📚 9C Homework Answers: Triangle Area, Sine and Cosine Rules | 9C 作业答案：三角形面积、正弦与余弦定理
This article provides worked solutions and full explanations for the type of questions found in Exercise 9C, which focuses on the sine rule, the cosine rule, and the area formula for non-right-angled triangles. Every answer below is written in the same order as a mark scheme would expect, so you can use it to check your homework, understand the methods, and prepare for exam questions on non-right-angled trigonometry.
本文提供 9C 练习中常见题型的分步答案与完整讲解，重点是非直角三角形的正弦定理、余弦定理和面积公式。下面的每一步都按评分标准期望的顺序书写，你可以用它核对作业、理解方法，并为非直角三角形三角函数的考试题做好准备。
1. Key Formulas for Exercise 9C | 9C 练习的核心公式
Before checking individual answers, make sure you know the three core formulas used throughout Exercise 9C. Everything in this exercise is a direct application of one of them.
在核对具体答案之前，请先掌握 9C 练习中反复使用的三个核心公式。本练习的所有题目都是这三个公式的直接应用。
Area of a triangle: A = ½ab sin C
Sine rule: a / sin A = b / sin B = c / sin C
Cosine rule: a² = b² + c² − 2bc cos A
is angle C. The sine rule can also be written upside down as sin A / a = sin B / b = sin C / c, which is often easier when you are finding an angle.
这里，a、b、c 是边长，A、B、C 是与这些边相对的角。面积公式使用的是边 a 和边 b 之间的夹角，也就是角 C。正弦定理也可以倒过来写成 sin A / a = sin B / b = sin C / c，当你要找的是角时，这种形式通常更好用。
A useful habit is to label the triangle first: mark every side with a small letter and the angle directly opposite it with the matching capital letter. Almost every mistake in Exercise 9C comes from pairing a side with the wrong angle, so this one habit prevents most lost marks.
一个很实用的习惯是先给三角形做标记：每条边用小写字母标注，与它正对的角用对应的大写字母标注。9C 里几乎所有的错误都来自把边和错误的角配对，所以养成这个习惯能避免大部分失分。
2. Choosing the Right Rule | 如何选择正确的定理
The most common exam complaint about this topic is "I did not know which rule to use." The decision depends only on what information the question gives you and what it asks for.
关于这个知识点，考试中最常见的抱怨是"我不知道该用哪条定理"。其实选择只取决于题目给了什么条件、要求你求什么。
| What you are given | 已知条件 | What to use | 应使用 |
|---|---|
| Two sides and the included angle | 两边及其夹角 | Cosine rule for the third side | 余弦定理求第三边 |
| Three sides | 三边 | Rearranged cosine rule for an angle | 余弦定理变形求角 |
| Two angles and any side | 两角及任一边 | Sine rule | 正弦定理 |
| Two sides and a non-included angle | 两边及非夹角 | Sine rule (watch for the ambiguous case) | 正弦定理（注意模糊情况） |
| Two sides and the included angle, asked for area | 两边及夹角，求面积 | A = ½ab sin C | A = ½ab sin C |
In short: if the angle sits between the two sides you know, reach for the cosine rule. If the angle is opposite one of the sides you know, reach for the sine rule.
简单来说：如果已知的角夹在两条已知边之间，就用余弦定理；如果已知的角与其中一条已知边相对，就用正弦定理。
3. Worked Example 1: Finding the Area of a Triangle | 例题 1：求三角形面积
8 cm, side b = 11 cm and angle C = 40°. This is the classic Exercise 9C opening question.
题目：已知三角形 ABC 中，边 a = 8 cm，边 b = 11 cm，角 C = 40°，求三角形面积。这是 9C 练习中最经典的开篇题型。
Substitute directly into the area formula. The two sides you use must be the ones that form the given angle, and here they are a and b, with the included angle C:
直接代入面积公式。使用的两条边必须是构成已知角的那两条边，这里就是 a 和 b，夹角为 C：
A = ½ × 8 × 11 × sin 40° = 44 × 0.642788 = 28.28 cm²
ce that ½ × 8 × 11 = 44 comes first, so you only need to evaluate one trigonometric value.
