The Law of Sines and Solving Triangles | 正弦定理与解三角形

📚 The Law of Sines and Solving Triangles | 正弦定理与解三角形

The Law of Sines is one of the most powerful tools in trigonometry. It connects the sides and angles of any triangle, allowing us to solve for unknown quantities with elegance and precision. This article explores the theorem, its proof, the ambiguous case, and its applications in A-Level mathematics.

正弦定理是三角学中最强大的工具之一。它将任意三角形的边与角联系起来，使我们能够简洁而精确地求解未知量。本文将深入探讨这一定理、其证明、模糊情形以及在A-Level数学中的应用。

1. The Law of Sines | 理解正弦定理

For any triangle ABC with side lengths a, b, c opposite angles A, B, C respectively, the Law of Sines states:

a ⁄ sin A = b ⁄ sin B = c ⁄ sin C = 2R

Here, R is the circumradius — the radius of the circle passing through all three vertices of the triangle. This single equation encapsulates the proportional relationship between each side and the sine of its opposite angle.

在任意三角形ABC中，设边长a、b、c分别对应角A、B、C，则正弦定理表述为：

a ⁄ sin A = b ⁄ sin B = c ⁄ sin C = 2R

其中R为外接圆半径，即经过三角形三个顶点的圆的半径。这唯一等式概括了每条边与其对角正弦之间的比例关系。

2. Proof of the Law of Sines | 正弦定理的证明

The proof begins by constructing the circumcircle of triangle ABC with center O and radius R. From the extended sine rule, the chord a subtends angle A at the circumference, giving the relationship a = 2R sin A. By repeating this reasoning for each side, we obtain the full theorem.

证明始于构造三角形ABC的外接圆，圆心为O、半径为R。根据圆周角定理，弦a在圆周上对应角A，由此得出关系式a = 2R sin A。对每条边重复这一推理，即可得到完整的定理。

An alternative proof uses the altitude from vertex C. Since h = b sin A = a sin B, dividing both sides by sin A sin B yields the equality a ⁄ sin A = b ⁄ sin B. This elegant approach avoids the circumcircle entirely.

另一种证明方法利用顶点C所作的高线。由于h = b sin A = a sin B，将两边同除以sin A sin B即可得到等式a ⁄ sin A = b ⁄ sin B。这种精巧的方法完全无需借助外接圆。

3. Solving Triangles: ASA and AAS | 解三角形：两角一边

When two angles and any one side are known (ASA or AAS), the triangle is uniquely determined. Since the sum of interior angles is always 180°, the third angle follows immediately. Then apply the Law of Sines to find the remaining sides.

当已知两角及任意一边（ASA或AAS）时，三角形被唯一确定。由于内角和恒为180°，第三个角可立即求出。随后应用正弦定理即可求得其余各边。

Find the third angle: C = 180° − A − B
Use a ⁄ sin A = b ⁄ sin B to find side b
Use a ⁄ sin A = c ⁄ sin C to find side c

求第三个角：C = 180° − A − B

利用a ⁄ sin A = b ⁄ sin B求边b

利用a ⁄ sin A = c ⁄ sin C求边c

Worked Example: In triangle ABC, A = 40°, B = 60°, a = 10 cm. First, C = 80°. Then b ⁄ sin 60° = 10 ⁄ sin 40°, giving b ≈ 13.47 cm. Similarly c ≈ 15.32 cm.

例题：在三角形ABC中，A = 40°，B = 60°，a = 10 cm。首先C = 80°。然后根据b ⁄ sin 60° = 10 ⁄ sin 40°，得b ≈ 13.47 cm。同理c ≈ 15.32 cm。

4. The Ambiguous Case (SSA) | 正弦定理的模糊情形

When we are given two sides and a non-included angle (SSA), the situation becomes delicate. There may be zero, one, or two valid triangles. This is famously known as the ambiguous case.

当已知两边及一非夹角（SSA）时，情况变得微妙起来。可能存在零个、一个或两个满足条件的三角形。这就是著名的模糊情形。

Given sides a, b and angle A, we examine the height h = b sin A:

If a < h: no triangle exists
If a = h: exactly one right triangle
If h < a < b: two possible triangles
If a ≥ b: exactly one triangle

已知边a、b及角A时，我们考察高h = b sin A：

若a < h：三角形不存在

若a = h：恰有一个直角三角形

若h < a < b：存在两个可能的三角形

若a ≥ b：恰有一个三角形

Example: Given b = 20, c = 15, B = 40°, we compute h = 15 sin 40° ≈ 9.64. Since 9.64 < 20 < 15 is false — actually check b vs c: here the given side b = 20 is greater than c = 15, so only one triangle exists. However, if the given side were smaller than the other side, two solutions would emerge.

