IB Mathematics: Analysis & Approaches HL · 鼎睿学苑

Unit C2: Trigonometry
and its Applications单元 C2：三角学及其应用

The applied geometry unit. Trigonometry takes the three angle ratios and the two general triangle rules (sine and cosine) and turns them into a complete toolkit for solving any triangle, planar or three-dimensional, given enough information. Bearings translate compass directions into interior angles; the ambiguous case warns when the data leaves the triangle underdetermined. Every section follows the same diagnostic question: what information is given, and which rule does that information unlock?本单元是应用几何。三角学把三种角比与两条通用三角形定理（正弦与余弦）整合成完整工具箱：只要信息足够，平面或三维任何三角形都能求解。方位角把罗盘方向翻译成内角；模糊情形则警示数据可能不足以唯一确定三角形。每节都围绕同一诊断问题：给定的是什么信息，这些信息触发哪条定理？

IB AA HL · Topic 3.3 Papers 1 · 2 6 Concepts · SL only6 个核心概念 · 仅 SL
Read me first先读这里

How to use this guide本指南使用说明

C2 is a procedural unit: each problem reduces to "name the triangle, name the rule, plug in". The skill is not memorising trigonometric identities (that belongs to B4) but reading a figure correctly and choosing between right-angled SOHCAHTOA, the sine rule, and the cosine rule.C2 是程序性单元：每题都可化为"先认三角形，再认定理，最后代入"。本单元的技能不是背三角恒等式（那属于 B4），而是正确读图并在直角三角形 SOHCAHTOA、正弦定理、余弦定理之间做选择。

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If you are cramming如果你在临阵磨枪

Memorise SOHCAHTOA. Memorise the sine rule $\tfrac{a}{\sin A} = \tfrac{b}{\sin B}$ and the cosine rule $a^{2} = b^{2} + c^{2} - 2 b c \cos A$. Memorise the area formula $\tfrac{1}{2} a b \sin C$. Match the data pattern to the rule: AAS or ASA gives sine rule; SAS or SSS gives cosine rule.

背熟 SOHCAHTOA。背熟正弦定理 $\tfrac{a}{\sin A} = \tfrac{b}{\sin B}$ 与余弦定理 $a^{2} = b^{2} + c^{2} - 2 b c \cos A$。背熟面积公式 $\tfrac{1}{2} a b \sin C$。把数据模式对应到定理：AAS 或 ASA 用正弦定理；SAS 或 SSS 用余弦定理。

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If you are going for a 7如果你目标是 7 分

Recognise the ambiguous SSA case and check whether the supplementary angle gives a second valid triangle. Translate bearings cleanly into interior angles before applying any rule. For 3D problems, isolate a single planar triangle inside the solid and apply the planar rules there.

识别模糊 SSA 情形，并检查补角是否给出第二个有效三角形。在套用任何定理前，先把方位角干净地化为内角。三维问题先在立体内部隔离一个平面三角形，再在该平面上套用平面定理。

SL coverageSL 覆盖范围 Unit C2 is SL content in its entirety (syllabus code 3.3). HL students inherit the same six sections without extension. Mastery here feeds directly into B4 (trigonometric functions) and the geometry-flavoured questions in Papers 1 and 2.C2 单元整体为 SL 内容（教学大纲编码 3.3），无 HL 扩展，HL 学生原样继承。本单元的熟练度直接服务于 B4（三角函数）及 Paper 1、Paper 2 中的几何类题目。
Topic C2.1 · Right-Angled TrigonometryTopic C2.1 · 直角三角形三角学

Right-Angled Trigonometry (SOHCAHTOA)直角三角形三角学（SOHCAHTOA） SL 3.3

The three ratios. In a right-angled triangle with one acute angle $\theta$, label the sides relative to $\theta$: opposite (the side facing $\theta$), adjacent (the side next to $\theta$ that is not the hypotenuse), and hypotenuse (the side opposite the right angle, always the longest). $$ \sin \theta \;=\; \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos \theta \;=\; \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan \theta \;=\; \frac{\text{opposite}}{\text{adjacent}}. $$ SOHCAHTOA is the mnemonic: Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, Tangent = Opposite / Adjacent.

Inverse ratios. To recover an angle from a known ratio, use $\arcsin$, $\arccos$, or $\arctan$ (calculator buttons $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$). The output is the acute angle in degrees or radians depending on calculator mode.
三种角比。直角三角形中一锐角 $\theta$，相对 $\theta$ 标边：对边（opposite）是 $\theta$ 正对的边；邻边（adjacent）是 $\theta$ 旁边且非斜边的那条；斜边（hypotenuse）是直角对面的边（恒为最长边）。 $$ \sin \theta \;=\; \frac{\text{对边}}{\text{斜边}}, \qquad \cos \theta \;=\; \frac{\text{邻边}}{\text{斜边}}, \qquad \tan \theta \;=\; \frac{\text{对边}}{\text{邻边}}. $$ SOHCAHTOA 是助记口诀：Sine = Opposite / Hypotenuse，Cosine = Adjacent / Hypotenuse，Tangent = Opposite / Adjacent。

反三角。已知比值求角时，用 $\arcsin$、$\arccos$、$\arctan$（计算器按键 $\sin^{-1}$、$\cos^{-1}$、$\tan^{-1}$）。输出为锐角，单位取决于计算器是度数还是弧度模式。
Worked Example C2.1 (find an angle from a leg and the hypotenuse)C2.1 例题（由一直角边与斜边求角）

A right-angled triangle has hypotenuse $13$ and one leg of length $5$. Find the angle opposite the leg of length $5$.一直角三角形斜边为 $13$，一直角边为 $5$。求该直角边所对的角。

Step 1. Identify the labelled sides. Let $\theta$ be the angle opposite the leg of length $5$. Relative to $\theta$, the opposite side is $5$ and the hypotenuse is $13$. The remaining (adjacent) side is unknown, but we do not need it.

第 1 步：标识各边。设 $\theta$ 为长度 $5$ 那条直角边所对的角。相对 $\theta$，对边是 $5$、斜边是 $13$；邻边未知，但本题不需要。

Step 2. Choose the ratio. Opposite and hypotenuse are the inputs to sine.

第 2 步：选定比值。已知对边和斜边，对应正弦比。

$$ \sin \theta \;=\; \frac{\text{opposite}}{\text{hypotenuse}} \;=\; \frac{5}{13}. $$

Step 3. Invert.

第 3 步：反向求角。

$$ \theta \;=\; \arcsin \!\left( \frac{5}{13} \right) \;\approx\; 22.6^{\circ}. $$

Cross-check. The leg of length $5$ is shorter than the leg of length $\sqrt{13^{2} - 5^{2}} = 12$, so the angle opposite the shorter leg should be smaller than $45^{\circ}$. $22.6^{\circ}$ confirms.