所以面积为 28.3 cm²（保留三位有效数字）。注意先算出 ½ × 8 × 11 = 44，这样你只需要计算一个三角函数值。
C = 40°: c² = 8² + 11² − 2 × 8 × 11 × cos 40° = 185 − 134.82 = 50.18, giving c ≈ 7.08 cm.
如果同一道题还要求第三条边 c，就要换成余弦定理，夹角仍是 C = 40°：c² = 8² + 11² − 2 × 8 × 11 × cos 40° = 185 − 134.82 = 50.18，得 c ≈ 7.08 cm。
gle should be opposite the shortest side, so c ≈ 7.08 cm is indeed the shortest side here.
有两个检查值得一做。第一，面积应小于 ½ × 8 × 11 = 44 cm²，因为 sin C 小于 1。第二，最小的角应对应最短的边，所以 c ≈ 7.08 cm 确实是这里最短的边。
4. Worked Example 2: Using the Sine Rule to Find a Side | 例题 2：用正弦定理求边
Question: in triangle ABC, angle A = 35°, angle B = 80° and side a = 7 cm. Find side b.
题目：三角形 ABC 中，角 A = 35°，角 B = 80°，边 a = 7 cm，求边 b。
Two angles and one side means the sine rule. Pair each side with the angle opposite it, then rearrange before substituting numbers, which reduces rounding errors:
两角加一边，说明要用正弦定理。把每条边与它对面的角配对，然后先变形再代入数字，这样可以减少舍入误差：
b = a sin B / sin A = 7 × sin 80° / sin 35° = 7 × 0.984808 / 0.573576 = 12.02 cm
r than angle A, side b must be longer than side a, and 12.0 cm is indeed longer than 7 cm.
所以 b ≈ 12.0 cm。因为角 B 大于角 A，边 b 必须比边 a 长，而 12.0 cm 确实大于 7 cm。
apply the sine rule again: c = 7 × sin 65° / sin 35° = 7 × 0.906308 / 0.573576 = 11.1 cm.
接下来求第三条边就很简单了。先用三角形内角和求出缺失的角：C = 180° − 35° − 80° = 65°。再套用正弦定理：c = 7 × sin 65° / sin 35° = 7 × 0.906308 / 0.573576 = 11.1 cm。
80° is the largest angle and b ≈ 12.0 cm is the longest side, so the answer is consistent.
任何三角形都可以快速自检：最长边必须对应最大的角。这里 80° 是最大的角，b ≈ 12.0 cm 是最长的边，答案自洽。
5. Worked Example 3: The Sine Rule and the Ambiguous Case | 例题 3：正弦定理与模糊情况
Question: in triangle ABC, a = 9 cm, b = 12 cm and angle A = 40°. Find angle B.
题目：三角形 ABC 中，a = 9 cm，b = 12 cm，角 A = 40°，求角 B。
Here the angle we know is opposite one of the known sides, so this is a sine rule question in its inverted form:
这里已知的角与其中一条已知边相对，所以这是正弦定理的倒写形式：
sin B = b sin A / a = 12 × sin 40° / 9 = 0.857051
ause sin B = sin (180° − B), there is a second possible value: B = 180° − 58.99° = 121.0°.
取反正弦得 B = 58.99°，约等于 59.0°。但由 sin B = sin (180° − B) 可知还存在第二个可能值：B = 180° − 58.99° = 121.0°。
° − 40° − 121.0° = 19.0°, which is also perfectly valid. So this question has two answers.
两种情况在几何上都成立。若 B = 59.0°，则 C = 180° − 40° − 59.0° = 81.0°，三角形合法。若 B = 121.0°，则 C = 180° − 40° − 121.0° = 19.0°，同样完全合法。所以本题有两个答案。
180° − 58.99° would be impossible to pair with it, and only the acute value would survive.
