Cosine Rule | 余弦定理

📚 Cosine Rule | 余弦定理

This revision note covers the cosine rule for Edexcel IGCSE Mathematics. You will learn the two main forms of the rule, when to apply it, how to use it to find a missing side or angle, and how to avoid common exam mistakes.

本篇复习笔记针对 Edexcel IGCSE 数学中的余弦定理。你将学习余弦定理的两种主要形式、何时使用它、如何用它求缺失的边或角，以及如何在考试中避免常见错误。

1. What Is the Cosine Rule? | 什么是余弦定理？

The cosine rule is a formula that works for any triangle, not only right-angled triangles. It connects the lengths of three sides of a triangle to the cosine of one of its angles.

余弦定理是一个适用于任意三角形的公式，而不仅仅是直角三角形。它把一个三角形的三条边长与其中一个角的余弦值联系起来。

The standard labelling convention is important. In triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.

标准标记方法非常重要。在三角形 ABC 中，边 a 是角 A 的对边，边 b 是角 B 的对边，边 c 是角 C 的对边。

For any triangle, the cosine rule can be written in two equivalent forms: one for finding a side, and one for finding an angle.

对于任意三角形，余弦定理可以写成两种等价形式：一种用于求边，另一种用于求角。

2. The Formula – Finding a Side | 公式：求边长

If you know two sides and the angle between them, which is often called the included angle, you can use the cosine rule to find the third side.

如果已知两条边和它们的夹角（通常称为夹边角），可以用余弦定理求出第三边。

a² = b² + c² − 2bc cos A

In this form, angle A is opposite side a, and sides b and c are the two sides that meet at angle A.

在此形式中，角 A 是边 a 的对角，而边 b 和 c 是角 A 的两条邻边。

You can also use the rule for any other vertex. For example, b² = a² + c² − 2ac cos B, and c² = a² + b² − 2ab cos C.

你也可以把公式用于其他顶点。例如，b² = a² + c² − 2ac cos B，以及 c² = a² + b² − 2ab cos C。

To find the missing side, substitute the known values, evaluate the expression, and then take the positive square root.

求缺失边时，先代入已知数值，计算整个表达式，最后取正平方根。

3. The Formula – Finding an Angle | 公式：求角度

If you know all three sides of a triangle, you can rearrange the cosine rule to find any angle.

如果已知三角形的三条边，你可以通过变形余弦定理来求任意一个角。

cos A = (b² + c² − a²) / (2bc)

Once you have substituted the three side lengths, use the inverse cosine function to find angle A.

在代入三条边长之后，使用反余弦函数求出角 A。

A = cos⁻¹( (b² + c² − a²) / (2bc) )

The angle version is very useful in IGCSE questions because it gives a direct way to test whether a triangle has an obtuse angle.

求角版本在 IGCSE 题目中非常有用，因为它能直接判断一个三角形是否含有钝角。

If the value of cos A is negative, then angle A is obtuse. If cos A is zero, then angle A is a right angle.

如果 cos A 的值为负数，则角 A 是钝角；如果 cos A 为零，则角 A 是直角。

4. When to Use the Cosine Rule vs the Sine Rule | 余弦定理与正弦定理的选用

In IGCSE trigonometry questions, you must choose between the sine rule and the cosine rule. The cosine rule is the best choice for the two cases below.

在 IGCSE 三角学题目中，你需要在正弦定理和余弦定理之间作出选择。在下面两种情况下，余弦定理是最佳选择。

Two sides and the included angle (SAS): Use a² = b² + c² − 2bc cos A to find the side opposite the included angle.

两边及其夹角（SAS）： 使用 a² = b² + c² − 2bc cos A 求夹角对边。
Three sides (SSS): Use cos A = (b² + c² − a²) / (2bc) to find any angle.

三边（SSS）： 使用 cos A = (b² + c² − a²) / (2bc) 求任意角。

The sine rule is more suitable when you know two angles and any side, or two sides and a non-included angle. However, the cosine rule is preferable in SAS and SSS situations because it avoids the ambiguous case.

正弦定理更适合已知两角和任意一边，或两边及其中一边的对角的情况。不过在 SAS 和 SSS 情况下，余弦定理更合适，因为它可以避免含糊解的情况。

5. Derivation and Why It Works | 推导与原理

A simple derivation places triangle ABC on a coordinate grid. Let A be at the origin, B be at (c, 0), and C be at (b cos A, b sin A).

一个简单的推导方法是把三角形 ABC 放在坐标系中。设 A 在原点，B 在 (c, 0)，C 在 (b cos A, b sin A)。

The distance between C and B is side a. Using the distance formula, we get the following chain of equations.

