📚 B9i Part 7: Trigonometric Ratios, Sine Rule and Cosine Rule | B9i 第7部分：三角函数比、正弦定理与余弦定理
In this B9i Part 7 revision note, we bring together the core trigonometric tools needed for IGCSE and A Level foundation work: the basic sine, cosine and tangent ratios, exact values, the sine and cosine rules, and the area of a triangle. These methods appear in both non-calculator and calculator papers, often hidden inside bearings and multi-step problems, and they are also the groundwork for later topics such as radian measure and trigonometric identities.
在 B9i 第7部分复习讲义中，我们汇总 IGCSE 和 A Level 基础阶段需要的核心三角工具：正弦、余弦与正切比，精确值，正弦定理与余弦定理，以及三角形面积公式。这些方法常出现在非计算器和计算器卷中，并经常隐藏在方位角和多步问题里，同时也是后续弧度制、三角恒等式等内容的必要铺垫。
1. Trigonometric Ratios in Right-Angled Triangles | 直角三角形中的三角比
Start every trigonometry question by labelling the triangle. The hypotenuse is always the longest side and it always lies opposite the right angle. Relative to a chosen angle θ, the side that faces θ is the opposite side, and the side that touches θ (but is not the hypotenuse) is the adjacent side.
每道三角题的起点都是给三角形标注名称。斜边永远是直角三角形中最长的边，且始终位于直角对面。相对于选定的角 θ，正对着 θ 的那条边称为对边，与 θ 相接（但不是斜边）的那条边称为邻边。
to the two sides it needs, in order: Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent.
标注完成后，三个基本三角比可以直接由定义写出。记忆口诀 SOH CAH TOA 把每个比与其所需的两条边一一对应：SOH 表示正弦等于对边比斜边，CAH 表示余弦等于邻边比斜边，TOA 表示正切等于对边比邻边。
tenuse cos θ = adjacent / hypotenuse tan θ = opposite / adjacent
ther. Second, tan θ = sin θ / cos θ, which is the definition of tangent in terms of the other two ratios and is used constantly in later identity work.
有两个辅助恒等式值得记住。第一，sin θ = cos(90° − θ)，因为一个锐角的对边正好是另一个锐角的邻边。第二，tan θ = sin θ / cos θ，这既是正切用另外两个比给出的定义，也是后续恒等式变形中反复使用的工具。
2. Exact Values for 30°, 45° and 60° | 30°、45° 和 60° 的精确值
Non-calculator papers require exact values rather than decimals. For 30° and 60°, start from an equilateral triangle of side 2 and cut it in half with a perpendicular bisector. The right-angled triangle you obtain has a hypotenuse of 2, a short leg of 1, and a third side of √3, found from Pythagoras: 2² − 1² = 3.
非计算器卷要求给出精确值而不是小数。30° 和 60° 可以从边长为 2 的等边三角形出发，用中垂线切成两半。得到的小直角三角形斜边为 2，短直角边为 1，第三条边由勾股定理求得：2² − 1² = 3，所以这条边等于 √3。
h both legs equal to 1. Pythagoras gives the hypotenuse: 1² + 1² = 2, so the hypotenuse is √2. The two acute angles are both 45°, which is exactly why sin 45° and cos 45° are equal.
45° 则使用两腰都为 1 的等腰直角三角形。由勾股定理得斜边：1² + 1² = 2，所以斜边为 √2。两个锐角都等于 45°，这也正是 sin 45° 与 cos 45° 相等的几何原因。
The table below summarises the results that must be recalled instantly, in both ratio and reciprocal form.
下表汇总了必须能够瞬间回忆出来的结果，包括比值形式和倒数形式。
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 = √3/3 |
| 45° | √2/2 | √2/2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
Notice the symmetry: the sine column runs 1/2, √2/2, √3/2 while the cosine column is the same list reversed. Any surd in a denominator should be rationalised, so 1/√3 is written as √3/3 and 2/√2 is written as √2.
