📚 178: Sine and Cosine Rules | 178：正弦与余弦定理
In IGCSE Mathematics, not every triangle is right-angled. When you encounter oblique triangles that lack a 90° angle, Pythagoras' theorem and basic SOHCAHTOA no longer apply. Instead, we turn to two powerful tools: the Sine Rule and the Cosine Rule. These rules allow us to find unknown sides and angles in any triangle, provided we have sufficient information. Mastering them is essential for success in the Edexcel IGCSE exam, as questions frequently appear in both calculator and non-calculator papers, often combined with area problems or real-life contexts such as navigation and bearings.
在IGCSE数学中，并非所有三角形都是直角三角形。当你遇到没有90°角的斜三角形时，勾股定理和基本的SOHCAHTOA便不再适用。此时我们需借助两个强有力的工具：正弦定理与余弦定理。只要已知足够的信息，这两个定理就能帮我们求出任意三角形中未知的边长和角度。掌握它们对在Edexcel IGCSE考试中取得好成绩至关重要，因为这类题目经常出现在计算器和非计算器试卷中，且常与面积问题或航海、方位角等实际情境结合考查。
1. Introduction to Non-Right Angled Triangles | 非直角三角形简介
In the Edexcel IGCSE specification, you are expected to solve problems involving triangles that are not right-angled. Unlike right-angled triangles where you can rely on trigonometric ratios (sin, cos, tan) defined for acute angles, generic triangles require a more flexible approach. The Sine Rule and Cosine Rule work for any triangle — acute, obtuse or even right-angled — and are derived from the geometric properties of triangles. Understanding when and how to use these rules is a core skill, and this article will guide you through each step with clear explanations and examples.
在Edexcel IGCSE课程标准中，你需要会解非直角三角形的问题。与直角三角形不同——在直角三角中你可以依赖为锐角定义的三角比（sin、cos、tan），一般三角形需要更灵活的方法。正弦定理和余弦定理适用于任意三角形——锐角、钝角甚至直角三角形——它们源自三角形的几何性质。理解何时以及如何使用这些定理是一项核心技能，本文将通过清晰的讲解与示例，引导你一步步掌握。
2. The Sine Rule – Statement | 正弦定理 – 定理表述
The Sine Rule states that in any triangle, the ratio of a side length to the sine of its opposite angle is constant. For a triangle with vertices A, B and C, sides a, b, c opposite to those angles respectively, we have:
正弦定理指出，在任意三角形中，边长与其对角的正弦值之比恒定。设三角形顶点为A、B、C，其对边分别为a、b、c，则有：
a / sin A = b / sin B = c / sin C
This form is most useful when you are looking for a missing side. Equivalently, you can invert the ratios to find a missing angle:
当你想求未知边长时，这一形式最有用。同理，将比值取倒数即可求未知角：
sin A / a = sin B / b = sin C / c
You only need two of the three ratios at any one time. The rule applies to all triangles, but you must label the triangle correctly first: side a is opposite angle A, side b opposite angle B, and side c opposite angle C. Edexcel often provides a diagram with labelled vertices; when it does not, draw your own and assign labels consistently.
每次你只需用到三个比例式中的两个。此定理适用于所有三角形，但你必须先正确标记三角形：边a对角A，边b对角B，边c对角C。Edexcel常提供带标记顶点的图示；若没有，则自己画图并统―标记。
3. Using the Sine Rule to Find a Missing Side | 用正弦定理求未知边
Suppose in triangle ABC, angle A = 40°, angle B = 60° and side a = 8 cm. Find side b.
假设在三角形ABC中，角A = 40°，角B = 60°，边a = 8 cm。求边b。
cm (to 2 decimal places). Always state the units and round appropriately as instructed in the question.
写出含两个相关比的正弦定理：a / sin A = b / sin B。代入已知值：8 / sin 40° = b / sin 60°。确保计算器处于角度模式。计算 sin 40° 与 sin 60°；则 b = (8 × sin 60°) / sin 40°。解得 b ≈ (8 × 0.8660) / 0.6428 ≈ 10.78 cm（保留两位小数）。务必标注单位并按题目要求适当四舍五入。
This method works whenever you know two angles and one side (AAS or ASA) — you can immediately find another side by cross-multiplying. Practice speed and accuracy, as these often form the first part of longer multi-step problems.
