The Sine Rule | 正弦定理

📚 The Sine Rule | 正弦定理

The sine rule is a powerful tool for solving any triangle, not just right-angled ones. It links sides and their opposite angles in a simple proportion, forming a cornerstone of A-Level trigonometry and appearing regularly in Edexcel exam questions on triangles, bearings, and practical problems.

正弦定理是解任意三角形的强大工具，不仅限于直角三角形。它将各边长与其对角的正弦值以简单比例相连，是 A-Level 三角学的基石，经常出现在 Edexcel 考试中关于三角形、方位角和实际应用的题目里。

1. What is the Sine Rule? | 什么是正弦定理？

In any triangle, the ratio of a side length to the sine of its opposite angle is constant. This is the essence of the sine rule. It works for acute, obtuse, and even right-angled triangles, giving a unified method to find missing measurements when you have certain known parts.

在任何三角形中，边长与其对角正弦的比值为定值。这就是正弦定理的核心。它适用于锐角、钝角甚至直角三角形，当你已知某些特定元素时，提供了一个统一的求解未知量的方法。

Specifically, if triangle ABC has sides a, b, c opposite angles A, B, C respectively, then a/sin A, b/sin B, and c/sin C are all equal. This equality allows you to set up equations to find an unknown side or angle.

具体来说，如果三角形 ABC 中边 a、b、c 分别对应角 A、B、C，那么 a/sin A、b/sin B 和 c/sin C 三者相等。这个等式能让你建立方程来求未知边或角。

2. Derivation of the Sine Rule | 正弦定理的推导

The sine rule can be derived from the basic definition of sine in a right-angled triangle. Consider triangle ABC and draw the altitude from vertex C to side AB (or its extension), creating two right-angled triangles. Let the height be h.

正弦定理可以由直角三角形中的正弦定义推导得出。考虑三角形 ABC，从顶点 C 向边 AB（或其延长线）作高，形成两个直角三角形。设这条高为 h。

h gives a sin B = b sin A, or a / sin A = b / sin B. Repeating the process with another altitude yields the full proportion including c / sin C.

在含边 a 的直角形中，有 sin B = h / a，故 h = a sin B。在含边 b 的直角形中，sin A = h / b，故 h = b sin A。令这两个 h 表达式相等，得到 a sin B = b sin A，即 a / sin A = b / sin B。用另一条高重复该步骤即可得出包含 c / sin C 的完整比例。

3. The Sine Rule Formula | 正弦定理公式

The standard formula is written as:

a / sin A = b / sin B = c / sin C

标准公式写作：

a / sin A = b / sin B = c / sin C

particularly useful when you are solving for an angle. Both forms are equivalent; choose the one that makes your unknown the numerator.

这也可以重排成 sin A / a = sin B / b = sin C / c 的形式，这在求角度时特别有用。两种形式等价；选择让所求未知量在分子的形式即可。

f the circumcircle. This is less common at A-Level but can appear in challenging problems.

有时你会看到公式扩展为包含外接圆的形式：a / sin A = 2R，其中 R 是外接圆半径。在 A-Level 中较少见，但可能出现在难题中。

4. Using the Sine Rule to Find an Unknown Side | 用正弦定理求未知边

When you know two angles and one side (AAS or ASA), you can find the remaining sides. First, calculate the third angle using the angle sum 180°. Then set up a proportion involving the known side-angle pair and the unknown side with its opposite angle.

当已知两角及其夹边或任意边（AAS 或 ASA），你可以求其余的边。首先用内角和 180° 求第三个角，然后建立一个包含已知边角对以及未知边与其对角的比例式。

a sin B) / sin A. Substitute the values and calculate with your calculator in degree mode.

例如，若 A = 40°，B = 75°，边 a = 12 cm，那么由 a / sin A = b / sin B 得 b = (a sin B) / sin A。代入数值，并在角度模式下用计算器计算。

Always check that your answer is sensible by comparing with the given information — the longest side is opposite the largest angle.

