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Solving Non-Right Triangles: Sine Rule and Cosine Rule | 非直角三角形求解:正弦定理与余弦定理

2026-08-30 15:54 · 本文部分内容由AI辅助创作

📚 Solving Non-Right Triangles: Sine Rule and Cosine Rule | 非直角三角形求解:正弦定理与余弦定理

For right-angled triangles, the trigonometric ratios sine, cosine and tangent are sufficient to find unknown sides and angles. But what happens when a triangle has no right angle? We need two powerful tools — the Sine Rule and the Cosine Rule — which extend trigonometry to any triangle.

对于直角三角形,正弦、余弦和正切这三种三角比足以求出未知的边和角。但如果三角形没有直角该怎么办?我们需要两个强大的工具——正弦定理和余弦定理——它们将三角学推广到了任意三角形。


1. When Do We Need the Sine and Cosine Rules? | 何时需要正弦定理与余弦定理

Any triangle can be solved (i.e., all sides and angles found) if we know three of its six parts, with at least one side. The choice of rule depends on which three parts are given.

任何三角形,只要已知六个要素中的三个(其中至少有一条边),就可以求解出全部边和角。选择哪个定理取决于已知的是哪三个要素。

In short: the Sine Rule works with angle–side pairs; the Cosine Rule works with side-side-side or side-angle-side.

简而言之:正弦定理处理“对角对边”的组合;余弦定理处理“边边边”或“边角边”的组合。


2. The Sine Rule (Law of Sines) | 正弦定理

For any triangle ABC with sides a, b, c opposite angles A, B, C respectively:

对于任意三角形 ABC,边 a、b、c 分别对角 A、B、C:

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

This elegant relation says that the ratio of a side to the sine of its opposite angle is constant throughout the triangle.

这一简洁的关系说明:三角形中每条边与其对角正弦之比在整个三角形中保持不变。

Equivalently, we may write it in reciprocal form:

等价地,也可以写成倒数形式:

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

The Sine Rule is used when we know:

正弦定理适用于已知:


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

The proof uses only the definition of sine in right triangles formed by an altitude.

推导只需利用“作高”得到的直角三角形中正弦的定义。

Draw altitude h from C to side AB (or its extension). In the right triangles formed:

从顶点 C 向 AB 边(或其延长线)作高 h。在形成的两个直角三角形中:

sin A = h / b , sin B = h / a

/ sin B. Similarly, repeating the argument with another altitude gives the full statement.

因此 h = b sin A = a sin B,从而得到 a / sin A = b / sin B。类似地,换一条高重复论证,便得到完整的定理。

fall outside the triangle, but the sine ratio remains valid because sin(180° − θ) = sin θ.

该推导还说明了为什么钝角的正弦也适用:高可能落在三角形外部,但由于 sin(180° − θ) = sin θ,正弦比值依然成立。


4. The Cosine Rule (Law of Cosines) | 余弦定理

For any triangle ABC:

对于任意三角形 ABC:

a² = b² + c² − 2bc · cos A

b² = a² + c² − 2ac · cos B

c² = a² + b² − 2ab · cos C

Notice the pattern: each side squared equals the sum of the squares of the other two sides, minus twice their product times the cosine of the included angle.

注意规律:任意一边的平方,等于另外两边平方之和,减去这两边乘积的 2 倍再乘以它们夹角的余弦。

The Cosine Rule is used when we know:

余弦定理适用于已知:

When finding an angle from three sides, rearrange:

当用三边求角时,可变形为:

cos A = (b² + c² − a²) / (2bc)


5. Derivation of the Cosine Rule | 余弦定理的推导

Place triangle ABC with C at the origin, side b along the x-axis. Then B has coordinates (a, 0) and A has coordinates (b cos C, b sin C).

将三角形 ABC 的顶点 C 放在原点,边 b 沿 x 轴方向。于是 B 的坐标为 (a, 0),A 的坐标为 (b cos C, b sin C)。

The distance between A and B is c. Using the distance formula:

A 与 B 之间的距离为 c。利用距离公式:

c² = (b cos C − a)² + (b sin C)²

Expanding:

展开:

c² = b² cos² C − 2ab cos C + a² + b² sin² C

Since cos² C + sin² C = 1:

因为 cos² C + sin² C = 1:

c² = a² + b² − 2ab cos C

te proof is concise and also shows the Pythagorean theorem as a special case when C = 90°.

