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Unit C2 · SolutionsUnit C2 · 解析

Trigonometry and its Applications · Solutions三角学及其应用 · 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2

Syllabus SL 3.3 (C2.1 to C2.6)考纲 SL 3.3（C2.1 至 C2.6）AA HL

C2 Toolkit (referenced throughout):C2 工具箱（全卷通用）：
PART I  ·  PAPER 1 SECTION A · SOLUTIONS第一部分  ·  第一卷 A 节 · 解析No calculator · 21 marks不可使用计算器 · 21 分

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1AC2.1 Right-Angled Trigonometry[4 marks]

Right triangle $PQR$ with right angle at $R$, $PQ = 13$, $QR = 5$. Find $PR$ exactly; write $\sin P,\cos P,\tan P$.直角三角形 $PQR$，直角在 $R$，$PQ = 13$，$QR = 5$。精确求 $PR$，并写出 $\sin P,\cos P,\tan P$。

Answers:答案：  (a) $PR = 12$  ·  (b) $\sin P = \tfrac{5}{13},\;\cos P = \tfrac{12}{13},\;\tan P = \tfrac{5}{12}$

(a) Pythagoras M1·A1

Hypotenuse $PQ = 13$, leg $QR = 5$. The remaining leg $PR$ satisfies $PR^{2} + QR^{2} = PQ^{2}$: $$ PR^{2} \;=\; 13^{2} - 5^{2} \;=\; 169 - 25 \;=\; 144 \;\Longrightarrow\; PR \;=\; 12. $$ (Recognise the Pythagorean triple $5{:}12{:}13$.)

(b) SOHCAHTOA at vertex $P$ A1·A1

At $P$: opposite $= QR = 5$, adjacent $= PR = 12$, hypotenuse $= PQ = 13$. Hence $$ \sin P = \tfrac{\text{opp}}{\text{hyp}} = \tfrac{5}{13}, \qquad \cos P = \tfrac{\text{adj}}{\text{hyp}} = \tfrac{12}{13}, \qquad \tan P = \tfrac{\text{opp}}{\text{adj}} = \tfrac{5}{12}. $$ Check: $\sin^{2}P + \cos^{2}P = \tfrac{25 + 144}{169} = 1$. $\checkmark$ Independent ratio check: $\tan P = \tfrac{\sin P}{\cos P} = \tfrac{5/13}{12/13} = \tfrac{5}{12}$. $\checkmark$
Identify "opposite" by walking away from the vertex. At any vertex of a right triangle, the "opposite" side is the leg you do not touch. Standing at $P$, you touch $PR$ (adjacent leg) and $PQ$ (hypotenuse); the side you do not touch is $QR$, the opposite. This rule eliminates the second most common Q1 mistake: swapping opposite and adjacent. The most common mistake is forgetting that the hypotenuse is always opposite the right angle, never opposite the angle you are computing — here $PQ = 13$ is the hypotenuse because the right angle is at $R$.

(a) 勾股定理 M1·A1

斜边 $PQ = 13$，一直角边 $QR = 5$。另一直角边 $PR$ 满足 $PR^{2} + QR^{2} = PQ^{2}$： $$ PR^{2} \;=\; 13^{2} - 5^{2} \;=\; 169 - 25 \;=\; 144 \;\Longrightarrow\; PR \;=\; 12. $$ （认出毕氏三元组 $5{:}12{:}13$。）

(b) 在顶点 $P$ 用 SOHCAHTOA A1·A1

在 $P$：对边 $= QR = 5$，邻边 $= PR = 12$，斜边 $= PQ = 13$。故 $$ \sin P = \tfrac{\mathrm{opp}}{\mathrm{hyp}} = \tfrac{5}{13}, \qquad \cos P = \tfrac{\mathrm{adj}}{\mathrm{hyp}} = \tfrac{12}{13}, \qquad \tan P = \tfrac{\mathrm{opp}}{\mathrm{adj}} = \tfrac{5}{12}. $$ 核对：$\sin^{2}P + \cos^{2}P = \tfrac{25 + 144}{169} = 1$。$\checkmark$ 独立比例核对：$\tan P = \tfrac{\sin P}{\cos P} = \tfrac{5/13}{12/13} = \tfrac{5}{12}$。$\checkmark$
用"远离顶点"判定"对边"。在直角三角形的任一顶点，"对边"就是你没碰到的那条边。站在 $P$，触及 $PR$（邻边）和 $PQ$（斜边）；未触及的 $QR$ 即对边。此规则消除 Q1 第二常见错误：对边/邻边互换。最常见错误是忘了斜边永远与直角相对，而非与待求角相对 — 此处直角在 $R$，故 $PQ = 13$ 才是斜边。
Q2MEDIUMPaper 1AC2.2 Sine Rule (AAS)[5 marks]

$\triangle ABC$ with $A = 40^{\circ}$, $B = 60^{\circ}$, $a = 8$. Find $C$; express $b$; show area $= \tfrac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}$.$\triangle ABC$ 中 $A = 40^{\circ}$、$B = 60^{\circ}$、$a = 8$。求 $C$；表示 $b$；证明面积 $= \tfrac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}$。

Answers:答案：  (a) $C = 80^{\circ}$  ·  (b) $b = \dfrac{8 \sin 60^{\circ}}{\sin 40^{\circ}}$, $k = 8$  ·  (c) shown已证

(a) Angle sum A1

$A + B + C = 180^{\circ}\;\Rightarrow\; C = 180^{\circ} - 40^{\circ} - 60^{\circ} = 80^{\circ}$.