当你已知两角一边（AAS或ASA）时，这种方法总是可行——通过交叉相乘你立刻能求得另一边。熟练且准确地计算很重要，因为这类题目通常构成较长多步问题的第一部分。
4. Using the Sine Rule to Find a Missing Angle | 用正弦定理求未知角
ngle. For example, in triangle ABC, a = 14 cm, b = 18 cm, and angle A = 50°. Find angle B.
若已知两边及一个非夹角（SSA），倒置的正弦定理可帮助求未知角。例如，在三角形ABC中，a = 14 cm，b = 18 cm，角A = 50°。求角B。
1 decimal place). Remember that the inverse sine function on your calculator gives the acute angle, but we must check for the ambiguous case as explored in the next section.
使用形式 sin A / a = sin B / b。代入：sin 50° / 14 = sin B / 18。于是 sin B = (18 × sin 50°) / 14。计算 sin 50° ≈ 0.7660，因此 sin B ≈ (18 × 0.7660) / 14 ≈ 0.9849。然后 B ≈ sin⁻¹(0.9849) ≈ 80.1°（保留一位小数）。切记，计算器给出的反正弦值是锐角，我们还需检查是否存在歧义情况，下一节将进行探讨。
5. The Ambiguous Case of the Sine Rule | 正弦定理的歧义情况
ut longer than the height of the triangle (i.e., h = b sin A), then two different triangles can satisfy the data. This is known as the ambiguous case.
当使用正弦定理由两边及一个非夹角（SSA）求角时，可能存在两个符合条件的三角形，因为正弦函数在第一和第二象限均为正。如果已知角是锐角，且该角的对边小于另一已知边但大于三角形的高（即 h = b sin A），那么就可能有两个不同的三角形满足条件。这就称为歧义情况。
b = 12 cm. Using the Sine Rule gives sin B = (12 × sin 30°) / 8 = 0.75, so B ≈ 48.6° or B' ≈ 180° − 48.6° = 131.4°. Both are valid as long as the sum of angles A + B' one or both solutions are acceptable. Edexcel exam questions that involve the ambiguous case often instruct you to find 'both possible values' or to determine which triangle fits a given diagram.
例如，在三角形ABC中，角A = 30°，a = 8 cm，b = 12 cm。应用正弦定理得 sin B = (12 × sin 30°) / 8 = 0.75，于是 B ≈ 48.6° 或 B' ≈ 180° − 48.6° = 131.4°。只要角A + B' < 180°，两者均有效，此处满足（30° + 131.4° = 161.4° < 180°）。你必须计算两种可能并根据问题情境判断是取其中一个解还是两者都取。涉及歧义情况的Edexcel考题通常会要求你找出“所有可能的值”，或确定哪个三角形符合给定的图形。
6. The Cosine Rule – Statement | 余弦定理 – 定理表述
The Cosine Rule links the three sides of a triangle with the cosine of one of its angles. For any triangle ABC, it is expressed as:
余弦定理将三角形的三边与其一个角的余弦联系起来。对任一三角形ABC，可表示为：
a² = b² + c² − 2bc cos A
Similarly, you can write it cyclically for other sides:
类似地，可循环写出其他边的表达式：
b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C
The Cosine Rule is a generalisation of Pythagoras' ² + c². You will typically use the Cosine Rule when you have two sides and the included angle (SAS) — to find the third side — g the formula is essential, but understanding its structure helps avoid careless mistakes.
余弦定理是勾股定理的推广；确实，当角A = 90°时，cos A = 0，公式简化为 a² = b² + c²。通常，当你已知两边及其夹角（SAS）——用来求第三边，或已知三边（SSS）而要求一个角时，就要用到余弦定理。熟记公式至关重要，但理解其结构有助于避免粗心错误。
7. Using the Cosine Rule to Find a Missing Side | 用余弦定理求未知边
Consider triangle ABC with b = 10 cm, c = 7 cm, and included angle A = 65°. Find side a.