注意其中的对称性：正弦一列依次是 1/2、√2/2、√3/2，而余弦一列正好是同一串数字的倒序。分母中若含根号必须有理化，因此 1/√3 要写成 √3/3，2/√2 要写成 √2。
3. Using Trigonometric Ratios to Find Sides and Angles | 用三角比求边和角
nuse is known and you need the opposite side, use sin θ = opposite / hypotenuse and rearrange by multiplying both sides by the hypotenuse: opposite = hypotenuse × sin θ.
当已知一个角与一条边，且要求对边时，应选择同时包含已知边与所求边的那个比。如果已知斜边而要求对边，就用 sin θ = 对边 / 斜边，两边同乘斜边完成变形：对边 = 斜边 × sin θ。
n and the hypotenuse is required, use cos θ = adjacent / hypotenuse to get hypotenuse = adjacent / cos θ. A common error is to multiply when division is needed, which produces an answer that is too small.
当未知量出现在分母上时，变形涉及除法而不是乘法。例如，已知邻边而要求斜边时，由 cos θ = 邻边 / 斜边 可得 斜边 = 邻边 / cos θ。常见错误是本该除却去乘，结果得到的答案明显偏小。
hen tan θ = 5/8 and θ = tan⁻¹(5/8) ≈ 32.0°. On most calculators this is the shift or second function of the tan key.
已知两条边求角时，要使用计算器上的反三角函数。若对边为 5 cm、邻边为 8 cm，则 tan θ = 5/8，θ = tan⁻¹(5/8) ≈ 32.0°。在多数计算器上，它是 tan 键的 shift 或第二功能。
θ = cos⁻¹(adjacent / hypotenuse) θ = tan⁻¹(opposite / adjacent)
Keep full calculator accuracy throughout a multi-step question and round only at the final line. Rounding an intermediate angle to one decimal place and then using it again can shift the final answer enough to lose the accuracy mark.
在多步题中应保留计算器的完整精度，只在最后一行取近似值。若把中间角先舍入到一位小数再继续使用，最终答案的偏差可能足以丢掉精度分。
4. The Sine Rule | 正弦定理
Right-angled triangles are only a special case. For any triangle, including obtuse ones, the sine rule relates each side to the sine of the angle opposite it. The sides are conventionally named a, b, c and the angles opposite them are A, B, C respectively.
直角三角形只是特殊情况。对于任意三角形（包括钝角三角形），正弦定理把每条边与其对角的正弦联系起来。通常把三边记作 a、b、c，它们对应的对角分别记作 A、B、C。
a / sin A = b / sin B = c / sin C
n the numerator avoids awkward rearranging, so use the first form for sides and the second form for angles.
求角时可以把定理倒过来写：sin A / a = sin B / b = sin C / c。把未知量放在分子上可以避免繁琐的变形，因此求边用第一种形式，求角用第二种形式。
the third follows from the angle sum of a triangle, A + B + C = 180°, before you apply the rule at all.
当已知两角及任意一边（AAS 或 ASA），或已知两边及其中一边的对角（SSA）时，就使用正弦定理。若已给出两个角，第三个角可以先由三角形内角和 A + B + C = 180° 求出，再套用定理。
° − 65° = 73°. Applying the rule, b = a sin B / sin A = 9.4 × sin 65° / sin 42° ≈ 12.7 cm.
例题：在三角形 ABC 中，A = 42°，B = 65°，a = 9.4 cm。则 C = 180° − 42° − 65° = 73°。套用定理得 b = a sin B / sin A = 9.4 × sin 65° / sin 42° ≈ 12.7 cm。
5. The Ambiguous Case of the Sine Rule | 正弦定理的歧义情形
se sine on a calculator returns only the acute angle. When the sine rule is used with two sides and a non-included angle, a second obtuse triangle may also satisfy the given data. This is called the ambiguous case.
由于 sin θ = sin(180° − θ)，计算器的反正弦只会返回锐角。当用正弦定理处理两边及其中一边的对角时，可能还存在第二个满足条件的钝角三角形。这就是所谓的歧义情形。
To check, calculate both candidates: θ and 180° − θ. Add each one to the known angle and keep only the candidate whose sum is less than 180°. If both sums are valid, the question genuinely has two possible triangles and both should be reported.