(b) Sine rule for $b$ M1·A1

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$. Substitute: $$ \frac{8}{\sin 40^{\circ}} \;=\; \frac{b}{\sin 60^{\circ}} \;\Longrightarrow\; b \;=\; \frac{8 \sin 60^{\circ}}{\sin 40^{\circ}}. $$ So $k = 8$.

(c) Area via $\tfrac{1}{2} a b \sin C$ M1·A1

Using sides $a, b$ and included angle $C = 80^{\circ}$: $$ \text{Area} \;=\; \tfrac{1}{2} a b \sin C \;=\; \tfrac{1}{2} \cdot 8 \cdot \frac{8 \sin 60^{\circ}}{\sin 40^{\circ}} \cdot \sin 80^{\circ} \;=\; \frac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}. \;\blacksquare $$
The sine rule is one equation, two unknowns — quote it that way. Examiners want to see both ratios written down before substitution: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ earns the M1 even if the arithmetic stalls. Writing "$b = \tfrac{a \sin B}{\sin A}$" with no setup costs the method mark. Second trap: candidates evaluate $\sin 40^{\circ}, \sin 60^{\circ}$ on Paper 1 and lose the exact-form A1. On a "Show that" with surds and sines, do not multiply out — the target form is the destination, leave it intact.

(a) 内角和 A1

$A + B + C = 180^{\circ}\;\Rightarrow\; C = 180^{\circ} - 40^{\circ} - 60^{\circ} = 80^{\circ}$。

(b) 正弦定理求 $b$ M1·A1

正弦定理：$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$。代入： $$ \frac{8}{\sin 40^{\circ}} \;=\; \frac{b}{\sin 60^{\circ}} \;\Longrightarrow\; b \;=\; \frac{8 \sin 60^{\circ}}{\sin 40^{\circ}}. $$ 故 $k = 8$。

(c) 用 $\tfrac{1}{2} a b \sin C$ 求面积 M1·A1

用 $a, b$ 及夹角 $C = 80^{\circ}$： $$ \text{Area} \;=\; \tfrac{1}{2} a b \sin C \;=\; \tfrac{1}{2} \cdot 8 \cdot \frac{8 \sin 60^{\circ}}{\sin 40^{\circ}} \cdot \sin 80^{\circ} \;=\; \frac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}. \;\blacksquare $$
正弦定理是一式两未知 — 引用时写完整。评卷期望看到代入前写出双侧比例：$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ 就拿 M1，即便后续算错。直接写 "$b = \tfrac{a \sin B}{\sin A}$" 不展开会丢方法分。第二陷阱：Paper 1 上把 $\sin 40^{\circ}, \sin 60^{\circ}$ 求出数值，会丢"精确形式" A1。"Show that"+ 根号 + 正弦时，不要乘开 — 目标式就是终点，原样保留。
Q3MEDIUMPaper 1AC2.3 Cosine Rule (SAS)[6 marks]

$\triangle ABC$ with $b = 5$, $c = 7$, $A = 60^{\circ}$. Find $a$ exactly, the exact area, and $bc\sin A$ in surd form.$\triangle ABC$ 中 $b = 5$、$c = 7$、$A = 60^{\circ}$。精确求 $a$、精确面积、$bc\sin A$（根号形式）。

Answers:答案：  (a) $a = \sqrt{39}$  ·  (b) $\text{Area} = \tfrac{35\sqrt{3}}{4}$  ·  (c) $bc \sin A = \tfrac{35\sqrt{3}}{2}$

(a) Cosine rule for $a$ M1·A1·A1

Cosine rule (side opposite the known angle): $a^{2} = b^{2} + c^{2} - 2bc\cos A$. With $\cos 60^{\circ} = \tfrac{1}{2}$: $$ a^{2} \;=\; 25 + 49 - 2 \cdot 5 \cdot 7 \cdot \tfrac{1}{2} \;=\; 74 - 35 \;=\; 39 \;\Longrightarrow\; a \;=\; \sqrt{39}. $$ ($a > 0$, so the positive root.)

(b) Area via $\tfrac{1}{2}bc\sin A$ M1·A1

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

(c) $bc \sin A$ A1

$bc \sin A = 2 \cdot \text{Area} = 2 \cdot \tfrac{35\sqrt{3}}{4} = \tfrac{35\sqrt{3}}{2}$.
Match the cosine rule to the angle, not to the side you want. The IB-standard form is $a^{2} = b^{2} + c^{2} - 2bc\cos A$: the side on the left is opposite the angle in the cosine. Candidates routinely write $b^{2} = a^{2} + c^{2} - 2ac\cos A$ when given angle $A$ and asked for $a$, which is the wrong rearrangement. Memorise the slogan: "angle on the right, opposite-side squared on the left". For exact-value problems on Paper 1, sub $\cos A$ before simplifying $b^{2} + c^{2}$ — this keeps the surd structure transparent and is the path the markscheme follows.