考虑三角形ABC，b = 10 cm，c = 7 cm，夹角 A = 65°。求边a。
0×7×cos 65° = 140 × 0.4226 (approx) = 59.164. So a² = 149 − 59.164 = 89.836. Therefore a = √89 the square root at the end and give your answer with the correct unit. The Edexcel mark scheme rewards clearly shown substitution and intermediate steps.
应用 a² = b² + c² − 2bc cos A。代入数值：a² = 10² + 7² − 2 × 10 × 7 × cos 65°。计算：100 + 49 = 149，2×10×7×cos 65° = 140 × 0.4226（约） = 59.164。因此 a² = 149 − 59.164 = 89.836。于是 a = √89.836 ≈ 9.48 cm（保留两位小数）。最后记得取平方根并附上正确单位。Edexcel评分方案很看重清晰写出的代入过程及中间步骤。
8. Using the Cosine Rule to Find a Missing Angle | 用余弦定理求未知角
When all three sides are known, rearrange the Cosine Rule to solve for an angle. For angle A, the formula becomes:
当三边均已知时，将余弦定理变形以求解角度。对于角A，公式为：
cos A = (b² + c² − a²) / (2bc)
mple, if a = 8 cm, b = 6 cm, c = 9 cm, find the largest angle (opposite the longest side c? Actually opposite side a = 8 is not the longest; we'll aim for angle A here). But let' s check if a triangle is acute or obtuse. A negative cosine directly indicates an obtuse angle.
那么角A = cos⁻¹[(b² + c² − a²) / (2bc)]。例如，已知 a = 8 cm，b = 6 cm，c = 9 cm，求最大角（最大边c的对角）。cos C = (a² + b² − c²) / (2ab) = (64 + 36 − 81) / (2×8×6) = 19/96 ≈ 0.1979。于是 C ≈ cos⁻¹(0.1979) ≈ 78.6°。这有助于判断三角形是锐角还是钝角。负的余弦值直接表明是钝角。
errors. Many students lose marks by writing cos A = (a² − b² − c²)/... which is incorrect.
Edexcel IGCSE题目常要求已知三边求三个角度。练习正确的公式变形，并注意括号内的运算顺序以避免符号错误。许多学生因误写为 cos A = (a² − b² − c²)/... 而失分。
9. Deciding Which Rule to Use | 判断使用哪一定理
A common exam pitfall is applying the wrong rule. Use this quick decision guide based on the given information:
考试中常见的陷阱是用错定理。可根据已知信息使用下面的快速决策指南：
• Two angles and any side (AAS or ASA) → use the Sine Rule to find another side.
• 已知两角任一边（AAS或ASA）→ 用正弦定理求另一边。
• cluded angle (SSA) → use the Sine Rule to find an angle, but check for the ambiguous case.
• 已知两边及一个非夹角（SSA）→ 用正弦定理求角，但需检验歧义情况。
• Two sides and the included angle (SAS) → use the Cosine Rule to find the third side.
• 已知两边及其夹角（SAS）→ 用余弦定理求第三边。
• Three sides (SSS) → use the Cosine Rule to find an angle.
• 已知三边（SSS）→ 用余弦定理求角。
Sometimes both rules can be used in succession within the same problem. For instance, after finding a side with the Cosine Rule, you might then use the Sine Rule to find an angle to avoid the ambiguous case when it does not apply. Being flexible and checking the given data before starting is a habit that saves time and reduces errors.
有时在同一个问题中需要先后使用两个定理。例如，用余弦定理求出边后，可能再用正弦定理求角，以避免非歧义情况下的顾虑。解题前先检查已知数据并灵活选择方法，这一习惯能节省时间并减少错误。
10. Area of a Triangle Using Sine | 用正弦求三角形面积
A closely related formula that appears alongside the Sine and Cosine Rules in IGCSE is the area of a triangle given two sides and the included angle:
在IGCSE中，与正弦定理和余弦定理相伴出现的一个紧密相关公式，就是已知两边及其夹角时的三角形面积公式：
Area = ½ ab sin C
ny past exam questions combine this area formula with the Sine Rule or Cosine Rule, first requiring you to find an angle or side before computing the area. Always specify the area unit, and round as required.