这一简洁的坐标法证明同时揭示:当 C = 90° 时,余弦定理就退化为勾股定理。


6. Applying the Sine Rule: Worked Example | 正弦定理的应用:例题

Example 1: In triangle ABC, angle A = 48°, angle B = 65°, and side a = 12 cm. Find side b.

例 1:在三角形 ABC 中,A = 48°,B = 65°,边 a = 12 cm。求边 b。

First find the missing angle: C = 180° − 48° − 65° = 67°. Using the Sine Rule:

先求第三个角:C = 180° − 48° − 65° = 67°。使用正弦定理:

a / sin A = b / sin B

12 / sin 48° = b / sin 65°

b = 12 × sin 65° / sin 48° ≈ 12 × 0.9063 / 0.7431 ≈ 14.6 cm

Notice that we used only the relevant pair of ratios; we did not need angle C to find b, but it is often useful for subsequent steps.

注意我们只用了相关的两对比例;求 b 并不需要 C,但在后续步骤中通常有用。


7. The Ambiguous Case (SSA) | 两边一对角(SSA)的歧义情况

When we know two sides and an angle opposite one of them, the Sine Rule may give zero, one, or two possible triangles.

当已知两边及其中一边的对角时,正弦定理可能给出零个、一个或两个可能的三角形。

This is known as the ambiguous case. For example, in triangle ABC, suppose we know a, b and angle A.

这就是所谓的“歧义情况”。例如在三角形 ABC 中,已知 a、b 和角 A。

Solving sin B = b sin A / a, we may find:

通过解 sin B = b sin A / a,我们可能得到:

Example 2: Given a = 7 cm, b = 9 cm, and angle A = 35°, find angle B.

例 2:已知 a = 7 cm,b = 9 cm,A = 35°,求角 B。

sin B = b sin A / a = 9 sin 35° / 7 ≈ 9 × 0.5736 / 7 ≈ 0.7375

ssible because C = 180° − 35° − 47.5° = 97.5° > 0 and C = 180° − 35° − 132.5° = 12.5° > 0.

因此 B₁ ≈ 47.5°,B₂ = 180° − 47.5° = 132.5°。两者都可行,因为对应角 C 分别为 97.5° 和 12.5°,均大于 0°。

Always check whether the obtuse solution is geometrically possible before accepting both.

在同时接受两个解之前,务必检查钝角解在几何上是否可行。


8. Applying the Cosine Rule: Worked Examples | 余弦定理的应用:例题

Example 3 (SAS): In triangle ABC, b = 8 cm, c = 11 cm, and angle A = 52°. Find side a.

例 3(边角边):在三角形 ABC 中,b = 8 cm,c = 11 cm,A = 52°。求边 a。

a² = b² + c² − 2bc cos A = 8² + 11² − 2 × 8 × 11 × cos 52°

a² = 64 + 121 − 176 × 0.6157 ≈ 185 − 108.36 ≈ 76.64

a ≈ √76.64 ≈ 8.75 cm

Example 4 (SSS): A triangle has sides 6 cm, 7 cm, 8 cm. Find the largest angle (opposite the side of 8 cm).

例 4(边边边):三角形的三边长为 6 cm、7 cm、8 cm。求最大角(对着 8 cm 的边)。

cos C = (a² + b² − c²) / (2ab) = (6² + 7² − 8²) / (2 × 6 × 7)

cos C = (36 + 49 − 64) / 84 = 21 / 84 = 0.25

C = arccos(0.25) ≈ 75.5°

The Cosine Rule is particularly suited for finding angles when all three sides are known, because the sine rule would require an extra step to avoid ambiguity.

当三边已知时,余弦定理特别适合求角,因为用正弦定理求角需要额外步骤并警惕两解歧义。


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9. Area of a Non-Right Triangle | 非直角三角形的面积

Closely related to these rules is the area formula:

与这两个定理密切相关的是面积公式:

Area = ½ · ab · sin C =12ab sin C

Area = ½ × a × b × sin C

In words: the area of any triangle equals half the product of two sides times the sine of the included angle.