(a) 余弦定理求 $a$ M1·A1·A1

余弦定理（左边是已知角的对边）：$a^{2} = b^{2} + c^{2} - 2bc\cos A$。$\cos 60^{\circ} = \tfrac{1}{2}$： $$ a^{2} \;=\; 25 + 49 - 2 \cdot 5 \cdot 7 \cdot \tfrac{1}{2} \;=\; 74 - 35 \;=\; 39 \;\Longrightarrow\; a \;=\; \sqrt{39}. $$ （$a > 0$ 故取正根。）

(b) 用 $\tfrac{1}{2}bc\sin A$ 求面积 M1·A1

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

(c) $bc \sin A$ A1

$bc \sin A = 2 \cdot \text{Area} = 2 \cdot \tfrac{35\sqrt{3}}{4} = \tfrac{35\sqrt{3}}{2}$。
余弦定理按"角"对，不是按"想求的边"对。IB 标准式 $a^{2} = b^{2} + c^{2} - 2bc\cos A$：左边的边对着余弦里的角。学生常在已知 $A$ 求 $a$ 时误写 $b^{2} = a^{2} + c^{2} - 2ac\cos A$，是错排。记口诀："右边角，左边对边平方"。Paper 1 精确值题中，先代 $\cos A$ 再简化 $b^{2} + c^{2}$ — 这样根号结构清晰，也是评卷标准路径。
Q4HARDPaper 1AC2.6 Ambiguous SSA Case[6 marks]

$\triangle ABC$ with $a = 8$, $b = 10$, $A = 30^{\circ}$. Show $\sin B = \tfrac{5}{8}$; find both $B$ values; verify each is valid and state corresponding $C$.$\triangle ABC$ 中 $a = 8$、$b = 10$、$A = 30^{\circ}$。证 $\sin B = \tfrac{5}{8}$；求两 $B$ 值；验证均有效并写出 $C$。

Answers:答案：  (b) $B_{1} \approx 38.7^{\circ},\; B_{2} \approx 141.3^{\circ}$  ·  (c) $C_{1} \approx 111.3^{\circ},\; C_{2} \approx 8.7^{\circ}$ (both valid均有效)

(a) Sine rule for $\sin B$ M1·A1

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\;\Rightarrow\;\sin B = \dfrac{b \sin A}{a} = \dfrac{10 \sin 30^{\circ}}{8} = \dfrac{10 \cdot \tfrac{1}{2}}{8} = \dfrac{5}{8}$. $\blacksquare$

(b) Two candidates for $B$ M1·A1

$B_{1} = \arcsin\tfrac{5}{8} \approx 38.7^{\circ}$ (acute, principal value); $B_{2} = 180^{\circ} - B_{1} \approx 141.3^{\circ}$ (obtuse). Both have the same sine, since $\sin\theta = \sin(180^{\circ} - \theta)$.

(c) Validity check and $C$ A1·R1

Each $B$ is valid iff $A + B < 180^{\circ}$ (so the third angle $C = 180^{\circ} - A - B$ is positive):
  • $B_{1} \approx 38.7^{\circ}$: $A + B_{1} \approx 68.7^{\circ} < 180^{\circ}$. $\checkmark$ $C_{1} = 180^{\circ} - 30^{\circ} - 38.7^{\circ} \approx 111.3^{\circ}$.
  • $B_{2} \approx 141.3^{\circ}$: $A + B_{2} \approx 171.3^{\circ} < 180^{\circ}$. $\checkmark$ $C_{2} = 180^{\circ} - 30^{\circ} - 141.3^{\circ} \approx 8.7^{\circ}$.
Hence the SSA data admit two distinct triangles. Geometrically: pivot side $b = 10$ from $A$ at angle $30^{\circ}$ to $AC$; the circle of radius $a = 8$ centred at $C$ crosses ray $AC$ at two points, giving two positions for $B$.
The SSA decision tree. Given two sides and a non-included angle (here $a$, $b$, $A$), the number of valid triangles depends on the ratio $\sin B = \tfrac{b \sin A}{a}$:
  • If $\sin B > 1$: no triangle (the side $a$ is too short to reach line $AC$).
  • If $\sin B = 1$: one right triangle ($B = 90^{\circ}$).
  • If $\sin B < 1$ and $a \ge b$: one triangle (the obtuse $B_{2}$ would force $A + B_{2} \ge 180^{\circ}$, rejected).
  • If $\sin B < 1$ and $a < b$: two triangles (this question: $a = 8 < 10 = b$).
Students who quote only $B_{1} = \arcsin\tfrac{5}{8}$ and stop lose the second triangle's marks; students who quote both without the validity check lose R1. The check "$A + B < 180^{\circ}$" is what separates a guess from a proof.