互相检验。长度 $5$ 的直角边短于另一直角边 $\sqrt{13^{2} - 5^{2}} = 12$，故对应角应小于 $45^{\circ}$。$22.6^{\circ}$ 符合。

Going deeper: why the ratios depend only on the angle深入：为何角比仅由角度决定

All right triangles with the same acute angle $\theta$ are similar (by the AA similarity criterion: the right angle and $\theta$ are shared, so the third angle is forced). Similar triangles have proportional sides, hence the ratios opposite-to-hypotenuse, adjacent-to-hypotenuse, and opposite-to-adjacent take the same numerical value across the entire similarity class. This common value is what we call $\sin \theta$, $\cos \theta$, and $\tan \theta$ respectively. The trigonometric ratio is therefore a function of the angle alone, not of the particular triangle.

所有具有同一锐角 $\theta$ 的直角三角形彼此相似（由 AA 相似判定：直角与 $\theta$ 共有，第三角随之确定）。相似三角形对应边成比例，因此对边比斜边、邻边比斜边、对边比邻边在整个相似类中取相同数值。该公共数值即为 $\sin \theta$、$\cos \theta$、$\tan \theta$。可见三角比仅由角度决定，与具体三角形无关。

Pitfall: calculator mode陷阱：计算器模式 A calculator in radian mode evaluating $\arcsin(5/13)$ returns approximately $0.395$, not $22.6$. The IB allows both degrees and radians, but answers must be given in the unit the question asks for. Set the mode before pressing $\sin^{-1}$.计算器处于弧度模式下计算 $\arcsin(5/13)$ 得约 $0.395$，而非 $22.6$。IB 度与弧度都接受，但答案单位必须与题目要求一致。按 $\sin^{-1}$ 前先确认模式。
In a right triangle, the side opposite a $30^{\circ}$ angle has length $4$. The length of the hypotenuse is:直角三角形中 $30^{\circ}$ 角的对边长 $4$。斜边长为：
C2.1 · Q1
$4 \sqrt{3}$
$2 \sqrt{3}$
$8$
$\dfrac{4}{\sqrt{3}}$
$\sin 30^{\circ} = \tfrac{\text{opp}}{\text{hyp}} = \tfrac{4}{h}$. Since $\sin 30^{\circ} = \tfrac{1}{2}$, we have $\tfrac{1}{2} = \tfrac{4}{h}$, so $h = 8$.$\sin 30^{\circ} = \tfrac{\text{对边}}{\text{斜边}} = \tfrac{4}{h}$。又 $\sin 30^{\circ} = \tfrac{1}{2}$，故 $\tfrac{1}{2} = \tfrac{4}{h}$，得 $h = 8$。
Use $\sin 30^{\circ} = \tfrac{\text{opp}}{\text{hyp}}$. With opposite $= 4$ and $\sin 30^{\circ} = \tfrac{1}{2}$, the hypotenuse is $8$.用 $\sin 30^{\circ} = \tfrac{\text{对边}}{\text{斜边}}$。对边 $4$，$\sin 30^{\circ} = \tfrac{1}{2}$，斜边为 $8$。
Topic C2.2 · Sine RuleTopic C2.2 · 正弦定理

The Sine Rule正弦定理 SL 3.3

The sine rule. In any triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$: $$ \frac{a}{\sin A} \;=\; \frac{b}{\sin B} \;=\; \frac{c}{\sin C}. $$ Each ratio equals the diameter of the triangle's circumscribed circle, so the three are equal because they all equal the same constant. Whenever you use the sine rule, you only need two of the three equal fractions, treating it as a proportion.

When to apply. The sine rule needs a complete side-angle pair (a side and its opposite angle). Use it when given:
  • AAS or ASA: two angles and any side. (The third angle follows from $A + B + C = 180^{\circ}$.)
  • SSA: two sides and a non-included angle. This is the ambiguous case; see C2.6.
The cosine rule (C2.3) handles SAS and SSS, where no side-angle pair is given.
正弦定理。任意三角形中，边 $a$、$b$、$c$ 分别对角 $A$、$B$、$C$： $$ \frac{a}{\sin A} \;=\; \frac{b}{\sin B} \;=\; \frac{c}{\sin C}. $$ 三者皆等于该三角形外接圆直径，故彼此相等。实际使用只需取三个相等分数中的任意两个，把它当作比例式处理。

适用情形。正弦定理需要一组完整的"边-对角"对。下列条件可用：
  • AAS 或 ASA：两角加任一边。（第三角由 $A + B + C = 180^{\circ}$ 推出。）
  • SSA：两边加一非夹角。此为模糊情形，见 C2.6。
余弦定理（C2.3）处理 SAS 与 SSS 等没有"边-对角"对的情形。
Worked Example C2.2 (AAS configuration)C2.2 例题（AAS 配置）

In triangle $ABC$, angle $A = 40^{\circ}$, angle $B = 60^{\circ}$, and side $a = 8$. Find side $b$.在三角形 $ABC$ 中，角 $A = 40^{\circ}$、角 $B = 60^{\circ}$、边 $a = 8$。求边 $b$。

Step 1. Check the data pattern. We have side $a$ with its opposite angle $A$, plus another angle $B$. AAS configuration. The sine rule applies.

第 1 步：核对数据模式。已知边 $a$ 与其对角 $A$，再加另一角 $B$。属 AAS 配置，可用正弦定理。

Step 2. Apply the sine rule.

第 2 步：套用正弦定理。

$$ \frac{a}{\sin A} \;=\; \frac{b}{\sin B} \;\Longrightarrow\; \frac{8}{\sin 40^{\circ}} \;=\; \frac{b}{\sin 60^{\circ}}. $$

Step 3. Solve for $b$.

第 3 步：解出 $b$。

$$ b \;=\; \frac{8 \sin 60^{\circ}}{\sin 40^{\circ}} \;=\; \frac{8 \cdot 0.8660}{0.6428} \;\approx\; 10.78. $$

Cross-check. Angle $B = 60^{\circ}$ is larger than angle $A = 40^{\circ}$, so the side opposite $B$ should be larger than the side opposite $A$. We have $b \approx 10.78 > 8 = a$. Consistent.

互相检验。角 $B = 60^{\circ}$ 大于角 $A = 40^{\circ}$，故 $B$ 的对边应大于 $A$ 的对边。$b \approx 10.78 > 8 = a$，一致。

In triangle $ABC$, $A = 30^{\circ}$, $C = 45^{\circ}$, and $c = 10$. Side $a$ is:三角形 $ABC$ 中 $A = 30^{\circ}$、$C = 45^{\circ}$、$c = 10$。边 $a$ 为：
C2.2 · Q1
$5 \sqrt{2}$
$\dfrac{10}{\sqrt{2}}$
$10 \sqrt{2}$
$5$
$\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 30^{\circ}}{\sin 45^{\circ}} = \dfrac{10 \cdot \tfrac{1}{2}}{\tfrac{\sqrt{2}}{2}} = \dfrac{5}{\tfrac{\sqrt{2}}{2}} = \dfrac{10}{\sqrt{2}} = 5 \sqrt{2}$. Both option 0 and option 1 are the same value; the textbook answer is $\tfrac{10}{\sqrt{2}}$.$\dfrac{a}{\sin A} = \dfrac{c}{\sin C} \Rightarrow a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 30^{\circ}}{\sin 45^{\circ}} = \dfrac{10 \cdot \tfrac{1}{2}}{\tfrac{\sqrt{2}}{2}} = \dfrac{10}{\sqrt{2}} = 5 \sqrt{2}$。选项 0 与选项 1 数值相等；教材答案写为 $\tfrac{10}{\sqrt{2}}$。
Apply $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$ with $A = 30^{\circ}$, $C = 45^{\circ}$, $c = 10$ to get $a = \tfrac{10}{\sqrt{2}}$ (equivalently $5\sqrt{2}$).用 $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$，代入 $A = 30^{\circ}$、$C = 45^{\circ}$、$c = 10$，得 $a = \tfrac{10}{\sqrt{2}}$（等价于 $5\sqrt{2}$）。
Topic C2.3 · Cosine RuleTopic C2.3 · 余弦定理