也就是说:任意三角形的面积等于两边之积的一半,再乘以它们夹角的正弦。

Example 5: Two sides of a triangle are 10 cm and 14 cm with included angle 40°. Find the area.

例 5:三角形的两边为 10 cm 和 14 cm,夹角为 40°。求面积。

Area = ½ × 10 × 14 × sin 40° = 70 × 0.6428 ≈ 45.0 cm²

This formula is useful in mechanics, physics and geometry when two sides and an included angle are easier to measure than a perpendicular height.

当两边及其夹角比垂线高度更容易测量时,这个公式在力学、物理和几何中都很有用。


10. Choosing the Right Rule: A Decision Strategy | 选择正确的定理:决策策略

The table below summarizes which rule to use depending on the given information.

下表总结了根据已知信息应选择哪个定理。

已知条件 Given 首选定理 First Choice 备注 Remark
两角一边 AAS / ASA 正弦定理 Sine Rule 先求第三个角,再求其他边
Find third angle first, then sides
两边一对角 SSA 正弦定理 Sine Rule ⚠ 警惕两解 ambiguity risk
两边及其夹角 SAS 余弦定理 Cosine Rule 直接求第三边 find third side directly
三边 SSS 余弦定理 Cosine Rule 先求最大角 avoid ambiguity

When in doubt, remember the triangle angle sum: A + B + C = 180°. This alone often decides whether a second solution is valid.

若犹豫不决,请记住三角形的内角和:A + B + C = 180°。仅这一条就常能判定第二个解是否有效。


11. Common Mistakes and Pitfalls | 常见错误与易错点


12. Complete Worked Example: Solving a Whole Triangle | 综合例题:求解整个三角形

Example 6: In triangle PQR, p = 15 cm, q = 20 cm, and angle P = 30°. Solve the triangle fully.

例 6:在三角形 PQR 中,p = 15 cm,q = 20 cm,角 P = 30°。完整地解这个三角形。

Step 1: Use the Sine Rule to find angle Q.

第一步:用正弦定理求角 Q。

sin Q = q sin P / p = 20 sin 30° / 15 = 20 × 0.5 / 15 ≈ 0.6667

8.2° gives R = 180° − 30° − 138.2° = 11.8°, which is valid. So two triangles are possible.

Q₁ ≈ 41.8°。另一个解 Q₂ = 180° − 41.8° = 138.2°,相应的 R = 11.8°,同样有效。因此这里有两个可能的三角形。

Step 2: For each case, find side r using the Sine Rule.

第二步:在每种情形下,用正弦定理求边 r。

Case 1: Q = 41.8°, R = 180° − 30° − 41.8° = 108.2°:

情形 1:Q = 41.8°,R = 108.2°:

r = p sin R / sin P = 15 × sin 108.2° / sin 30° ≈ 15 × 0.9500 / 0.5 ≈ 28.5 cm

Case 2: Q = 138.2°, R = 11.8°:

情形 2:Q = 138.2°,R = 11.8°:

r = 15 × sin 11.8° / sin 30° ≈ 15 × 0.2045 / 0.5 ≈ 6.1 cm

Step 3: Verify with the angle sum and, if possible, the Cosine Rule on one of the results.

第三步:用内角和验证;如有可能,再用余弦定理验证其中一个结果。

This example demonstrates exactly why the SSA case is called ambiguous: the given data can describe two different triangles.

这个例子精确说明了为什么 SSA 情形被称为“歧义”:同样的数据可能描述两个不同的三角形。


13. Summary and Examination Tips | 总结与考试建议

Mastering the Sine Rule and Cosine Rule gives you a complete toolkit for solving any triangle, including those with obtuse angles.

熟练掌握正弦定理与余弦定理,你便拥有了求解任何三角形(包括含钝角的三角形)的完整工具包。

With practice, deciding between the Sine Rule and the Cosine Rule becomes automatic, and the ambiguous case loses its mystery.

多加练习后,在正弦定理与余弦定理之间做选择会变成一种本能,两解情况也不再神秘。


Published by TutorHao | Mathematics Revision Series | aleveler.com

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