(a) 正弦定理求 $\sin B$ M1·A1

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\;\Rightarrow\;\sin B = \dfrac{b \sin A}{a} = \dfrac{10 \sin 30^{\circ}}{8} = \dfrac{10 \cdot \tfrac{1}{2}}{8} = \dfrac{5}{8}$。$\blacksquare$

(b) $B$ 的两个候选值 M1·A1

$B_{1} = \arcsin\tfrac{5}{8} \approx 38.7^{\circ}$（锐角，主值）；$B_{2} = 180^{\circ} - B_{1} \approx 141.3^{\circ}$（钝角）。两者正弦相同，因 $\sin\theta = \sin(180^{\circ} - \theta)$。

(c) 有效性核对与 $C$ A1·R1

每个 $B$ 有效当且仅当 $A + B < 180^{\circ}$（使第三角 $C = 180^{\circ} - A - B$ 为正）：
  • $B_{1} \approx 38.7^{\circ}$：$A + B_{1} \approx 68.7^{\circ} < 180^{\circ}$。$\checkmark$ $C_{1} = 180^{\circ} - 30^{\circ} - 38.7^{\circ} \approx 111.3^{\circ}$。
  • $B_{2} \approx 141.3^{\circ}$：$A + B_{2} \approx 171.3^{\circ} < 180^{\circ}$。$\checkmark$ $C_{2} = 180^{\circ} - 30^{\circ} - 141.3^{\circ} \approx 8.7^{\circ}$。
故 SSA 数据给出两个不同三角形。几何上：以 $A$ 为顶点把 $b = 10$ 与 $AC$ 成 $30^{\circ}$ 摆放；以 $C$ 为心、$a = 8$ 为半径的圆与射线 $AC$ 交两点，给出 $B$ 的两个位置。
SSA 决策树。给定两边与非夹角（此处 $a$、$b$、$A$），有效三角形数依比值 $\sin B = \tfrac{b \sin A}{a}$ 而定：
  • $\sin B > 1$：无解（$a$ 太短，够不到 $AC$ 边）。
  • $\sin B = 1$：唯一直角三角形（$B = 90^{\circ}$）。
  • $\sin B < 1$ 且 $a \ge b$：唯一三角形（钝角 $B_{2}$ 会使 $A + B_{2} \ge 180^{\circ}$，舍）。
  • $\sin B < 1$ 且 $a < b$：两个三角形（本题 $a = 8 < 10 = b$）。
仅写 $B_{1} = \arcsin\tfrac{5}{8}$ 就收笔会丢第二个三角形的分；两值都写却不核对有效性会丢 R1。"$A + B < 180^{\circ}$"这步把"猜"变成"证"。
PART II  ·  PAPER 1 SECTION B · SOLUTIONS第二部分  ·  第一卷 B 节 · 解析No calculator · 11 marks不可使用计算器 · 11 分

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BC2.3 + C2.4 Area, Cosine Rule, Inradius, Circumradius[11 marks]

$\triangle ABC$ with $a = 5$, $b = 8$, $C = 60^{\circ}$. Find area; find $c$; semiperimeter $s$; inradius $r = \tfrac{\text{Area}}{s}$; circumradius $R = \tfrac{abc}{4 \cdot \text{Area}}$.$\triangle ABC$ 中 $a = 5$、$b = 8$、$C = 60^{\circ}$。求面积、$c$、半周长 $s$、内切圆半径 $r = \tfrac{\text{Area}}{s}$、外接圆半径 $R = \tfrac{abc}{4 \cdot \text{Area}}$。

Answers:答案：  (a) $10\sqrt{3}$  ·  (b) $c = 7$  ·  (c) $s = 10$  ·  (d) $r = \sqrt{3}$  ·  (e) $R = \tfrac{7\sqrt{3}}{3}$

(a) Area $= \tfrac{1}{2}ab\sin C$ M1·A1

$\text{Area} = \tfrac{1}{2} \cdot 5 \cdot 8 \cdot \sin 60^{\circ} = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$.

(b) Cosine rule for $c$ M1·A1·A1

$c^{2} = a^{2} + b^{2} - 2ab\cos C = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \tfrac{1}{2} = 89 - 40 = 49 \;\Rightarrow\; c = 7$.

(c) Perimeter and semiperimeter A1

Perimeter $= 5 + 8 + 7 = 20$; $s = \tfrac{20}{2} = 10$.

(d) Inradius M1·A1

$r = \dfrac{\text{Area}}{s} = \dfrac{10\sqrt{3}}{10} = \sqrt{3}$.