只有当某个候选值使三个角之和超过 180° 时才需要舍弃它。例如若角 A 是 70°，那么 B = 180° − 58.99° 就无法与之共存，此时只保留锐角解。
6. Worked Example 4: The Cosine Rule for a Missing Side | 例题 4：用余弦定理求边
Question: in triangle ABC, b = 6 cm, c = 9 cm and angle A = 115°. Find side a.
题目：三角形 ABC 中，b = 6 cm，c = 9 cm，角 A = 115°，求边 a。
Two sides and the included angle: cosine rule. The side you are finding must be the one opposite the known angle, so a is opposite A:
两边及其夹角：用余弦定理。所求的边必须在已知角的对面，所以 a 与 A 相对：
a² = 6² + 9² − 2 × 6 × 9 × cos 115° = 117 + 45.64 = 162.64
ificant figures. Note that cos 115° is negative, so the term −2bc cos A actually adds length, which is why a is longer than either b or c.
所以 a ≈ 12.75 cm，保留三位有效数字为 12.8 cm。注意 cos 115° 是负数，所以 −2bc cos A 这一项实际上增加了长度，这正是 a 比 b、c 都长的原因。
This is a very useful check for obtuse triangles. If you know the included angle is obtuse and your answer for the opposite side comes out shorter than both given sides, you have almost certainly entered the angle wrongly or used the wrong calculator mode.
这是检验钝角三角形的一个很有用的方法。如果你已知夹角是钝角，而求出的对边却比两条已知边都短，那几乎可以肯定你输错了角或用了错误的计算器模式。
7. Worked Example 5: The Cosine Rule Rearranged for an Angle | 例题 5：余弦定理变形求角
a triangle has sides a = 7 cm, b = 8 cm and c = 11 cm. Find all three angles and the area.
题目：一个三角形的三边为 a = 7 cm，b = 8 cm，c = 11 cm，求三个角及面积。
With three sides and no angle, you must rearrange the cosine rule to make the angle the subject:
已知三边而无任何角时，必须把余弦定理变形，让角成为主项：
cos A = (b² + c² − a²) / (2bc)
eating for B: cos B = (49 + 121 − 64) / (2 × 7 × 11) = 106 / 154 = 0.688312, so B = 46.5°.
代入得 cos A = (64 + 121 − 49) / (2 × 8 × 11) = 136 / 176 = 0.772727，所以 A = 39.4°。再求 B：cos B = (49 + 121 − 64) / (2 × 7 × 11) = 106 / 154 = 0.688312，所以 B = 46.5°。
4 − 121) / (2 × 7 × 8) = −8 / 112 = −0.071429, so C = 94.1°. The negative cosine confirms that C is obtuse, which fits, since 11 cm is the longest side.
求 C 时你有两种选择：用内角和，C = 180° − 39.4° − 46.5° = 94.1°；或者第三次使用余弦定理。用余弦定理得 cos C = (49 + 64 − 121) / (2 × 7 × 8) = −8 / 112 = −0.071429，所以 C = 94.1°。余弦值为负，说明 C 是钝角，这与 11 cm 是最长边相吻合。
hat form angle C: A = ½ × 7 × 8 × sin 94.1° = 28 × 0.99744 = 27.9 cm². Checking with Heron's formula gives the same value, which is a good confirmation that every step was correct.
现在求面积，使用构成角 C 的两条边：A = ½ × 7 × 8 × sin 94.1° = 28 × 0.99744 = 27.9 cm²。用海伦公式检验会得到相同结果，这很好地确认了每一步都正确。
8. Area Without the Included Angle: Heron's Formula | 不含夹角时求面积：海伦公式
Some 9C extension questions give you three sides and ask for the area, but no angle. You have two routes, and both earn full marks.
9C 的部分拓展题会给出三边并要求面积，但不给任何角。你有两条路可走，两者都能得满分。
rule, then apply A = ½ab sin C with the two sides that form that angle. Route two is Heron's formula, which uses the semi-perimeter s:
方法一是先用余弦定理变形求出一个角，再用构成该角的两条边代入 A = ½ab sin C。方法二是海伦公式，需要
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图