检验方法是同时算出两个候选值：θ 与 180° − θ。把它们分别与已知角相加，只保留和小于 180° 的那个；若两个和都小于 180°，则题目确实存在两个三角形，两个答案都应写出。
A second consistency check uses side lengths: the longest side must face the largest angle. If your calculated angles suggest that the smallest side faces the largest angle, something has gone wrong.
第二个一致性检验利用边与角的大小关系：最长边必须对应最大角。如果算出的角显示最短边对应了最大角，说明过程中出现了错误。
6. The Cosine Rule | 余弦定理
The cosine rule is the generalisation of Pythagoras' theorem to triangles that are not right-angled. It connects all three sides with one angle, and it is the tool to reach for when the sine rule cannot be applied.
余弦定理是勾股定理在非直角三角形上的推广。它把三条边与其中一个角联系在一起，是正弦定理无法使用时的首选工具。
a² = b² + c² − 2bc cos A
In words: the square of the side opposite angle A equals the sum of the squares of the other two sides minus twice their product times the cosine of the included angle. Notice the pattern: the side on the left is the one opposite the angle on the right.
用文字表述：A 角的对边平方，等于另外两边的平方和减去这两边乘积的两倍再乘以夹角 A 的余弦。注意其中的对应关系：等号左边的边，正是等号右边那个角的对边。
When the three sides are known and an angle is required, rearrange the rule to make the cosine the subject. Dividing by 2bc gives the standard angle form.
当已知三边而要求角时，把定理变形，使余弦成为主元。两边同除以 2bc 就得到求角的标准形式。
cos A = (b² + c² − a²) / (2bc)
k, put A = 90°. Then cos A = 0 and the rule collapses to a² = b² + c², which is Pythagoras' theorem. This confirms that the cosine rule really is the more general statement.
可以做个检验：令 A = 90°，此时 cos A = 0，定理退化为 a² = b² + c²，正是勾股定理。这说明余弦定理确实是更一般的表述。
If instead all three sides were 7, 9 and 8, you would find the angle with cos A = (7² + 9² − 8²) / (2 × 7 × 9) = 66/126 ≈ 0.5238, giving A ≈ 58.4°.
例题：三角形两边 b = 7 cm、c = 9 cm，夹角 A = 58°。则 a² = 7² + 9² − 2 × 7 × 9 × cos 58° = 49 + 81 − 126 × 0.5299 ≈ 63.23，所以 a ≈ 7.95 cm。若反过来已知三边为 7、9、8，则求角时用 cos A = (7² + 9² − 8²) / (2 × 7 × 9) = 66/126 ≈ 0.5238，得 A ≈ 58.4°。
7. Choosing the Correct Rule | 如何选择正确的定理
Most marks in this topic are lost at the decision stage rather than in the arithmetic. Read what the question gives you and match it to one of the standard patterns before writing anything down.
本专题失分大多发生在判断阶段，而不是计算阶段。动笔之前先看清题目给了什么，把它与下列标准情形对应起来。
| Given information | 已知条件 | Method | 方法 |
|---|---|
| Right angle + one side + one angle | 直角三角形 + 一边 + 一角 | SOH CAH TOA | 基本三角比 |
| Two angles + any side (AAS / ASA) | 两角 + 任意一边 | Sine rule for a side | 正弦定理求边 |
| Two sides + non-included angle (SSA) | 两边 + 其中一边的对角 | Sine rule for an angle | 正弦定理求角 |
| Two sides + included angle (SAS) | 两边 + 夹角 | Cosine rule for a side | 余弦定理求边 |
| Three sides (SSS) | 三边 | Cosine rule for an angle | 余弦定理求角 |
| Two sides + included angle, area asked | 两边 + 夹角，求面积 | Area = ½ ab sin C | 面积公式 |
A useful tie-breaker: if the angle you are working with sits between the two known sides, use the cosine rule; if it does not, use the sine rule. If a triangle contains a right angle, the simpler SOH CAH TOA route is almost always faster.
一个实用的判断准则：若所用角恰好夹在两条已知边之间，用余弦定理；若不夹在中间，用正弦定理。若三角形中含直角，通常用更简单的 SOH CAH TOA 更快。
8. Area of a Triangle | 三角形面积公式
eight needs a perpendicular height, which is rarely given in examination triangles. The trigonometric version uses two sides and the angle between them, so no height is required.