(e) Circumradius M1·A1·A1

$$ R \;=\; \frac{abc}{4 \cdot \text{Area}} \;=\; \frac{5 \cdot 8 \cdot 7}{4 \cdot 10\sqrt{3}} \;=\; \frac{280}{40\sqrt{3}} \;=\; \frac{7}{\sqrt{3}} \;=\; \frac{7\sqrt{3}}{3}. $$ (Rationalised: multiply by $\tfrac{\sqrt{3}}{\sqrt{3}}$.) Numerically $R \approx 4.04$, and the inradius $r = \sqrt{3} \approx 1.73$ satisfies $r < R$ as required by Euler's inequality $R \ge 2r$ ($4.04 \ge 3.46$). $\checkmark$
Cross-check with the extended sine rule. The HL identity $\dfrac{c}{\sin C} = 2R$ gives an independent route to $R$: $2R = \dfrac{7}{\sin 60^{\circ}} = \dfrac{7}{\sqrt{3}/2} = \dfrac{14}{\sqrt{3}} = \dfrac{14\sqrt{3}}{3}$, so $R = \dfrac{7\sqrt{3}}{3}$. $\checkmark$ Whenever a Paper 1B problem hands you "$R = \tfrac{abc}{4 \cdot \text{Area}}$" and you have a side opposite a nice angle, the extended sine rule is faster and self-checking. The two formulas are equivalent via the area identity Area $= \tfrac{abc}{4R}$.

(a) 面积 $= \tfrac{1}{2}ab\sin C$ M1·A1

$\text{Area} = \tfrac{1}{2} \cdot 5 \cdot 8 \cdot \sin 60^{\circ} = 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3}$。

(b) 余弦定理求 $c$ M1·A1·A1

$c^{2} = a^{2} + b^{2} - 2ab\cos C = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \tfrac{1}{2} = 89 - 40 = 49 \;\Rightarrow\; c = 7$。

(c) 周长与半周长 A1

周长 $= 5 + 8 + 7 = 20$；$s = \tfrac{20}{2} = 10$。

(d) 内切圆半径 M1·A1

$r = \dfrac{\text{Area}}{s} = \dfrac{10\sqrt{3}}{10} = \sqrt{3}$。

(e) 外接圆半径 M1·A1·A1

$$ R \;=\; \frac{abc}{4 \cdot \text{Area}} \;=\; \frac{5 \cdot 8 \cdot 7}{4 \cdot 10\sqrt{3}} \;=\; \frac{280}{40\sqrt{3}} \;=\; \frac{7}{\sqrt{3}} \;=\; \frac{7\sqrt{3}}{3}. $$ （有理化：乘 $\tfrac{\sqrt{3}}{\sqrt{3}}$。）数值上 $R \approx 4.04$，内切圆 $r = \sqrt{3} \approx 1.73$ 满足 $r < R$，与 Euler 不等式 $R \ge 2r$ 吻合（$4.04 \ge 3.46$）。$\checkmark$
用扩展正弦定理交叉验证。HL 恒等式 $\dfrac{c}{\sin C} = 2R$ 给出独立路径：$2R = \dfrac{7}{\sin 60^{\circ}} = \dfrac{7}{\sqrt{3}/2} = \dfrac{14}{\sqrt{3}} = \dfrac{14\sqrt{3}}{3}$，故 $R = \dfrac{7\sqrt{3}}{3}$。$\checkmark$ Paper 1B 凡给出 "$R = \tfrac{abc}{4 \cdot \text{Area}}$" 且某边对一个"好角"时，扩展正弦定理更快且自校。两式经面积恒等式 Area $= \tfrac{abc}{4R}$ 等价。
PART III  ·  PAPER 2 · SOLUTIONS第三部分  ·  第二卷 · 解析Calculator · 18 marks可使用计算器 · 18 分

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2C2.5 Bearings (2D Navigation)[8 marks]

Ship: $A \to B$ on bearing $060^{\circ}$, $100$ km; $B \to C$ on bearing $150^{\circ}$, $80$ km. Show $\widehat{ABC} = 90^{\circ}$; find $AC$; find bearing of $C$ from $A$.船：$A \to B$ 方位角 $060^{\circ}$、$100$ km；$B \to C$ 方位角 $150^{\circ}$、$80$ km。证 $\widehat{ABC} = 90^{\circ}$；求 $AC$；求 $C$ 相对 $A$ 的方位角。

Answers:答案：  (a) shown已证  ·  (b) $AC = 20\sqrt{41} \approx 128$ km  ·  (c) bearing方位角 $\approx 098.7^{\circ}$

(a) Interior angle at $B$ M1·A1·A1

At $B$, draw the north line. The direction from $B$ back to $A$ is the reverse of the bearing $060^{\circ}$, namely $060^{\circ} + 180^{\circ} = 240^{\circ}$. The direction from $B$ onward to $C$ is the bearing $150^{\circ}$. The interior angle $\widehat{ABC}$ is the angle between rays $BA$ and $BC$: $$ \widehat{ABC} \;=\; 240^{\circ} - 150^{\circ} \;=\; 90^{\circ}. \;\blacksquare $$

(b) Pythagoras for $AC$ M1·A1

With $\widehat{ABC} = 90^{\circ}$, $\triangle ABC$ is right-angled at $B$ with legs $AB = 100$, $BC = 80$: $$ AC \;=\; \sqrt{100^{2} + 80^{2}} \;=\; \sqrt{10000 + 6400} \;=\; \sqrt{16400}. $$ Simplify the surd: $16400 = 400 \cdot 41$, so $\sqrt{16400} = 20\sqrt{41}$. Decimal: $20\sqrt{41} \approx 20 \cdot 6.4031 \approx 128.06 \approx 128$ km (3 sf).