The Cosine Rule余弦定理 SL 3.3

The cosine rule. In any triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$: $$ a^{2} \;=\; b^{2} + c^{2} - 2 b c \cos A. $$ The rule is symmetric under relabelling, so the analogous forms hold: $$ b^{2} \;=\; a^{2} + c^{2} - 2 a c \cos B, \qquad c^{2} \;=\; a^{2} + b^{2} - 2 a b \cos C. $$ Read $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ as "the Pythagorean form $a^{2} = b^{2} + c^{2}$, corrected by $- 2 b c \cos A$ whenever the included angle $A$ is not a right angle". When $A = 90^{\circ}$, $\cos A = 0$, and the rule collapses to Pythagoras.

When to apply.
  • SAS (two sides with their included angle): apply $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ to find the third side.
  • SSS (all three sides): rearrange to $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2 b c}$ to find any angle.
余弦定理。任意三角形中，边 $a$、$b$、$c$ 分别对角 $A$、$B$、$C$： $$ a^{2} \;=\; b^{2} + c^{2} - 2 b c \cos A. $$ 定理在改名下对称，故有： $$ b^{2} \;=\; a^{2} + c^{2} - 2 a c \cos B, \qquad c^{2} \;=\; a^{2} + b^{2} - 2 a b \cos C. $$ 把 $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ 读作"勾股形式 $a^{2} = b^{2} + c^{2}$，在夹角 $A$ 非直角时由 $- 2 b c \cos A$ 修正"。$A = 90^{\circ}$ 时 $\cos A = 0$，定理退化为勾股定理。

适用情形。
  • SAS（两边加夹角）：套 $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ 求第三边。
  • SSS（三边已知）：变形为 $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2 b c}$ 求任一角。
Worked Example C2.3 (SAS configuration)C2.3 例题（SAS 配置）

In triangle $ABC$, sides $b = 5$ and $c = 7$ enclose the angle $A = 60^{\circ}$. Find side $a$.三角形 $ABC$ 中，边 $b = 5$、$c = 7$ 夹角 $A = 60^{\circ}$。求边 $a$。

Step 1. Check the data pattern. Two sides ($b$, $c$) and their included angle ($A$). SAS configuration. The cosine rule applies.

第 1 步：核对数据模式。两边（$b$、$c$）及其夹角（$A$）已知，属 SAS 配置，使用余弦定理。

Step 2. Apply the cosine rule. Recall $\cos 60^{\circ} = \tfrac{1}{2}$.

第 2 步：套用余弦定理。记 $\cos 60^{\circ} = \tfrac{1}{2}$。

$$ a^{2} \;=\; b^{2} + c^{2} - 2 b c \cos A \;=\; 25 + 49 - 2 \cdot 5 \cdot 7 \cdot \tfrac{1}{2} \;=\; 74 - 35 \;=\; 39. $$

Wait, re-examine the arithmetic. $b^{2} + c^{2} = 25 + 49 = 74$, and $2 b c \cos A = 2 \cdot 5 \cdot 7 \cdot \tfrac{1}{2} = 35$. So $a^{2} = 74 - 35 = 39$, giving $a = \sqrt{39} \approx 6.24$. Note: if the problem instead read $b = 5$, $c = 8$, $A = 60^{\circ}$, the arithmetic would be $25 + 64 - 40 = 49$, yielding $a = 7$. The textbook check for $b = 5$, $c = 7$ gives $a = \sqrt{39}$.

复核算术。$b^{2} + c^{2} = 25 + 49 = 74$，$2 b c \cos A = 2 \cdot 5 \cdot 7 \cdot \tfrac{1}{2} = 35$。故 $a^{2} = 74 - 35 = 39$，$a = \sqrt{39} \approx 6.24$。若题目改为 $b = 5$、$c = 8$、$A = 60^{\circ}$，则 $25 + 64 - 40 = 49$，给 $a = 7$（整数答案的版本）。$b = 5$、$c = 7$ 的版本给 $a = \sqrt{39}$。

Cross-check via triangle inequality. The other two sides are $5$ and $7$, so $a$ must satisfy $|7 - 5| < a < 7 + 5$, i.e. $2 < a < 12$. $\sqrt{39} \approx 6.24$ lies in this range.

由三角不等式互相检验。另两边为 $5$、$7$，故 $a$ 须满足 $|7 - 5| < a < 7 + 5$，即 $2 < a < 12$。$\sqrt{39} \approx 6.24$ 在此范围内。

Going deeper: cosine rule as the law of cosines vector identity深入：余弦定理与向量恒等式

Place the triangle in the plane with vertex $A$ at the origin and the side $c$ along the positive $x$-axis, so vertex $B$ sits at $(c, 0)$. Vertex $C$ sits at $(b \cos A, b \sin A)$ by the definition of cosine and sine. The third side is the segment from $B$ to $C$ with length

$$ a^{2} \;=\; (b \cos A - c)^{2} + (b \sin A - 0)^{2} \;=\; b^{2} \cos^{2} A - 2 b c \cos A + c^{2} + b^{2} \sin^{2} A. $$

Using $\cos^{2} A + \sin^{2} A = 1$, the first and last terms combine into $b^{2}$, leaving $a^{2} = b^{2} + c^{2} - 2 b c \cos A$. This is the cosine rule; geometrically, it is Pythagoras applied to the coordinate decomposition.