例题：已知b = 20，c = 15，B = 40°，计算得h = 15 sin 40° ≈ 9.64。因为9.64 < 20且20 > 15，即给定边大于另一边，所以仅有一个三角形存在。但若给定边小于另一边，则会出现两个解。

5. Finding Angles Using the Law of Sines | 利用正弦定理求角

When solving for an angle, we must be cautious. The sine function is positive in both the first and second quadrants, so sin θ = sin(180° − θ). This means two angles between 0° and 180° share the same sine value.

在求解角度时必须格外小心。正弦函数在第一和第二象限均为正值，因此sin θ = sin(180° − θ)。这意味着在0°到180°之间有两个角具有相同的正弦值。

To determine which solution is valid, check whether the computed angle plus any known angle exceeds 180°. If it does, reject that solution. Always verify with the triangle sum rule.

要判断哪个解有效，需检查所求角与已知角之和是否超过180°。若超过，则舍去该解。务必用三角形内角和定理进行验证。

Example: In triangle ABC, a = 8, b = 10, A = 50°. Then sin B = 10 sin 50° ⁄ 8 ≈ 0.9575, so B ≈ 73.2° or 106.8°. If C = 180° − 50° − 73.2° = 56.8°, valid. If B = 106.8°, then C = 23.2°, also valid! Both triangles exist.

例题：在三角形ABC中，a = 8，b = 10，A = 50°。则sin B = 10 sin 50° ⁄ 8 ≈ 0.9575，故B ≈ 73.2°或106.8°。若取B = 73.2°，则C = 56.8°，有效；若取B = 106.8°，则C = 23.2°，同样有效！两个三角形均存在。

6. Area of a Triangle | 三角形面积公式

The Law of Sines leads directly to a beautiful area formula. For any triangle, the area K is given by:

K = ½ ab sin C = ½ bc sin A = ½ ca sin B

This formula requires only two sides and the included angle. It is particularly useful when altitude is difficult to compute directly.

正弦定理直接引出一个优美的面积公式。任意三角形的面积K为：

K = ½ ab sin C = ½ bc sin A = ½ ca sin B

该公式只需两边及其夹角。在难以直接计算高线时尤为实用。

Example: Two sides of a triangle are 6 cm and 9 cm with an included angle of 35°. The area is K = ½ × 6 × 9 × sin 35° ≈ 15.49 cm².

例题：三角形两边长分别为6 cm和9 cm，夹角为35°。其面积为K = ½ × 6 × 9 × sin 35° ≈ 15.49 cm²。

7. The Circumradius Formula | 外接圆半径公式

The extended form of the Law of Sines states that each ratio equals 2R, where R is the circumradius. This gives us a direct method to find R:

R = a ⁄ (2 sin A)

This result connects trigonometry with circle geometry. If a side and its opposite angle are known, the circumradius is immediately determined.

正弦定理的扩展形式表明每个比值均等于2R，其中R为外接圆半径。这给了我们直接求R的方法：

R = a ⁄ (2 sin A)

这一结论将三角学与圆几何联系起来。若已知一边及其对角，外接圆半径即可立即求得。

Example: In triangle ABC, a = 14 cm, A = 70°. Then R = 14 ⁄ (2 sin 70°) ≈ 7.45 cm.

例题：在三角形ABC中，a = 14 cm，A = 70°。则R = 14 ⁄ (2 sin 70°) ≈ 7.45 cm。

8. Real-World Applications | 实际应用

The Law of Sines is indispensable in surveying, navigation, and astronomy. Surveyors use it to measure distances across rivers or lakes without crossing them. Navigators apply it to determine positions from celestial angles.

正弦定理在测量、航海和天文学中不可或缺。测量员利用它测量无法直接跨越的河流或湖泊的距离。航海员利用天体角度确定位置。

Surveying Example: To measure the distance across a river, a surveyor stands at point A and measures angle A to two reference points B and C on the opposite bank. If AB = 50 m, AC = 40 m, and angle A = 60°, the distance BC is not directly found by sine law (since BC is opposite A, but we need the included angle) — instead we apply the cosine law: BC² = AB² + AC² − 2(AB)(AC)cos A ≈ 2100, so BC ≈ 45.8 m.

