📚 Using the Sine and Cosine Rules | 正弦定理与余弦定理的应用
When a triangle does not contain a right angle, the familiar trigonometric ratios of SOH CAH TOA are no longer directly applicable. Instead, we turn to the sine and cosine rules, two powerful theorems that allow us to solve for unknown sides and angles in any triangle. These rules form the backbone of triangle trigonometry in the IB curriculum, enabling students to tackle problems ranging from pure geometry to real-world navigation and surveying.
当一个三角形不包含直角时，我们熟悉的 SOH CAH TOA 三角比就不能直接使用了。此时，我们需要用到正弦定理和余弦定理——这两个强大的定理能帮助我们求解任意三角形中的未知边和角。在 IB 课程中，这些规则是三角形三角学的核心，让学生能够处理从纯几何到现实世界中的导航和测量等各种问题。
1. Introduction to Non-Right-Angled Triangles | 非直角三角形简介
Any triangle that is not right-angled is referred to as an oblique triangle. To solve such a triangle means to find all its side lengths and interior angles, given a sufficient set of information. Unlike right-angled triangles, where SOH CAH TOA suffices, oblique triangles require either the law of sines (sine rule) or the law of cosines (cosine rule), or a combination of both. The choice of rule depends on what is known about the triangle: side-side-side (SSS), side-angle-side (SAS), angle-side-angle (ASA), angle-angle-side (AAS), or the ambiguous side-side-angle (SSA).
任何非直角三角形都称为斜三角形。解斜三角形意味着，在给定足够信息的条件下，求出所有边长和内角。与只需 SOH CAH TOA 的直角三角形不同，斜三角形需要使用正弦定理、余弦定理，或两者结合。选择哪个定理取决于已知条件：边边边 (SSS)、边角边 (SAS)、角边角 (ASA)、角角边 (AAS)，或存在歧义的边边角 (SSA)。
2. The Sine Rule | 正弦定理
The sine rule states that in any triangle ABC, with sides a opposite angle A, b opposite B, and c opposite C, the ratio of a side length to the sine of its opposite angle is constant. The formula is:
正弦定理指出，在任意三角形 ABC 中，边 a 对角 A、边 b 对角 B、边 c 对角 C，边长与其对角的正弦值之比是一个常数。公式如下：
a / sin A = b / sin B = c / sin C
en more convenient when solving for angles. The constant value equals the diameter of the circumscribed circle of the triangle, 2R. This relationship is particularly useful when we are given two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).
等价地，也可写成 sin A / a = sin B / b = sin C / c，在求角度时往往更方便。这个常数值等于三角形外接圆的直径 2R。当我们已知两角一边 (AAS 或 ASA) 或两边及一个非夹角 (SSA) 时，这个关系式尤其有用。
3. Proof of the Sine Rule | 正弦定理的证明
viding by b sin A sin C. Another geometric proof drops an altitude from a vertex to the opposite side and uses the definition of sine in the resulting right triangles. Both proofs highlight the rule's generality.
一个巧妙的证明利用了三角形的面积。面积可用三种方式表达：面积 = ½ ab sin C = ½ bc sin A = ½ ac sin B。将前两个表达式相等得到 ab sin C = bc sin A，除以 b sin A sin C 后化简即得 a / sin A = c / sin C。另一种几何证明是从一个顶点向对边作高，并在形成的直角三角形中使用正弦定义。两种证明都彰显了该定理的普适性。
4. Using the Sine Rule: Finding Sides | 使用正弦定理求边长
missing sides. For example, suppose A = 35°, B = 72°, and side a = 12 cm. First, C = 180° – 35° – = 12 × sin 73° / sin 35° ≈ 20.1 cm. This process is straightforward and avoids ambiguity.
若已知两角一边 (AAS 或 ASA)，可先用三角形内角和为 180° 求出第三个角。然后直接应用正弦定理求出未知边长。例如，设 A = 35°，B = 72°，边 a = 12 cm。首先，C = 180° – 35° – 72° = 73°。由 a / sin A = b / sin B 得 b = 12 × sin 72° / sin 35° ≈ 19.8 cm。类似地，c = 12 × sin 73° / sin 35° ≈ 20.1 cm。这一过程直接明了，没有歧义。
5. Using the Sine Rule: Finding Angles | 使用正弦定理求角度
= 10. Then sin B = 10 × sin 40° / 8 ≈ 0.8035, giving B ≈ 53.5°. However, because sin (180° – θ) = sin θ, there is a potential second solution B' s case, which must be checked for validity by ensuring the angle sum does not exceed 180°.
若已知两边及一个非夹角 (SSA)，可利用正弦定理求出一个角。写出比例式 sin B / b = sin A / a，解出 sin B。例如，已知 A = 40°，a = 8，b = 10，则 sin B = 10 × sin 40° / 8 ≈ 0.8035，得 B ≈ 53.5°。但由于 sin (180° – θ) = sin θ，可能存在第二个解 B' ≈ 126.5°。这就引出了著名的歧义情况，必须检验角之和是否超过 180°，以确定解的有效性。
6. The Ambiguous Case (SSA) | 歧义情况 (SSA)
is shorter than the other given side, yet longer than the altitude from the unknown angle' e obtuse given angle A ≥ 90°, only one triangle is possible when a > b, and none if a ≤ b.