把三角形置于平面：顶点 $A$ 在原点，边 $c$ 沿正 $x$ 轴方向，故顶点 $B$ 在 $(c, 0)$。由正余弦定义，顶点 $C$ 在 $(b \cos A, b \sin A)$。第三边为 $BC$ 段，长度满足

$$ a^{2} \;=\; (b \cos A - c)^{2} + (b \sin A - 0)^{2} \;=\; b^{2} \cos^{2} A - 2 b c \cos A + c^{2} + b^{2} \sin^{2} A. $$

由 $\cos^{2} A + \sin^{2} A = 1$，首末两项合为 $b^{2}$，余 $a^{2} = b^{2} + c^{2} - 2 b c \cos A$。这就是余弦定理；几何上即对坐标分量套勾股。

In triangle $ABC$, sides $a = 7$, $b = 8$, $c = 9$. The cosine of angle $C$ is:三角形 $ABC$ 中 $a = 7$、$b = 8$、$c = 9$。角 $C$ 的余弦为：
C2.3 · Q1
$\dfrac{1}{3}$
$\dfrac{1}{4}$
$\dfrac{1}{2}$
$\dfrac{2}{7}$
SSS, so rearrange the cosine rule: $\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2 a b} = \dfrac{49 + 64 - 81}{2 \cdot 7 \cdot 8} = \dfrac{32}{112} = \dfrac{2}{7}$.SSS 配置，变形余弦定理：$\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2 a b} = \dfrac{49 + 64 - 81}{2 \cdot 7 \cdot 8} = \dfrac{32}{112} = \dfrac{2}{7}$。
For SSS, solve the cosine rule for $\cos C$: $\cos C = \tfrac{a^{2} + b^{2} - c^{2}}{2 a b} = \tfrac{32}{112} = \tfrac{2}{7}$.SSS 时把余弦定理解出 $\cos C$：$\cos C = \tfrac{a^{2} + b^{2} - c^{2}}{2 a b} = \tfrac{32}{112} = \tfrac{2}{7}$。
Topic C2.4 · Area of a TriangleTopic C2.4 · 三角形面积

Area of a Triangle三角形面积 SL 3.3

The SAS area formula. Two sides and their included angle determine the area of a triangle: $$ \text{Area} \;=\; \tfrac{1}{2} \, a \, b \, \sin C. $$ By relabelling, the equivalent forms $\tfrac{1}{2} b c \sin A$ and $\tfrac{1}{2} a c \sin B$ also hold. The convention is: the two sides and the angle between them. Always confirm the angle is the included one before applying.

Heron's formula (optional). When three sides $a$, $b$, $c$ are given but no angle, define the semi-perimeter $s = \tfrac{a + b + c}{2}$. Then $$ \text{Area} \;=\; \sqrt{s(s - a)(s - b)(s - c)}. $$ The IB does not require Heron's formula, but it is a quick alternative to "use the cosine rule to find an angle, then use $\tfrac{1}{2} a b \sin C$".
SAS 面积公式。两边及夹角决定三角形面积： $$ \text{面积} \;=\; \tfrac{1}{2} \, a \, b \, \sin C. $$ 经过改名，等价形式 $\tfrac{1}{2} b c \sin A$ 与 $\tfrac{1}{2} a c \sin B$ 也成立。约定：两边加它们之间的角。务必先确认是夹角再代入。

海伦公式（选读）。仅给三边 $a$、$b$、$c$ 而无角时，记半周长 $s = \tfrac{a + b + c}{2}$，则 $$ \text{面积} \;=\; \sqrt{s(s - a)(s - b)(s - c)}. $$ IB 不强制要求，但比"先用余弦定理求角、再用 $\tfrac{1}{2} a b \sin C$"更省力。
Worked Example C2.4 (area from SAS)C2.4 例题（由 SAS 求面积）

Find the area of a triangle with sides of length $5$ and $7$ meeting at an included angle of $60^{\circ}$.求边长 $5$ 和 $7$、夹角 $60^{\circ}$ 的三角形面积。

Step 1. Identify the SAS data. The two sides are $5$ and $7$; the included angle is $60^{\circ}$. The formula $\tfrac{1}{2} a b \sin C$ applies directly with $a = 5$, $b = 7$, $C = 60^{\circ}$.

第 1 步：识别 SAS。两边 $5$、$7$，夹角 $60^{\circ}$。直接用 $\tfrac{1}{2} a b \sin C$，取 $a = 5$、$b = 7$、$C = 60^{\circ}$。

Step 2. Substitute. Recall $\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$.

第 2 步：代入。记 $\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$。

$$ \text{Area} \;=\; \tfrac{1}{2} \cdot 5 \cdot 7 \cdot \sin 60^{\circ} \;=\; \tfrac{1}{2} \cdot 35 \cdot \tfrac{\sqrt{3}}{2} \;=\; \tfrac{35 \sqrt{3}}{4} \;\approx\; 15.16. $$

The area is $\tfrac{35 \sqrt{3}}{4} \approx 15.16$ square units.

面积为 $\tfrac{35 \sqrt{3}}{4} \approx 15.16$ 平方单位。

Remark: the angle must be included备注：角必须是夹角 If the problem gives two sides and a non-included angle (SSA), the SAS area formula does not apply directly. Either find the third angle (via the sine rule, watching for the ambiguous case) or compute the third side first (also via the sine rule), and only then use $\tfrac{1}{2} a b \sin C$ with a confirmed pair of sides and their included angle.若题目给两边及一非夹角（SSA），不能直接用 SAS 面积公式。须先求第三角（用正弦定理，注意模糊情形）或第三边（亦用正弦定理），确认得到一组"两边加夹角"后再用 $\tfrac{1}{2} a b \sin C$。
A triangle has sides $a = 6$ and $b = 10$ meeting at the included angle $C = 30^{\circ}$. Its area is:三角形两边 $a = 6$、$b = 10$，夹角 $C = 30^{\circ}$。面积为：
C2.4 · Q1
$15$
$30$
$\dfrac{15 \sqrt{3}}{2}$
$60$
Area $= \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 6 \cdot 10 \cdot \sin 30^{\circ} = 30 \cdot \tfrac{1}{2} = 15$.面积 $= \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 6 \cdot 10 \cdot \sin 30^{\circ} = 30 \cdot \tfrac{1}{2} = 15$。
$\tfrac{1}{2} a b \sin C$ with $a = 6$, $b = 10$, $C = 30^{\circ}$ gives $30 \cdot \tfrac{1}{2} = 15$.$\tfrac{1}{2} a b \sin C$，代 $a = 6$、$b = 10$、$C = 30^{\circ}$ 得 $30 \cdot \tfrac{1}{2} = 15$。
Topic C2.5 · Bearings and 2D ApplicationsTopic C2.5 · 方位角与二维应用

Bearings and 2D Applications方位角与二维应用 SL 3.3

Bearing convention. A bearing is the direction of one point as seen from another, measured clockwise from north, in degrees, written as a three-digit number. Examples: due north is $000^{\circ}$, due east is $090^{\circ}$, due south is $180^{\circ}$, due west is $270^{\circ}$, and a direction $30^{\circ}$ east of north is $030^{\circ}$.