测量实例：为测量河流宽度，测量员站在A点，测量到对岸两个参考点B和C的角度。若AB = 50 m，AC = 40 m，角A = 60°，则距离BC不能直接用正弦定理求得（BC面对角A，但缺少夹角），而应使用余弦定理：BC² = AB² + AC² − 2(AB)(AC)cos A ≈ 2100，故BC ≈ 45.8 m。

This example reminds us to choose the correct tool: the sine law for known angles and opposite sides, the cosine law for two sides and an included angle.

此例提醒我们选择正确的工具：已知角及其对边时用正弦定理，已知两边及夹角时用余弦定理。

9. Common Mistakes | 常见错误

Students often misapply the Law of Sines in several ways. Being aware of these pitfalls can save precious marks in examinations.

学生在应用正弦定理时常犯几类错误。了解这些陷阱可以在考试中节省宝贵的分数。

Mismatching sides and angles: The side a must be opposite angle A, not adjacent to it.
Ignoring the ambiguous case: Always check whether a second solution exists for SSA.
Using degrees instead of radians or vice versa without proper calculator settings.
Forgetting to verify angle sums: Reject any angle combination exceeding 180°.

错配边角：边a必须与角A相对，而非相邻。

忽略模糊情形：务必检查SSA情形是否存在第二个解。

混淆角度单位：使用度数或弧度时必须确保计算器设置一致。

忘记验证角度和：任何超过180°的角度组合都应舍去。

10. Worked Exam Question | 典型考题解析

Let us solve a complete problem step by step, the way you should present it in an exam.

让我们逐步完整求解一道题，按照考试中应有的格式书写。

Question: In triangle PQR, PQ = 18 cm, QR = 12 cm, and angle P = 35°. Find angle R, giving your answer to one decimal place.

题目：在三角形PQR中，PQ = 18 cm，QR = 12 cm，角P = 35°。求角R，精确到一位小数。

Solution: Label sides: r = QR = 12 cm (opposite R), p = PQ = 18 cm (opposite P). By the Law of Sines:

r ⁄ sin R = p ⁄ sin P

Substituting: 12 ⁄ sin R = 18 ⁄ sin 35°. Thus sin R = 12 sin 35° ⁄ 18 ≈ 0.3824. Therefore R ≈ 22.5° or 157.5°. Since angle P = 35°, the angle sum gives 35° + 157.5° = 192.5° > 180°, so the obtuse solution is rejected. Hence R ≈ 22.5°.

解答：标记边：r = QR = 12 cm（对角R），p = PQ = 18 cm（对角P）。根据正弦定理：

r ⁄ sin R = p ⁄ sin P

代入：12 ⁄ sin R = 18 ⁄ sin 35°。故sin R = 12 sin 35° ⁄ 18 ≈ 0.3824。因此R ≈ 22.5°或157.5°。由于角P = 35°，内角和35° + 157.5° = 192.5° > 180°，故舍去钝角解。因此R ≈ 22.5°。

11. Practice Problems | 练习与巩固

Attempt these problems independently, then check your answers below.

请独立完成以下练习，然后对照下方答案检查。

Problem 1	In triangle ABC, A = 48°, B = 73°, c = 21 cm. Find side a.
Problem 2	In triangle XYZ, x = 9, y = 11, X = 50°. How many triangles exist?
Problem 3	Find the area of a triangle with sides 8 cm and 14 cm and included angle 110°.

Answers: 1) a ≈ 15.93 cm 2) Two triangles 3) K ≈ 52.58 cm²

答案：1) a ≈ 15.93 cm 2) 两个三角形 3) K ≈ 52.58 cm²

12. Summary | 要点总结

The Law of Sines is a cornerstone of triangle trigonometry. Master its statement, understand its ambiguous case, and choose it wisely alongside the Cosine Law. With consistent practice, solving triangles becomes second nature.

正弦定理是三角形三角学的基石。掌握其公式、理解其模糊情形，并善于与余弦定理配合使用。通过持续练习，解三角形定将成为得心应手的技能。

Situation	Tool to Use
Two angles and one side	Law of Sines (unique solution)
Two sides and a non-included angle	Law of Sines (check ambiguous case)
Two sides and the included angle	Cosine Law
Three sides	Cosine Law

情形：两角一边 → 使用正弦定理（唯一解）

情形：两边及一非夹角 → 使用正弦定理（检查模糊情形）

情形：两边及夹角 → 使用余弦定理

情形：三边 → 使用余弦定理

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The Law of Sines and Solving Triangles | 正弦定理与解三角形