当已知角为锐角且该角的对边比另一已知边短，但又大于从未知角顶点所作的高时，就会出现歧义情况。数学上，设 h = b sin A，则：若 a < h，无解；若 a = h，得到一个直角三角形；若 h < a < b，可能有两个不同的三角形（锐角解和钝角解）；若 a ≥ b，只有一个三角形（钝角解不可能）。若已知角 A ≥ 90°，则仅当 a > b 时存在一个三角形，若 a ≤ b 则无解。
7. The Cosine Rule | 余弦定理
The cosine rule is a generalization of Pythagoras' theorem for any triangle. It relates the lengths of three sides to the cosine of one of its angles. The three symmetrical formulas are:
余弦定理是勾股定理在任意三角形中的推广。它将三条边长与其中一个角的余弦联系起来。三个对称的公式如下：
a² = b² + c² – 2bc cos A
b² = a² + c² – 2ac cos B
c² = a² + b² – 2ab cos C
These formulas allow us to find the third side when two sides and the included angle (SAS) are known, or to find any angle when all three sides (SSS) are given. Note that when the angle is 90°, the cosine term vanishes and the rule reduces to Pythagoras' theorem.
这些公式让我们能够在已知两边及其夹角 (SAS) 时求出第三边，或在已知三边 (SSS) 时求出任意一个角。注意，当角为 90° 时，余弦项消失，定理就退化为勾股定理。
8. Proof of the Cosine Rule | 余弦定理的证明
laces triangle ABC on a coordinate plane with point C at the origin and side a along the x‑ tes of B are (a, 0) and A are (b cos C, b sin C). The distance formula gives c² = (b cos C – a)² + (b sin C – 0)² = b² cos² C + a² – 2ab cos C + b² sin² C = a² + b² (cos² C + sin² C) – 2ab cos C = a² + b² – , and trigonometry. An alternative proof using the altitude of a triangle works similarly.
标准证明方法是将三角形 ABC 放在坐标平面上，令点 C 在原点，边 a 沿 x 轴方向。则 B 点坐标为 (a, 0)，A 点坐标为 (b cos C, b sin C)。由距离公式得 c² = (b cos C – a)² + (b sin C – 0)² = b² cos² C + a² – 2ab cos C + b² sin² C = a² + b² (cos² C + sin² C) – 2ab cos C = a² + b² – 2ab cos C。这一证明巧妙地将代数、几何与三角学联系起来。利用三角形高的另一种证明方法也能得到相同结果。
9. Using the Cosine Rule: Finding Sides | 使用余弦定理求边长
e third side directly. For example, let b = 5 cm, c = 7 cm, and A = 60°. Then a² = 5² + 7² – 2 × 5 × 7 × cos 60° = 25 + 49 – 70 × 0.5 = 74 – 35 = 39, so a = √39 d yields a unique solution, as the triangle is uniquely determined by SAS. Always ensure the angle used is between the two known sides.
当已知两边及其夹角 (SAS) 时，余弦定理可直接求出第三边。例如，设 b = 5 cm，c = 7 cm，A = 60°，则 a² = 5² + 7² – 2 × 5 × 7 × cos 60° = 25 + 49 – 70 × 0.5 = 74 – 35 = 39，故 a = √39 ≈ 6.24 cm。这种方法直接且给出唯一解，因为 SAS 唯一确定了一个三角形。务必确保所用的角是两条已知边的夹角。
10. Using the Cosine Rule: Finding Angles | 使用余弦定理求角度
known, the cosine rule is rearranged to solve for an angle. For instance, cos A = (b² + c² – a²) / (2bc). Suppose a triangle has sides a = 8, b = 9, and c = 11. Then cos A = (9² + 11² – 8²) / (2 × 9 × 11) = (81 + 121 – ase entirely because the cosine function uniquely determines an angle between 0° and 180°.
当三边 (SSS) 已知时，可将余弦定理变形以求角度。例如，cos A = (b² + c² – a²) / (2bc)。假设三角形边长为 a = 8，b = 9，c = 11。则 cos A = (9² + 11² – 8²) / (2 × 9 × 11) = (81 + 121 – 64) / 198 = 138 / 198 ≈ 0.69697，得 A ≈ 45.8°。类似地可求出其他角。这种方法完全避免了歧义情况，因为余弦函数在 0° 到 180° 之间能唯一确定一个角。
11. Area of a Triangle Using Sine | 用正弦求三角形面积
2°, then Area = ½ × 6.2 × 8.5 × sin 72° ≈ 25.1 m². In IB problems, this formula frequently appears in conjunction with the sine and cosine rules to compute areas of composite figures.
任意三角形的面积可以不通过高，而利用两边及其夹角来计算。公式为 面积 = ½ ab sin C，其中 a 和 b 是两条边，C 是它们的夹角。当高不直接已知时，这尤其有用。例如，若 a = 6.2 m，b = 8.5 m，C = 72°，则面积 = ½ × 6.2 × 8.5 × sin 72° ≈ 25.1 m²。在 IB 考题中，该公式常与正弦和余弦定理一起使用，来计算组合图形的面积。
12. Combined Applications and Problem-Solving | 综合应用与解题技巧
ded angle of 100° (the difference in bearings), then apply the cosine rule: d² = 15² + 20² – nformation, deciding which rule to apply, and checking for ambiguous cases when using SSA.
现实情境往往需要同时使用这两个定理。例如，在导航中，一艘船以 040° 的方位角航行 15 km，再以 120° 的方位角航行 20 km。要求出起点到终点的直线距离，我们构建一个两边为 15 和 20、夹角为 100°（方位角之差）的三角形，然后用余弦定理：d² = 15² + 20² – 2×15×20×cos 100°。在更复杂的情境中，如测量学中的三角测量，我们会把图形分解成多个子三角形，角度用正弦定理求解，边长用余弦定理求解。成功解题的关键在于：仔细识别已知信息，决定应用哪个定理，并在使用 SSA 时检验歧义情况。
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