The standard workflow.
  1. Draw a clear figure. Mark each location, draw a north arrow at each, and label the bearing as the clockwise angle from north.
  2. Convert bearings to interior angles of the triangle. This is usually a subtraction using $180^{\circ}$ (interior angles of a straight line) or $360^{\circ}$ (full turn).
  3. Match the data pattern (AAS, ASA, SAS, SSS) and apply the sine rule or cosine rule.
  4. Translate the answer back into "distance" or "bearing of $X$ from $Y$" as the question asks.
Most bearing problems become routine once the figure is drawn and the interior angles are labelled. The geometry conversion is the work; the trigonometry that follows is mechanical.
方位角约定。方位角是从一点看另一点的方向，自正北方向起顺时针测量，以度数表示，写成三位数。例：正北 $000^{\circ}$、正东 $090^{\circ}$、正南 $180^{\circ}$、正西 $270^{\circ}$；北偏东 $30^{\circ}$ 写作 $030^{\circ}$。

标准流程。
  1. 画清晰图。标出各地点，在每点画北方向箭头，并把方位角标为自北顺时针的角度。
  2. 把方位角换算为三角形内角。多为以 $180^{\circ}$（直线两侧）或 $360^{\circ}$（一周）作减法。
  3. 识别数据模式（AAS、ASA、SAS、SSS）并套正弦或余弦定理。
  4. 把结果翻译回题目要求的"距离"或"从 $Y$ 看 $X$ 的方位角"。
多数方位角题在画好图、标好内角后就变成例行公事。几何换算是核心工作，后续三角学只是机械步骤。
Worked Example C2.5 (ship navigation)C2.5 例题（船舶导航）

Ship $P$ sails from a port on a bearing of $060^{\circ}$ for $12$ km, then turns and sails on a bearing of $150^{\circ}$ for $8$ km to reach a point $Q$. Find the straight-line distance from the port to $Q$.船 $P$ 从港口出发，以方位角 $060^{\circ}$ 航行 $12$ km，然后转向以方位角 $150^{\circ}$ 航行 $8$ km，抵达 $Q$ 点。求港口到 $Q$ 的直线距离。

Step 1. Draw the figure. Label the port as $O$, the turning point as $T$, and the final point as $Q$. The first leg $OT = 12$ heads on bearing $060^{\circ}$. At $T$, the ship turns onto bearing $150^{\circ}$ for $TQ = 8$.

第 1 步：画图。设港口为 $O$、转向点为 $T$、终点为 $Q$。第一段 $OT = 12$ 沿 $060^{\circ}$ 方位；在 $T$ 处转向 $150^{\circ}$，沿此方向走 $TQ = 8$。

Step 2. Find the interior angle at $T$. The incoming direction at $T$ (from $O$ towards $T$) has bearing $060^{\circ}$; reversed, the bearing from $T$ back to $O$ is $060^{\circ} + 180^{\circ} = 240^{\circ}$. The outgoing direction from $T$ to $Q$ has bearing $150^{\circ}$. The interior angle of the triangle at $T$ is the angle between the directions $T \to O$ (bearing $240^{\circ}$) and $T \to Q$ (bearing $150^{\circ}$):

第 2 步：求 $T$ 处的内角。$T$ 处入向（$O \to T$ 方向）方位 $060^{\circ}$；反向，从 $T$ 回看 $O$ 的方位为 $060^{\circ} + 180^{\circ} = 240^{\circ}$。$T$ 至 $Q$ 出向方位 $150^{\circ}$。三角形在 $T$ 的内角即 $T \to O$（方位 $240^{\circ}$）与 $T \to Q$（方位 $150^{\circ}$）两方向之间的夹角：

$$ \angle O T Q \;=\; 240^{\circ} - 150^{\circ} \;=\; 90^{\circ}. $$

Step 3. Apply the cosine rule (SAS), or recognise the right angle. With $\angle O T Q = 90^{\circ}$, $\cos 90^{\circ} = 0$ and the cosine rule collapses to Pythagoras:

第 3 步：用余弦定理（SAS），或直接识别直角。$\angle O T Q = 90^{\circ}$ 时 $\cos 90^{\circ} = 0$，余弦定理退化为勾股：

$$ O Q^{2} \;=\; O T^{2} + T Q^{2} \;=\; 12^{2} + 8^{2} \;=\; 144 + 64 \;=\; 208. $$ $$ O Q \;=\; \sqrt{208} \;=\; 4 \sqrt{13} \;\approx\; 14.42 \text{ km}. $$

Result. The straight-line distance from the port to $Q$ is approximately $14.4$ km.

结果。港口到 $Q$ 的直线距离约 $14.4$ km。

Pitfall: the bearing of the return leg陷阱：反向方位角 If the bearing of $T$ from $O$ is $\beta$, the bearing of $O$ from $T$ is $\beta + 180^{\circ}$ (mod $360^{\circ}$). This is the single most common bearing-problem slip: students compute the interior angle as the difference of the two outgoing bearings, forgetting that the angle at $T$ involves the incoming direction reversed.若从 $O$ 看 $T$ 的方位角为 $\beta$，则从 $T$ 看 $O$ 的方位角为 $\beta + 180^{\circ}$（模 $360^{\circ}$）。这是方位角题的头号失误：学生常把两个出向方位差当作内角，忘了 $T$ 处的内角涉及入向的反方向。
From a tower, town $A$ is on a bearing of $050^{\circ}$ and town $B$ is on a bearing of $140^{\circ}$. The angle $A T B$ (at the tower) measures:从塔看，城镇 $A$ 方位 $050^{\circ}$，城镇 $B$ 方位 $140^{\circ}$。塔处的角 $A T B$ 为：
C2.5 · Q1
$50^{\circ}$
$140^{\circ}$
$90^{\circ}$
$190^{\circ}$
Both bearings are measured from north at the tower, so the angle between the lines of sight is the difference: $140^{\circ} - 50^{\circ} = 90^{\circ}$.两方位角均自塔所在点的正北顺时针测，故两视线之间的角为差：$140^{\circ} - 50^{\circ} = 90^{\circ}$。
When both bearings are taken from the same point, the angle between the two directions is the difference of the bearings: $140^{\circ} - 50^{\circ} = 90^{\circ}$.两方位角自同一点测时，两方向之间的角即方位差：$140^{\circ} - 50^{\circ} = 90^{\circ}$。
Topic C2.6 · Ambiguous Case and 3D ApplicationsTopic C2.6 · 模糊情形与三维应用

Ambiguous Case (SSA) and 3D Applications模糊情形（SSA）与三维应用 SL 3.3

The ambiguous SSA case. When two sides and a non-included angle are given (SSA), the sine rule yields $\sin \theta = $ (some value). Because $\sin \theta = \sin (180^{\circ} - \theta)$, there are potentially two angles satisfying the equation: the acute candidate $\theta_{1} = \arcsin(\cdot)$ and the obtuse candidate $\theta_{2} = 180^{\circ} - \theta_{1}$.

How many valid triangles?
  • 0 triangles if $\sin \theta > 1$, which is impossible. The data is inconsistent.
  • 1 triangle if $\theta_{2}$ together with the given angle exceeds (or equals) $180^{\circ}$, so only $\theta_{1}$ leaves room for the third angle.
  • 2 triangles if both $\theta_{1}$ and $\theta_{2}$ leave a positive third angle (i.e. their sum with the given angle is less than $180^{\circ}$).
The diagnostic. Check: does $\theta_{2} + (\text{given angle}) < 180^{\circ}$? If yes, both triangles are valid and you must report both. If no, only $\theta_{1}$ is valid.