一定要对照已知信息检查答案是否合理——最长的边对角最大的角。

5. Using the Sine Rule to Find an Unknown Angle | 用正弦定理求未知角

the form sin B / b = sin A / a, then solve for sin B and apply the inverse sine function.

要求角度，你需要知道两边以及其中一边的对角（SSA）。使用 sin B / b = sin A / a 的形式，然后解出 sin B，并应用反正弦函数。

: the inverse sine on your calculator gives an acute angle, but there may be a second possibility.

例如，已知 a = 18 cm，b = 14 cm，A = 60°，可写出 sin B / 14 = sin 60° / 18，得 sin B = (14 × sin 60°) / 18。然后 B = sin⁻¹(结果)。注意：计算器上的反正弦给出锐角，但可能存在第二个解。

6. The Ambiguous Case of the Sine Rule | 正弦定理的多解情况

hen you are given two sides and a non-included acute angle (SSA). Because sin θ = sin(180° – ). Both could be valid parts of a triangle, provided the angle sum remains less than 180°.

多解情况（模糊情况）出现在已知两边和一个非夹角的锐角（SSA）时。由于 sin θ = sin(180° – θ)，方程 sin B = 0.8 在 0° 到 180° 范围内可能有两个角度：一个锐角（比如 53.1°）及其补角（126.9°）。只要三角形内角和不超 180°，两者都可能有效。

Key fact: If the known angle A is acute and the side opposite it (a) is shorter than the other given side (b), there may be two solutions. Always test the supplementary angle: check whether A + (180° – B₁) < 180°. If yes, the second triangle exists. If not, only the acute angle works.

关键点：若已知角 A 是锐角，且其对角 a 小于另一已知边 b，则可能有两解。一定要测试补角：判断 A + (180° – B₁) < 180° 是否成立。若成立，第二个三角形存在；若不成立，只有锐角解。

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If a ≥ b, or A is obtuse, the ambiguous case cannot occur — there is at most one triangle.

若 a ≥ b，或 A 为钝角，则不可能出现多解——最多一个三角形。

7. Worked Example: Finding a Side | 例题：求边长

angle P = 42°, angle Q = 78°, and side p = 8.5 cm (opposite P). Find side q (opposite Q).

题目：在三角形 PQR 中，角 P = 42°，角 Q = 78°，边 p = 8.5 cm（对 P）。求边 q（对 Q）。

Solution: First, note that angle R = 180° – sin 42° ≈ 0.6691. Thus q ≈ (8.5 × 0.9781) / 0.6691 ≈ 8.314 / 0.6691 ≈ 12.4 cm (to 3 s.f.).

解答：首先，角 R = 180° – (42° + 78°) = 60°。用正弦定理：p / sin P = q / sin Q → 8.5 / sin 42° = q / sin 78°。整理得 q = (8.5 × sin 78°) / sin 42°。计算：sin 78° ≈ 0.9781，sin 42° ≈ 0.6691。因此 q ≈ (8.5 × 0.9781) / 0.6691 ≈ 8.314 / 0.6691 ≈ 12.4 cm（取三位有效数字）。

8. Worked Example: Angle and Ambiguous Case | 例题：求角度及多解分析

a = 9 cm, b = 13 cm, and angle A = 35°. Find angle B and determine all possible triangles.

题目：三角形 ABC 中，a = 9 cm, b = 13 cm，角 A = 35°。求角 B 并确定所有可能的三角形。

0.8285. Then B₁ = sin⁻¹(0.8285) ≈ 56.0° (to 1 d.p.). The supplementary angle is B₂ = 180° – 56.0° = 124.0°.

解答：由 sin B / b = sin A / a 得 sin B = (b sin A) / a = (13 × sin 35°) / 9。sin 35° ≈ 0.5736，故 sin B ≈ (13 × 0.5736) / 9 = 7.4568 / 9 ≈ 0.8285。于是 B₁ = sin⁻¹(0.8285) ≈ 56.0°（1位小数）。其补角为 B₂ = 180° – 56.0° = 124.0°。

24.0° = 159.0° < 180°, so a second triangle is possible. In this triangle, angle C₂ = 180° – .0°. Both solutions are valid. Notice that because a < b, the ambiguous case was expected.