同样可以写成 Area = ½ bc sin A 或 Area = ½ ac sin B。当你不知道垂直高度时，这个公式非常有用。例如，一个三角形 a = 12 cm，b = 9 cm，夹角 C = 70°，面积 = ½ × 12 × 9 × sin 70° ≈ 0.5 × 108 × 0.9397 ≈ 50.7 cm²。许多历年真题将这个面积公式与正弦或余弦定理结合考查，往往需要你先求出一个角或一条边，再计算面积。务必注明面积单位，并按要求四舍五入。
11. Worked Exam-style Example | 考试题型精讲
angle R = 42°. (a) Find side r. (b) Find angle P. (c) Calculate the area of the triangle.
在三角形PQR中，p = 15 cm，q = 10 cm，角 R = 42°。(a) 求边r。(b) 求角 P。(c) 计算三角形的面积。
225 + 100 − 300 × 0.7431 ≈ 325 − 222.93 = 102.07. Then r ≈ √102.07 ≈ 10.1 cm (to 1 d.p.).
(a) 已知两边及其夹角 → 余弦定理：r² = p² + q² − 2pq cos R = 225 + 100 − 300 × 0.7431 ≈ 325 − 222.93 = 102.07。于是 r ≈ √102.07 ≈ 10.1 cm（1位小数）。
angle R = 42°, P must be acute because sum of angles would be >180° if obtuse. So P = 83.0°.
(b) 求角P，现在可用正弦定理，因为我们已有边r对角R、边p对角P。用 R / r → sin P = (p sin R) / r = (15 × sin 42°) / 10.103 ≈ (15 × 0.6691) / 10.103 ≈ 0.9927。于是 P ≈ sin⁻¹(0.9927) ≈ 83.0°（1位小数）。补角 180° − 83.0° = 97.0°，但由于角R = 42°，取钝角时内角和将超过180°，故P必须为锐角。所以 P = 83.0°。
(c) Area = ½ p q sin R = ½ × 15 × 10 × sin 42° = 75 × 0.6691 ≈ 50.2 cm² (1 d.p.).
(c) 面积 = ½ p q sin R = ½ × 15 × 10 × sin 42° = 75 × 0.6691 ≈ 50.2 cm²（1位小数）。
This example shows how the rules interlink; mastery of such multi-step problems is crucial for achieving top grades.
此例展示了定理之间的相互关联；掌握这类多步骤题目对取得高分至关重要。
12. Exam Tips and Common Mistakes | 考试技巧与常见错误
Top IGCSE performers pay close attention to the following pitfalls:
表现最佳的IGCSE考生会密切关注以下易错点：
• Forgetting to check the ambiguous case in SSA situations — if the question asks for 'the two possible values', drawing a simple sketch can help visualise the second possible triangle.
• 在SSA情形中忘记检查歧义情况——如果题目要求“给出来两个可能的值”，画一个简图有助于想象第二个可能三角形。
• Using the wrong rearrangement of the Cosine Rule when solving for an angle — always check that you have a minus sign between the two sides on the numerator: (b² + c² − a²).
• 求角时使用了错误的余弦定理变形——务必确认分子中两项之间有减号：(b² + c² − a²)。
• Degrees vs. radians — ensure the calculator is in degree mode at all times, except when the problem explicitly states radians (rare in IGCSE).
• 角度制与弧度制混淆——确保计算器始终处于度数模式，除非题目明确指出使用弧度（IGCSE中罕见）。
• Rounding too early — keep intermediate values in the calculator's memory and only round the final answer as specified. Edexcel is strict about accuracy marks.
• 过早四舍五入——中间结果保留在计算器内存中，仅对最终答案按要求舍入。Edexcel对精度扣分很严格。
• Labelling errors — always label the triangle systematically, especially when a diagram is not given. A consistent labelling (A opposite a, etc.) prevents substitution mistakes.
• 标记错误——始终系统地标记三角形，特别是当题目没有给出图形时。一致的标记（A对a等）可避免代换错误。
By consistently practising these rules and following a structured approach, the Sine and Cosine Rules will become an area of strength rather than anxiety.
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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