(c) Bearing of $C$ from $A$ M1·A1·A1

At $A$, the bearing of $B$ is $060^{\circ}$. The angle $\widehat{BAC}$ inside the right triangle satisfies $\tan(\widehat{BAC}) = \dfrac{BC}{AB} = \dfrac{80}{100} = 0.8$, so $\widehat{BAC} = \arctan 0.8 \approx 38.66^{\circ}$. From $A$, ray $AC$ is rotated $\widehat{BAC}$ clockwise from ray $AB$ (since the leg $BC$ continues clockwise of $AB$ at $B$). Hence $$ \text{bearing of } C \text{ from } A \;=\; 060^{\circ} + 38.66^{\circ} \;=\; 098.66^{\circ} \;\approx\; 098.7^{\circ}. $$ (Three-digit format: write as $098.7^{\circ}$, not $98.7^{\circ}$.)
The "north-line + reverse bearing" trick. Interior angles of bearings triangles never come for free. The reliable procedure: (1) at each waypoint, draw a north arrow; (2) annotate the bearing into the waypoint as "$\theta + 180^{\circ}$ from north" — this is the reverse bearing; (3) the interior angle equals "reverse-bearing-in minus bearing-out", reduced mod $360^{\circ}$ and taken to be at most $180^{\circ}$. Here $240^{\circ} - 150^{\circ} = 90^{\circ}$. Skipping the north line is the single biggest source of wrong-sign / wrong-angle answers in Paper 2 bearings problems. Second hazard: bearings are always three digits, clockwise from north: $098.7^{\circ}$ keeps the leading zero, never $98.7^{\circ}$ — an A1 in itself.

(a) $B$ 处内角 M1·A1·A1

在 $B$ 画"北"线。$B$ 返向 $A$ 的方向为 $060^{\circ}$ 的反向 $060^{\circ} + 180^{\circ} = 240^{\circ}$；$B$ 续向 $C$ 的方向为 $150^{\circ}$。内角 $\widehat{ABC}$ 为射线 $BA$ 与 $BC$ 的夹角： $$ \widehat{ABC} \;=\; 240^{\circ} - 150^{\circ} \;=\; 90^{\circ}. \;\blacksquare $$

(b) 用勾股求 $AC$ M1·A1

$\widehat{ABC} = 90^{\circ}$，故 $\triangle ABC$ 在 $B$ 处直角，两腿 $AB = 100$、$BC = 80$： $$ AC \;=\; \sqrt{100^{2} + 80^{2}} \;=\; \sqrt{10000 + 6400} \;=\; \sqrt{16400}. $$ 化简根号：$16400 = 400 \cdot 41$，故 $\sqrt{16400} = 20\sqrt{41}$。十进制：$20\sqrt{41} \approx 20 \cdot 6.4031 \approx 128.06 \approx 128$ km（3 位有效数字）。

(c) $C$ 相对 $A$ 的方位角 M1·A1·A1

在 $A$，$B$ 的方位角为 $060^{\circ}$。直角三角形内的角 $\widehat{BAC}$ 满足 $\tan(\widehat{BAC}) = \dfrac{BC}{AB} = \dfrac{80}{100} = 0.8$，故 $\widehat{BAC} = \arctan 0.8 \approx 38.66^{\circ}$。射线 $AC$ 相对 $AB$ 向顺时针偏转 $\widehat{BAC}$（因 $BC$ 边相对 $AB$ 顺时针延续），故 $C$ 相对 $A$ 的方位角： $$ 060^{\circ} + 38.66^{\circ} \;=\; 098.66^{\circ} \;\approx\; 098.7^{\circ}. $$ （三位数格式：$098.7^{\circ}$，不是 $98.7^{\circ}$。）
"北线 + 反方位角" 技巧。方位角题的内角从来不是免费给的。可靠步骤：(1) 每个测站画北箭头；(2) 入测站的方位角标记为"自北 $\theta + 180^{\circ}$"，即反方位角；(3) 内角 $=$ "入站反方位 $-$ 出站方位"，按 $360^{\circ}$ 取模，取不超过 $180^{\circ}$ 的值。此处 $240^{\circ} - 150^{\circ} = 90^{\circ}$。漏掉北线是 Paper 2 方位题最大失分点。第二陷阱：方位角必须三位数、自北顺时针：写 $098.7^{\circ}$ 保留前导零，绝不写 $98.7^{\circ}$ — 这本身就是一个 A1。
Q7HARDPaper 2C2.6 3D Trigonometry (Square Pyramid)[10 marks]

Right pyramid: square base $PQRS$ side $6$, apex $T$ above centre $O$, slant edge $10$. Find $OP$ and height $TO$; angle $TP$ to base; apex angle $\widehat{PTQ}$; dihedral $\widehat{TMO}$ where $M$ is midpoint of $PQ$.正四棱锥：底面 $PQRS$ 边长 $6$，顶 $T$ 在中心 $O$ 上方，斜棱 $10$。求 $OP$ 与高 $TO$；$TP$ 与底面所成角；顶角 $\widehat{PTQ}$；二面角 $\widehat{TMO}$（$M$ 为 $PQ$ 中点）。