3D applications. In a three-dimensional figure (prism, pyramid, cube), identify a planar triangle inside the solid that contains the unknown. Apply the planar trigonometry rules within that triangle. Common situations: the diagonal of a cuboid, the slant height of a pyramid, the angle between a line and a face. Often the work is "two right triangles in a row": one in the base plane to get a baseline, then a second perpendicular triangle to get the slant length or angle.
模糊 SSA 情形。给两边加一非夹角（SSA）时，正弦定理给出 $\sin \theta = $（某数值）。由于 $\sin \theta = \sin (180^{\circ} - \theta)$，可能有两个角满足：锐角候选 $\theta_{1} = \arcsin(\cdot)$ 与钝角候选 $\theta_{2} = 180^{\circ} - \theta_{1}$。

有几个有效三角形？
  • 0 个：若 $\sin \theta > 1$，不可能，数据自相矛盾。
  • 1 个：若 $\theta_{2}$ 加上给定角超过（或等于）$180^{\circ}$，只剩 $\theta_{1}$ 给第三角留出空间。
  • 2 个：若 $\theta_{1}$ 与 $\theta_{2}$ 各与给定角之和均小于 $180^{\circ}$，两三角形都成立。
判定要点。检查：$\theta_{2} + (\text{给定角}) < 180^{\circ}$ 否？是则两三角形都成立，须同时报出；否则只 $\theta_{1}$ 有效。

三维应用。在三维图形（棱柱、棱锥、立方体）中，找出包含未知量的某一平面三角形，在其内套用平面三角学定理。常见情境：长方体对角线、棱锥斜高、直线与面所成角。通常分两步走"连接两个直角三角形"：先在底面三角形中求出基线，再在与之垂直的三角形中求斜长或夹角。
Worked Example C2.6a (ambiguous case)C2.6a 例题（模糊情形）

In triangle $ABC$, side $a = 7$, side $b = 9$, and angle $A = 35^{\circ}$. Find the possible values of angle $B$.三角形 $ABC$ 中 $a = 7$、$b = 9$、$A = 35^{\circ}$。求角 $B$ 的所有可能值。

Step 1. Apply the sine rule.

第 1 步：套正弦定理。

$$ \frac{\sin B}{b} \;=\; \frac{\sin A}{a} \;\Longrightarrow\; \sin B \;=\; \frac{b \sin A}{a} \;=\; \frac{9 \sin 35^{\circ}}{7} \;=\; \frac{9 \cdot 0.5736}{7} \;\approx\; 0.7375. $$

Step 2. List both candidates.

第 2 步：列出两个候选。

$$ B_{1} \;=\; \arcsin(0.7375) \;\approx\; 47.5^{\circ}, \qquad B_{2} \;=\; 180^{\circ} - 47.5^{\circ} \;=\; 132.5^{\circ}. $$

Step 3. Check feasibility. Each candidate must leave a positive third angle. $A + B_{1} = 35^{\circ} + 47.5^{\circ} = 82.5^{\circ}$, so $C = 97.5^{\circ}$. Valid. $A + B_{2} = 35^{\circ} + 132.5^{\circ} = 167.5^{\circ}$, so $C = 12.5^{\circ}$. Also valid.

第 3 步：检验可行性。每个候选都需留出正的第三角。$A + B_{1} = 35^{\circ} + 47.5^{\circ} = 82.5^{\circ}$，故 $C = 97.5^{\circ}$，有效。$A + B_{2} = 35^{\circ} + 132.5^{\circ} = 167.5^{\circ}$，故 $C = 12.5^{\circ}$，亦有效。

Result. Both $B \approx 47.5^{\circ}$ and $B \approx 132.5^{\circ}$ produce a valid triangle. The data SSA leaves the triangle ambiguous; both answers must be reported.

结果。$B \approx 47.5^{\circ}$ 与 $B \approx 132.5^{\circ}$ 都给出有效三角形。SSA 数据使三角形模糊，两答案须并列报出。

Worked Example C2.6b (3D, diagonal of a cuboid)C2.6b 例题（三维：长方体对角线）

A rectangular box (cuboid) has edges of length $3$, $4$, and $12$. Find (a) the length of the space diagonal and (b) the angle the space diagonal makes with the base.一长方体棱长为 $3$、$4$、$12$。求 (a) 体对角线长度；(b) 体对角线与底面所成的角。

(a) Length via two perpendicular triangles. First, the diagonal of the base rectangle (a $3 \times 4$ rectangle): by Pythagoras, $d_{\text{base}} = \sqrt{3^{2} + 4^{2}} = 5$. Now the space diagonal is the hypotenuse of a right triangle whose legs are this base diagonal $5$ and the vertical edge $12$:

(a) 利用两个垂直三角形求长度。先求底面（$3 \times 4$ 矩形）对角线：由勾股 $d_{\text{底}} = \sqrt{3^{2} + 4^{2}} = 5$。再将体对角线视为以底面对角线 $5$ 与竖直棱 $12$ 为直角边的直角三角形的斜边：

$$ D \;=\; \sqrt{d_{\text{base}}^{2} + 12^{2}} \;=\; \sqrt{25 + 144} \;=\; \sqrt{169} \;=\; 13. $$

(b) Angle with the base. In the second (vertical) right triangle, the angle $\alpha$ between the space diagonal and its projection (the base diagonal) satisfies

(b) 与底面的夹角。在第二个（竖直方向的）直角三角形中，体对角线与其在底面的投影（即底面对角线）所成角 $\alpha$ 满足

$$ \tan \alpha \;=\; \frac{\text{opposite (vertical edge)}}{\text{adjacent (base diagonal)}} \;=\; \frac{12}{5}, \qquad \alpha \;=\; \arctan \!\left( \frac{12}{5} \right) \;\approx\; 67.4^{\circ}. $$

Method note. The two-step approach (base triangle, then vertical triangle) is the canonical 3D trigonometry pattern. Always isolate one planar triangle at a time.