C 到 B 的距离就是边 a。使用距离公式，我们得到下面的等式链。

a² = (b cos A − c)² + (b sin A)²

= b² cos²A − 2bc cos A + c² + b² sin²A

= b²(cos²A + sin²A) + c² − 2bc cos A

= b² + c² − 2bc cos A

This derivation uses the identity cos²A + sin²A = 1, which is valid for any angle A.

这个推导使用了恒等式 cos²A + sin²A = 1，该恒等式对任意角 A 都成立。

Understanding the derivation helps you remember why the formula includes the term −2bc cos A and why it reduces to Pythagoras’ theorem when A = 90°.

理解推导过程有助于你记住公式中为什么包含 −2bc cos A 这一项，以及为什么当 A = 90° 时它会退化为勾股定理。

6. Worked Example 1: Finding a Side | 例题1：求边长

In triangle ABC, side AB = 7 cm, side AC = 9 cm, and angle A = 120°. Find the length of side BC.

在三角形 ABC 中，边 AB = 7 cm，边 AC = 9 cm，角 A = 120°。求边 BC 的长度。

Let a = BC, b = AC = 9 cm, and c = AB = 7 cm. Since the angle between b and c is A, use the side version of the cosine rule.

设 a = BC，b = AC = 9 cm，c = AB = 7 cm。因为 b 与 c 之间的夹角是 A，所以使用余弦定理的边形式。

a² = 9² + 7² − 2 × 9 × 7 × cos 120°

Since cos 120° = −0.5, the calculation becomes 81 + 49 − 126 × (−0.5) = 130 + 63 = 193.

因为 cos 120° = −0.5，所以计算为 81 + 49 − 126 × (−0.5) = 130 + 63 = 193。

a = √193 ≈ 13.9 cm

The length of side BC is approximately 13.9 cm.

边 BC 的长度约为 13.9 cm。

7. Worked Example 2: Finding an Angle | 例题2：求角度

Triangle ABC has sides a = 9 cm, b = 8 cm, and c = 7 cm. Find the largest angle.

三角形 ABC 的边为 a = 9 cm，b = 8 cm，c = 7 cm。求最大的角。

The largest angle is opposite the longest side, so it is angle A, opposite side a = 9 cm.

最大的角一定在最长的边对面，所以它是边 a = 9 cm 的对角 A。

cos A = (8² + 7² − 9²) / (2 × 8 × 7)

Substituting the values gives cos A = (64 + 49 − 81) / 112 = 32 / 112 = 2/7.

代入数值得到 cos A = (64 + 49 − 81) / 112 = 32 / 112 = 2/7。

A = cos⁻¹(2/7) ≈ 73.4°

The largest angle is approximately 73.4°.

最大的角约为 73.4°。

8. Worked Example 3: Worded Problem | 例题3：应用题

Two boats leave a harbour at the same time. Boat X travels 12 km on a bearing of 035°, and boat Y travels 15 km on a bearing of 110°. Find the distance between the two boats at the end of their journeys.

两艘船同时离开港口。X 船沿方位角 035° 行驶了 12 km，Y 船沿方位角 110° 行驶了 15 km。求两艘船在行驶结束时的距离。

The bearing difference is 110° − 35° = 75°, so the angle between the two paths is 75°.

方位角差为 110° − 35° = 75°，所以两条航行路线之间的夹角为 75°。

Let d be the distance between the two boats. The two known paths are the sides adjacent to the angle 75°, so use the cosine rule.

设 d 为两艘船之间的距离。两条已知路线是夹角 75° 的两条邻边，因此使用余弦定理。

d² = 12² + 15² − 2 × 12 × 15 × cos 75°

Since cos 75° ≈ 0.2588, we have d² = 144 + 225 − 360 × 0.2588 ≈ 369 − 93.17 = 275.83.

因为 cos 75° ≈ 0.2588，所以 d² = 144 + 225 − 360 × 0.2588 ≈ 369 − 93.17 = 275.83。

d ≈ √275.83 ≈ 16.6 km

The distance between the two boats is approximately 16.6 km.

两艘船之间的距离约为 16.6 km。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Many students lose marks on cosine rule questions because of small but avoidable errors. Check the list below before you move on.

很多学生在余弦定理题目中失分，往往是因为一些细小但可以避免的错误。继续学习之前，请先检查下面的列表。

Using the wrong form of the formula: Use the side form when finding a side and the angle form when finding an angle.

用错公式形式： 求边时用边形式，求角时用角形式。
Not matching sides and angles correctly: Side a must be opposite angle A. Relabel the triangle carefully before substituting.

边与角对应错误： 边 a 必须是角 A 的对边。代入前先仔细标记三角形。
Forgetting to take the square root: If a² is calculated, side a is √a², not a².