Answers:答案：  (a) $OP = 3\sqrt{2},\; TO = \sqrt{82} \approx 9.06$  ·  (b) $\approx 64.9^{\circ}$  ·  (c) $\widehat{PTQ} \approx 34.9^{\circ}$  ·  (d) $\widehat{TMO} \approx 71.7^{\circ}$

(a) Half-diagonal $OP$ and height $TO$ M1·A1·A1

The diagonal of a square of side $6$ is $6\sqrt{2}$ (Pythagoras: $\sqrt{6^{2} + 6^{2}}$). Centre $O$ is the midpoint, so $OP = \tfrac{1}{2} \cdot 6\sqrt{2} = 3\sqrt{2}$. $\triangle TOP$ is right-angled at $O$ (apex directly above centre). Hypotenuse $TP = 10$, leg $OP = 3\sqrt{2}$: $$ TO^{2} \;=\; TP^{2} - OP^{2} \;=\; 100 - 18 \;=\; 82 \;\Longrightarrow\; TO \;=\; \sqrt{82} \;\approx\; 9.06 \text{ cm}. $$

(b) Angle slant-edge to base M1·A1

The angle between line $TP$ and the base is the angle $\widehat{TPO}$ in right triangle $TOP$: $$ \tan(\widehat{TPO}) \;=\; \frac{TO}{OP} \;=\; \frac{\sqrt{82}}{3\sqrt{2}} \;=\; \frac{\sqrt{41}}{3} \;\approx\; 2.134, \quad \widehat{TPO} \;\approx\; 64.9^{\circ}. $$

(c) Apex angle $\widehat{PTQ}$ via cosine rule M1·A1·A1

Triangle $TPQ$ has $TP = TQ = 10$ and $PQ = 6$ (a base edge). Cosine rule with angle $\widehat{PTQ}$ opposite $PQ$: $$ \cos(\widehat{PTQ}) \;=\; \frac{TP^{2} + TQ^{2} - PQ^{2}}{2 \cdot TP \cdot TQ} \;=\; \frac{100 + 100 - 36}{200} \;=\; \frac{164}{200} \;=\; 0.82. $$ Hence $\widehat{PTQ} = \arccos 0.82 \approx 34.9^{\circ}$. Alternative via isoceles split: drop altitude from $T$ to midpoint $M$ of $PQ$, giving right triangle $TMP$ with $TP = 10$, $PM = 3$; then $\sin(\tfrac{1}{2}\widehat{PTQ}) = \tfrac{3}{10} = 0.3$, so $\tfrac{1}{2}\widehat{PTQ} \approx 17.46^{\circ}$ and $\widehat{PTQ} \approx 34.9^{\circ}$. $\checkmark$

(d) Dihedral $\widehat{TMO}$ M1·A1

$M$ = midpoint of $PQ$. $TM$ is the slant height of face $TPQ$ (from apex to base edge); $OM$ is the in-base perpendicular from centre to that edge.
  • $TM$: in isosceles $\triangle TPQ$, the altitude from $T$ to $PQ$ has foot $M$. $TM^{2} = TP^{2} - PM^{2} = 100 - 3^{2} = 91$, so $TM = \sqrt{91}$.
  • $OM = 3$ (half the base side, since $O$ is centre and $M$ is midpoint of a side).
  • $TO = \sqrt{82}$ from (a), perpendicular to the base.
$\triangle TMO$ is right-angled at $O$ (the height $TO$ is perpendicular to $OM$ which lies in the base): $$ \tan(\widehat{TMO}) \;=\; \frac{TO}{OM} \;=\; \frac{\sqrt{82}}{3} \;\approx\; 3.018, \quad \widehat{TMO} \;\approx\; 71.7^{\circ}. $$
"Angle of a line to a plane" $=$ angle to its projection. In (b), the angle the slant edge $TP$ makes with the base is not $\widehat{TPQ}$ or any face angle; it is the angle between $TP$ and its perpendicular projection onto the base, which is the segment $OP$. The rule: drop a perpendicular from the line's free endpoint to the plane (here $T \to O$), and the right triangle thus formed has the wanted angle at the line's foot in the plane (here $P$). Similarly in (d), the dihedral angle between two planes meeting along an edge $PQ$ is found by erecting perpendiculars to that edge inside each plane — $MO$ inside the base, $MT$ inside the face — and measuring the angle between them at the common foot $M$. Get the "perpendicular to the common edge" instinct and 3D trig collapses to a single right triangle every time.