常见公式 面积 = ½ × 底 × 高 需要一条垂线高度，而考试中的三角形很少直接给出高度。三角形式的面积公式改用两条边及其夹角，因此完全不需要高度。
Area = ½ ab sin C
Here C must be the angle included between sides a and b. Using a non-included angle is the single most common mistake with this formula, and it is easy to avoid by drawing the triangle and shading the angle you intend to use.
其中 C 必须是边 a 与边 b 的夹角。使用非夹角是这条公式最常见的错误，画图并给打算使用的角打上阴影就能轻松避免。
Two special cases follow from this formula. An equilateral triangle of side s has area (√3 ng C = 60°. A right-angled triangle with legs p and q gives area ½pq, because sin 90° = 1.
由此可推出两个特例。边长为 s 的等边三角形面积为 (√3/4)s²，只需令 C = 60° 即得。两直角边为 p、q 的直角三角形面积为 ½pq，因为 sin 90° = 1。
that side in Area = ½ ab sin C, then use the cosine rule on the answer to find a diagonal.
面积公式也常与其他定理配合使用。在多步题中，你可能先用正弦定理求出一条缺失的边，再用它代入 面积 = ½ ab sin C，最后对结果使用余弦定理求出某条对角线。
9. Bearings and Real-World Triangle Problems | 方位角与实际三角问题
A bearing is measured from north, clockwise, and is always written with three digits, so 60° is written as 060°. Because the measurement always starts at north and always turns clockwise, bearings are consistent between observers, which is exactly why navigation uses them.
方位角从正北方向开始按顺时针量取，并且始终写成三位数字，因此 60° 写作 060°。由于始终从正北起量、始终顺时针转动，不同观测者得到的方位角一致，这正是导航采用方位角的原因。
Three facts solve almost every bearing triangle. North lines are parallel, so alternate and corresponding angles apply. A bearing of 180° is a reversal of direction. And the angle between two bearings taken from the same point is simply the difference between them.
三条事实几乎能解决所有方位角三角形问题：正北方向线互相平行，因此同位角与内错角成立；相差 180° 表示方向相反；从同一点量得的两条方位角之间的夹角就是它们的差。
is 070°, then the bearing of A from B is 070° + 180° = 250°. If the original bearing exceeds 180°, subtract instead: from 250° back to 070°.
若已知从 A 看 B 的方位角，要求从 B 看 A 的方位角，就加或减 180°，并把结果调回 000° 至 360° 区间。若从 A 看 B 是 070°，则从 B 看 A 是 070° + 180° = 250°。若原方位角大于 180°，则改为减去 180°：从 250° 回到 070°。
note that the turn angle between the two legs is 150° − 060° = 90°, so the triangle has two known sides and an included angle. That is two sides and an included angle, so the cosine rule gives the answer directly.
一道典型题目是：一艘船沿方位角 060° 航行 8 km，随后转向沿方位角 150° 航行 11 km。求它到出发点的距离。首先注意到两段航程之间的转角为 150° − 060° = 90°，因此三角形中已知两边及其夹角。这正是两边加夹角的情形，直接使用余弦定理即可求出答案。
10. Exam Technique and Common Errors | 考试技巧与常见错误
Check that your calculator is in degree mode. Every rule in this topic assumes degrees unless the question explicitly uses radians, and a calculator left in radian mode will return answers that are wildly wrong rather than slightly wrong.
先确认计算器处于角度制（DEG）模式。本专题所有定理默认使用角度，除非题目明确说明使用弧度；若计算器停留在弧度制，答案会完全错误，而不是只有轻微偏差。
Do not round until the final line. Store intermediate values in the calculator memory or in an unrounded variable, and round only the value you write down. Standard practice is to give three significant figures unless the question specifies otherwise.
在最后一行之前不要取近似值。把中间结果存入计算器内存或保留未舍入的变量，只对最终写下的数值取近似。除非题目另有要求，通常保留三位有效数字。
Watch the order of operations in the cosine rule. The term 2bc cos A must be subtracted after the two squares are added, and the calculator will not do this for you if you type the expression carelessly. Writing the bracket explicitly avoids the problem.
注意余弦定理中的运算顺序。必须先求两个
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