验证 B₂：A + B₂ = 35° + 124.0° = 159.0° < 180°，因此第二个三角形可能。此三角形中 C₂ = 180° – 159.0° = 21.0°。两个解均有效。注意因为 a < b，发生多解情况符合预期。

9. Connection to Area of a Triangle | 正弦定理与三角形面积

ght = b sin C. Although the sine rule itself does not directly give area, combining it with the area formula can solve problems where you know two sides and the included angle, or find an angle from area and then use the sine rule for sides.

面积公式 面积 = ½ ab sin C 与正弦定理密切相关。它起源于同样的作高构造：面积 = ½ × 底 × 高，其中高 = b sin C。尽管正弦定理本身不直接求面积，但将它与面积公式结合，可解决已知两边及其夹角，或从面积求角度后再用正弦定理求边的问题。

24 = ½ × 10 × 8 × sin C, then use that angle in the sine rule to find the remaining sides.

例如，已知面积 = 24 cm²，a = 10 cm，b = 8 cm，可从 24 = ½ × 10 × 8 × sin C 求出 sin C，然后利用该角通过正弦定理求其余边。

10. Real-world Applications | 实际应用

The sine rule is widely used in navigation, surveying, and engineering. In bearings problems, a triangle is often formed by two journeys from a point or between three points. The sine rule then allows you to find unknown distances or angles when the triangle is not right-angled.

正弦定理在导航、测量和工程中应用广泛。在方位角问题中，从一点出发的两段航程或三点之间常构成三角形。当三角形不是直角时，正弦定理便可用于求未知距离或角度。

For instance, a ship sails 15 km on bearing 045°, then changes course to 120° and sails 20 km. The distance from the starting point can be found by first working out the interior angle between the two legs (75°) and then applying the sine rule or cosine rule, depending on the given information.

例如，一艘船以方位角 045° 航行 15 km，然后转至 120° 航行 20 km。先求出两段航程间的内角（75°），再根据已知信息选用正弦定理或余弦定理求离起点的距离。

11. Common Mistakes | 常见错误

One of the most frequent errors is mismatching sides and angles: always pair each side with the sine of its opposite angle. Another is using the sine rule when you actually need the cosine rule — remember, sine rule requires at least one complete side–angle pair opposite each other.

最常见的错误之一是边角不对应：一定要将每个边与其对角的正确正弦配对。另一个错误是在本应用余弦定理时却用了正弦定理——记住，正弦定理要求至少有一对对边对角完整已知。

Students sometimes forget to check the ambiguous case in SSA situations, leading to lost solutions. Also, ensure your calculator is in degree mode, and avoid rounding intermediate values too early to prevent loss of accuracy.

学生有时会忘记在 SSA 情况下检查多解，从而丢失解。此外，确保计算器处于角度模式，并避免过早舍去中间值，以防精度损失。

12. Exam Tips | 考试技巧

udes your unknown: if finding a side, use a / sin A = b / sin B; if finding an angle, use sin B / b = sin A / a.

在 Edexcel 考试中，首先清晰地标记三角形并写下所有已知信息。将正弦定理写成直接包含所求量的形式：若求边，用 a / sin A = b / sin B；若求角度，用 sin B / b = sin A / a。

Always consider the ambiguous case when given SSA with an acute angle. State clearly “Case 1” and “Case 2” if two triangles exist, and check whether each case satisfies the angle sum. Use sensible accuracy (3 significant figures unless stated otherwise) and state final answers with units.

当已知 SSA 且为锐角时，永远要考虑多解。如果有两个三角形，请清楚地写出“情况1”和“情况2”，并检查每个情况是否满足内角和。使用合理的精度（除非另有说明，通常保留三位有效数字），并在最终答案中标出单位。

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