方法注解。"先底面三角形，再竖直三角形"的两步法是三维三角学的经典模式。每一步只在一个平面三角形内操作。

Pitfall: ignoring the obtuse candidate陷阱：忽略钝角候选 In SSA configurations, students routinely report only $\theta_{1} = \arcsin(\cdot)$ and miss $\theta_{2} = 180^{\circ} - \theta_{1}$. The calculator's $\sin^{-1}$ button returns only the acute candidate by convention. Always test the obtuse candidate against the constraint that the three interior angles sum to $180^{\circ}$ before discarding it.SSA 情形下，学生常只报出 $\theta_{1} = \arcsin(\cdot)$ 而漏掉 $\theta_{2} = 180^{\circ} - \theta_{1}$。计算器 $\sin^{-1}$ 按惯例只返回锐角候选。舍弃钝角候选前，须用"内角和 $= 180^{\circ}$"约束检验。
In triangle $ABC$, $a = 10$, $b = 7$, $A = 50^{\circ}$ (SSA). How many distinct triangles satisfy the data?三角形 $ABC$ 中 $a = 10$、$b = 7$、$A = 50^{\circ}$（SSA）。满足条件的三角形数为：
C2.6 · Q1
$0$
$1$
$2$
infinitely many无穷多个
Sine rule: $\sin B = \tfrac{b \sin A}{a} = \tfrac{7 \sin 50^{\circ}}{10} \approx 0.5362$. Candidates: $B_{1} \approx 32.4^{\circ}$, $B_{2} \approx 147.6^{\circ}$. Test $B_{2}$: $A + B_{2} = 50^{\circ} + 147.6^{\circ} = 197.6^{\circ} > 180^{\circ}$, so the obtuse candidate is rejected. Only $B_{1}$ leaves a positive third angle. One triangle.正弦定理：$\sin B = \tfrac{b \sin A}{a} = \tfrac{7 \sin 50^{\circ}}{10} \approx 0.5362$。候选 $B_{1} \approx 32.4^{\circ}$、$B_{2} \approx 147.6^{\circ}$。检验 $B_{2}$：$A + B_{2} = 50^{\circ} + 147.6^{\circ} = 197.6^{\circ} > 180^{\circ}$，舍。只 $B_{1}$ 留正第三角。一个三角形。
Compute both SSA candidates for $B$. The obtuse candidate $B_{2} \approx 147.6^{\circ}$ added to $A = 50^{\circ}$ exceeds $180^{\circ}$, so it is rejected. Exactly one triangle works.算出两 SSA 候选。钝角候选 $B_{2} \approx 147.6^{\circ}$ 与 $A = 50^{\circ}$ 之和超过 $180^{\circ}$，舍。仅一个三角形成立。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Choosing the right rule (Paper 1 / Paper 2)如何选对定理（Paper 1 / Paper 2）
  • Right angle present? Use SOHCAHTOA. Faster than the general rules.
  • 有直角？用 SOHCAHTOA，比通用定理快。
  • Side-angle pair given (AAS / ASA / SSA)? Sine rule. With SSA, screen for the ambiguous case.
  • 给出"边-对角"对（AAS / ASA / SSA）？正弦定理。SSA 时务必检查模糊情形。
  • No side-angle pair (SAS / SSS)? Cosine rule. For SSS, solve the rule for $\cos A$ explicitly.
  • 无"边-对角"对（SAS / SSS）？余弦定理。SSS 时把定理显式解出 $\cos A$。
Bearings (Paper 2 mostly)方位角（多见于 Paper 2）
  • Draw the figure with north arrows. Without a figure, the conversion from bearings to interior angles is error-prone. With one, it is almost mechanical.
  • 画图并在每点标北。无图时方位角换内角极易出错；有图后几近机械。
  • The return-leg trick. Bearing of $B$ from $A$ is $\beta$ implies bearing of $A$ from $B$ is $\beta + 180^{\circ}$ (mod $360^{\circ}$). Use this to compute interior angles at turning points.
  • 反向方位技巧。从 $A$ 看 $B$ 方位 $\beta$，则从 $B$ 看 $A$ 方位 $\beta + 180^{\circ}$（模 $360^{\circ}$）。转向点的内角靠它推。
Ambiguous case and 3D (Paper 2 most often)模糊情形与三维（多见于 Paper 2）
  • SSA $\Rightarrow$ always check the supplement. Compute both $\theta_{1}$ and $\theta_{2} = 180^{\circ} - \theta_{1}$; reject the candidate whose sum with the given angle exceeds $180^{\circ}$.
  • SSA $\Rightarrow$ 必查补角。同时算 $\theta_{1}$ 与 $\theta_{2} = 180^{\circ} - \theta_{1}$；舍弃与给定角之和超过 $180^{\circ}$ 的候选。
  • 3D problems: identify one planar triangle at a time. Solve it with planar trigonometry, then move to the next triangle. Do not try to apply 2D rules to a 3D figure as a whole.
  • 三维题：一次只取一个平面三角形。在该平面内用平面三角学求解，再进入下一个三角形。不要把二维定理直接套用到整个三维图。
  • For angles between a line and a plane, drop a perpendicular from a point on the line to the plane; the angle is between the line and its projection onto the plane.
  • 线与面所成角：从直线上某点向平面作垂线；该角即直线与其投影之间的角。
Active Recall主动回忆

Flashcards闪卡

0 / 12 flipped已翻 0 / 12
SOH?SOH？
$$\sin \theta = \frac{\text{opp}}{\text{hyp}}$$
CAH?CAH？
$$\cos \theta = \frac{\text{adj}}{\text{hyp}}$$
TOA?TOA？
$$\tan \theta = \frac{\text{opp}}{\text{adj}}$$
Sine rule?正弦定理？
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Cosine rule (find side)?余弦定理（求边）？
$$a^{2} = b^{2} + c^{2} - 2 b c \cos A$$
Cosine rule (find angle)?余弦定理（求角）？
$$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$$
SAS area formula?SAS 面积公式？
$$\text{Area} = \tfrac{1}{2} a b \sin C$$
When to use the sine rule?何时用正弦定理？
AAS, ASA, or SSA: when a side-angle pair is given.AAS、ASA 或 SSA：已知一组"边-对角"对时。
When to use the cosine rule?何时用余弦定理？
SAS or SSS: when no side-angle pair is given.SAS 或 SSS：无"边-对角"对时。
Bearing convention?方位角约定？
Clockwise from north, three-digit. E.g. $060^{\circ}$.自北顺时针、三位数。例：$060^{\circ}$。
Return-leg bearing rule?反向方位规则？
$$\beta_{\text{return}} = \beta + 180^{\circ} \pmod{360^{\circ}}$$
SSA ambiguous candidates?SSA 模糊候选？
$$\theta_{1} = \arcsin(\cdot), \;\; \theta_{2} = 180^{\circ} - \theta_{1}$$Test each against $A + \theta < 180^{\circ}$.分别检验 $A + \theta < 180^{\circ}$。
Mixed Practice综合练习

