid triangles if the geometry allows them.
当未知量是角，且已知两边及其中一边的对角时，一定要检验补角 180° − θ；如果几何条件允许，就把两个成立的三角形都写出来。
ea result Area = ½ a b sin C, which combines with the sine rule to give Area = abc / (4R).
最后记住配套的面积公式 Area = ½ a b sin C，它与正弦定理结合可得 Area = abc / (4R)。
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