忘记开平方： 如果算出 a²，边长 a 是 √a²，而不是 a²。
Calculator in the wrong mode: Always set your calculator to degree mode for IGCSE geometry questions unless a radian measure is given.

计算器模式错误： 在 IGCSE 几何题中，除非题目使用弧度制，否则一定要把计算器设为角度模式。
Ignoring the negative value of cos: When cos C is negative, the angle is obtuse. Do not assume every triangle angle must be acute.

忽略 cos 的负值： 当 cos C 为负数时，角 C 是钝角。不要假设三角形的每个角都是锐角。

10. Exam-Style Practice Questions | 模拟练习

Try these three questions on your own, then check your answers below. They are written in a style similar to Edexcel IGCSE extended paper questions.

请先独立完成下面三个问题，再核对答案。它们的题型与 Edexcel IGCSE 扩展卷题目相似。

Question \| 题目	Answer \| 答案
In triangle ABC, AB = 6 cm, AC = 8 cm, and angle A = 50°. Find BC. 在三角形 ABC 中，AB = 6 cm，AC = 8 cm，角 A = 50°。求 BC。	BC² = 6² + 8² − 2 × 6 × 8 × cos 50° ≈ 38.28 BC ≈ 6.19 cm \| BC ≈ 6.19 cm
Triangle XYZ has sides XY = 9 cm, YZ = 7 cm, and XZ = 6 cm. Find angle Y. 三角形 XYZ 的三条边为 XY = 9 cm，YZ = 7 cm，XZ = 6 cm。求角 Y。	cos Y = (9² + 7² − 6²) / (2 × 9 × 7) = 94 / 126 ≈ 0.7460 Y ≈ 41.8° \| Y ≈ 41.8°
A triangle has sides 5 cm, 7 cm, and 10 cm. Find the largest angle. 一个三角形的三条边为 5 cm、7 cm、10 cm。求最大角。	cos θ = (5² + 7² − 10²) / (2 × 5 × 7) = −26 / 70 θ = cos⁻¹(−0.3714) ≈ 111.8° \| θ = cos⁻¹(−0.3714) ≈ 111.8°

Question | 题目

Answer | 答案

In triangle ABC, AB = 6 cm, AC = 8 cm, and angle A = 50°. Find BC.

在三角形 ABC 中，AB = 6 cm，AC = 8 cm，角 A = 50°。求 BC。

BC² = 6² + 8² − 2 × 6 × 8 × cos 50° ≈ 38.28

BC ≈ 6.19 cm | BC ≈ 6.19 cm

Triangle XYZ has sides XY = 9 cm, YZ = 7 cm, and XZ = 6 cm. Find angle Y.

三角形 XYZ 的三条边为 XY = 9 cm，YZ = 7 cm，XZ = 6 cm。求角 Y。

cos Y = (9² + 7² − 6²) / (2 × 9 × 7) = 94 / 126 ≈ 0.7460

Y ≈ 41.8° | Y ≈ 41.8°

A triangle has sides 5 cm, 7 cm, and 10 cm. Find the largest angle.

一个三角形的三条边为 5 cm、7 cm、10 cm。求最大角。

cos θ = (5² + 7² − 10²) / (2 × 5 × 7) = −26 / 70

θ = cos⁻¹(−0.3714) ≈ 111.8° | θ = cos⁻¹(−0.3714) ≈ 111.8°

11. Summary and Revision Checklist | 总结与复习清单

Use the cosine rule when you know two sides and the included angle, or when you know all three sides. Always label the triangle clearly first.

已知两边及其夹角，或已知三边时，使用余弦定理。一定要先清楚地标记三角形。

For a missing side, use a² = b² + c² − 2bc cos A. For a missing angle, use cos A = (b² + c² − a²) / (2bc).

求缺失边时，使用 a² = b² + c² − 2bc cos A；求缺失角时，使用 cos A = (b² + c² − a²) / (2bc)。

Remember that the longest side is opposite the largest angle, and a negative cosine indicates an obtuse angle.

记住，最长的边对着最大的角，而负的余弦值表示钝角。

Check your calculator mode, take the square root when finding a side, and always round your final answer to a sensible degree of accuracy.

检查计算器模式，求边时记得开平方，并始终把最终答案四舍五入到合理的精确度。

With practice, you will quickly recognise when a cosine rule question appears on the Edexcel IGCSE paper. Keep your working clear and ordered to maximise your marks.

多加练习后，你就能迅速识别 Edexcel IGCSE 试卷中的余弦定理题目。保持步骤清晰有序，才能最大限度得分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Cosine Rule | 余弦定理