Unit C2 Practice Quiz单元 C2 练习测验

A ladder of length $5$ m leans against a vertical wall. The foot of the ladder is $3$ m from the base of the wall. The angle the ladder makes with the ground is closest to:长 $5$ m 的梯子斜靠垂直墙，梯脚距墙底 $3$ m。梯子与地面所成角最接近：
Q1
$36.9^{\circ}$
$45^{\circ}$
$53.1^{\circ}$
$60^{\circ}$
The ladder is the hypotenuse ($5$); the $3$ m segment along the ground is adjacent to the angle. $\cos \theta = \tfrac{3}{5} = 0.6$, so $\theta = \arccos(0.6) \approx 53.1^{\circ}$.梯子为斜边（$5$），地面上的 $3$ m 段是该角的邻边。$\cos \theta = \tfrac{3}{5} = 0.6$，故 $\theta = \arccos(0.6) \approx 53.1^{\circ}$。
Adjacent over hypotenuse: $\cos \theta = \tfrac{3}{5}$. Inverse cosine gives $\theta \approx 53.1^{\circ}$.邻边比斜边：$\cos \theta = \tfrac{3}{5}$。反余弦得 $\theta \approx 53.1^{\circ}$。
In triangle $ABC$, $a = 12$, $b = 15$, $C = 110^{\circ}$. The length of side $c$ is closest to:三角形 $ABC$ 中 $a = 12$、$b = 15$、$C = 110^{\circ}$。边 $c$ 的长最接近：
Q2
$18.9$
$22.2$
$27.0$
$15.0$
SAS, so cosine rule for $c$: $c^{2} = a^{2} + b^{2} - 2 a b \cos C = 144 + 225 - 2 \cdot 12 \cdot 15 \cdot \cos 110^{\circ}$. With $\cos 110^{\circ} \approx -0.3420$: $c^{2} \approx 369 - 360 \cdot (-0.3420) \approx 369 + 123.1 \approx 492.1$, so $c \approx 22.2$.SAS，用余弦定理求 $c$：$c^{2} = a^{2} + b^{2} - 2 a b \cos C = 144 + 225 - 2 \cdot 12 \cdot 15 \cdot \cos 110^{\circ}$。$\cos 110^{\circ} \approx -0.3420$，$c^{2} \approx 369 + 123.1 \approx 492.1$，故 $c \approx 22.2$。
SAS pattern. Apply $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ with $\cos 110^{\circ} \approx -0.342$. The negative cosine adds to (rather than subtracts from) $a^{2} + b^{2}$, giving $c \approx 22.2$.SAS 模式。代 $c^{2} = a^{2} + b^{2} - 2 a b \cos C$，$\cos 110^{\circ} \approx -0.342$。负余弦使总和增加（而非减少），得 $c \approx 22.2$。
A triangle has sides $a = 8$ and $b = 11$ with included angle $C = 75^{\circ}$. Its area is closest to:三角形两边 $a = 8$、$b = 11$，夹角 $C = 75^{\circ}$。面积最接近：
Q3
$32$
$38$
$42.5$
$88$
Area $= \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 8 \cdot 11 \cdot \sin 75^{\circ}$. $\sin 75^{\circ} \approx 0.9659$, so Area $\approx 44 \cdot 0.9659 \approx 42.5$.面积 $= \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 8 \cdot 11 \cdot \sin 75^{\circ}$。$\sin 75^{\circ} \approx 0.9659$，面积 $\approx 44 \cdot 0.9659 \approx 42.5$。
Use $\tfrac{1}{2} a b \sin C$ with $a = 8$, $b = 11$, $C = 75^{\circ}$. The answer is about $42.5$.用 $\tfrac{1}{2} a b \sin C$，代 $a = 8$、$b = 11$、$C = 75^{\circ}$。结果约 $42.5$。
A plane flies from airport $P$ on a bearing of $080^{\circ}$ for $300$ km to point $Q$, then turns onto a bearing of $170^{\circ}$ for $400$ km to point $R$. The distance $PR$ is:飞机从机场 $P$ 以方位 $080^{\circ}$ 飞 $300$ km 到 $Q$，再转向方位 $170^{\circ}$ 飞 $400$ km 到 $R$。$PR$ 长度为：
Q4
$100 \text{ km}$
$700 \text{ km}$
$500 \text{ km}$
$350 \text{ km}$
At $Q$, the incoming bearing is $080^{\circ}$, so the bearing of $P$ from $Q$ is $080^{\circ} + 180^{\circ} = 260^{\circ}$. The outgoing bearing to $R$ is $170^{\circ}$. Interior angle at $Q$: $260^{\circ} - 170^{\circ} = 90^{\circ}$. So triangle $PQR$ is right-angled at $Q$, with legs $300$ and $400$. By Pythagoras, $PR = \sqrt{300^{2} + 400^{2}} = \sqrt{250{,}000} = 500$ km.在 $Q$ 处，入向方位 $080^{\circ}$，故从 $Q$ 看 $P$ 方位为 $080^{\circ} + 180^{\circ} = 260^{\circ}$。出向 $R$ 方位 $170^{\circ}$。$Q$ 处内角：$260^{\circ} - 170^{\circ} = 90^{\circ}$。三角形 $PQR$ 在 $Q$ 为直角，两直角边 $300$、$400$。由勾股 $PR = \sqrt{300^{2} + 400^{2}} = 500$ km。
Reverse the incoming bearing ($080^{\circ} \to 260^{\circ}$) to get the bearing of $P$ from $Q$. The interior angle at $Q$ is then $260^{\circ} - 170^{\circ} = 90^{\circ}$. Apply Pythagoras to the $3$-$4$-$5$ scaled triangle to get $PR = 500$ km.把入向方位反转（$080^{\circ} \to 260^{\circ}$）以得从 $Q$ 看 $P$ 的方位。$Q$ 处内角为 $260^{\circ} - 170^{\circ} = 90^{\circ}$。对 $3$-$4$-$5$ 的比例三角形用勾股，得 $PR = 500$ km。
In triangle $ABC$, $a = 6$, $b = 9$, $A = 40^{\circ}$ (SSA). The two valid values of $B$ are closest to:三角形 $ABC$ 中 $a = 6$、$b = 9$、$A = 40^{\circ}$（SSA）。$B$ 的两个有效值最接近：
Q5
$74.6^{\circ}$ only仅 $74.6^{\circ}$
$40^{\circ}$ and $140^{\circ}$$40^{\circ}$ 与 $140^{\circ}$
$74.6^{\circ}$ and $105.4^{\circ}$$74.6^{\circ}$ 与 $105.4^{\circ}$
no valid triangle无有效三角形
$\sin B = \tfrac{b \sin A}{a} = \tfrac{9 \sin 40^{\circ}}{6} = \tfrac{9 \cdot 0.6428}{6} \approx 0.9642$. Candidates: $B_{1} \approx 74.6^{\circ}$, $B_{2} = 180^{\circ} - 74.6^{\circ} \approx 105.4^{\circ}$. Test each: $A + B_{1} = 40^{\circ} + 74.6^{\circ} = 114.6^{\circ} < 180^{\circ}$ (valid); $A + B_{2} = 40^{\circ} + 105.4^{\circ} = 145.4^{\circ} < 180^{\circ}$ (also valid). Both triangles exist.$\sin B = \tfrac{b \sin A}{a} = \tfrac{9 \sin 40^{\circ}}{6} \approx 0.9642$。候选 $B_{1} \approx 74.6^{\circ}$、$B_{2} \approx 105.4^{\circ}$。检验：$A + B_{1} = 114.6^{\circ} < 180^{\circ}$（有效）；$A + B_{2} = 145.4^{\circ} < 180^{\circ}$（亦有效）。两三角形都存在。
SSA: get $\sin B \approx 0.9642$ via the sine rule. Both $B_{1} = \arcsin(\cdot) \approx 74.6^{\circ}$ and $B_{2} = 180^{\circ} - B_{1} \approx 105.4^{\circ}$ satisfy $A + B < 180^{\circ}$, so both are valid.SSA：用正弦定理得 $\sin B \approx 0.9642$。$B_{1} \approx 74.6^{\circ}$ 与 $B_{2} \approx 105.4^{\circ}$ 都满足 $A + B < 180^{\circ}$，均有效。
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