(a) 对角之半 $OP$ 与高 $TO$ M1·A1·A1

边长 $6$ 的正方形对角线为 $6\sqrt{2}$（勾股：$\sqrt{6^{2} + 6^{2}}$）。中心 $O$ 即中点，故 $OP = \tfrac{1}{2} \cdot 6\sqrt{2} = 3\sqrt{2}$。 $\triangle TOP$ 在 $O$ 处直角（顶点正在中心上方）。斜边 $TP = 10$、一腿 $OP = 3\sqrt{2}$： $$ TO^{2} \;=\; TP^{2} - OP^{2} \;=\; 100 - 18 \;=\; 82 \;\Longrightarrow\; TO \;=\; \sqrt{82} \;\approx\; 9.06 \text{ cm}. $$

(b) 斜棱与底面所成角 M1·A1

$TP$ 与底面所成角即直角 $\triangle TOP$ 中的 $\widehat{TPO}$： $$ \tan(\widehat{TPO}) \;=\; \frac{TO}{OP} \;=\; \frac{\sqrt{82}}{3\sqrt{2}} \;=\; \frac{\sqrt{41}}{3} \;\approx\; 2.134, \quad \widehat{TPO} \;\approx\; 64.9^{\circ}. $$

(c) 顶角 $\widehat{PTQ}$（余弦定理） M1·A1·A1

$\triangle TPQ$ 中 $TP = TQ = 10$、$PQ = 6$（底边）。余弦定理（$\widehat{PTQ}$ 对边 $PQ$）： $$ \cos(\widehat{PTQ}) \;=\; \frac{TP^{2} + TQ^{2} - PQ^{2}}{2 \cdot TP \cdot TQ} \;=\; \frac{100 + 100 - 36}{200} \;=\; \frac{164}{200} \;=\; 0.82. $$ 故 $\widehat{PTQ} = \arccos 0.82 \approx 34.9^{\circ}$。等腰拆分法核对：从 $T$ 向 $PQ$ 中点 $M$ 作垂线，得直角三角形 $TMP$，$TP = 10$、$PM = 3$；故 $\sin(\tfrac{1}{2}\widehat{PTQ}) = \tfrac{3}{10} = 0.3$，$\tfrac{1}{2}\widehat{PTQ} \approx 17.46^{\circ}$，$\widehat{PTQ} \approx 34.9^{\circ}$。$\checkmark$

(d) 二面角 $\widehat{TMO}$ M1·A1

$M$ 为 $PQ$ 中点。$TM$ 即面 $TPQ$ 的斜高（顶点到底边）；$OM$ 即底面内从中心到该边的垂线。
  • $TM$：等腰 $\triangle TPQ$ 中，$T$ 到 $PQ$ 的高足为 $M$。$TM^{2} = TP^{2} - PM^{2} = 100 - 3^{2} = 91$，故 $TM = \sqrt{91}$。
  • $OM = 3$（$O$ 为中心、$M$ 为一边中点，距为半边长）。
  • $TO = \sqrt{82}$（(a) 中所得，垂直底面）。
$\triangle TMO$ 在 $O$ 处直角（高 $TO$ 垂直于底面内的 $OM$）： $$ \tan(\widehat{TMO}) \;=\; \frac{TO}{OM} \;=\; \frac{\sqrt{82}}{3} \;\approx\; 3.018, \quad \widehat{TMO} \;\approx\; 71.7^{\circ}. $$
"线与面所成角" $=$ 线与其投影的夹角。(b) 中斜棱 $TP$ 与底面所成角不是 $\widehat{TPQ}$ 或任何面角，而是 $TP$ 与其在底面正投影的夹角，即与 $OP$ 的夹角。规则：从线在外的端点向面作垂线（此处 $T \to O$），所得直角三角形里所求角位于线在面上的脚处（此处 $P$）。同理 (d) 中两面沿公共棱 $PQ$ 相交的二面角，需在两面内分别作垂直于该棱的线 — 底面内的 $MO$、面内的 $MT$ — 在公共脚 $M$ 处度量夹角。养成"先作公共棱的垂线"的习惯，3D 三角每次都塌缩为一个直角三角形。

Examiner Watch-List · C2 Cross-Question Pitfalls评卷雷区 · C2 跨题易错点

Rule-quoting discipline M1 hygiene

Every M1 in C2 is gated by writing the rule before substitution. "Sine rule: $\tfrac{a}{\sin A} = \tfrac{b}{\sin B}$, so $\ldots$" earns the M1; jumping straight to "$b = \tfrac{8 \sin 60^{\circ}}{\sin 40^{\circ}}$" does not. Same for cosine rule, $\tfrac{1}{2}ab\sin C$, Pythagoras, and SOHCAHTOA.

Exact vs. decimal A1 hygiene

On Paper 1 (no calc), keep $\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$, $\sqrt{39}$, $20\sqrt{41}$ in surd form — never decimalise. On Paper 2 (calc), give decimals to the requested precision (angles 1 dp, lengths 3 sf) unless the question says "exact".

Bearings format notation A1

Three digits, zero-padded, clockwise from north: $060^{\circ}$, $098.7^{\circ}$, $240^{\circ}$ — never $60^{\circ}$ or $98.7^{\circ}$. Diagrams with north arrows at every station prevent the second-most-common bearings error (reverse-bearing confusion).

SSA: two triangles or one? R1 hygiene

When given two sides and a non-included angle, after computing $\sin B$, always test: (i) is $\sin B \le 1$? (ii) is the opposite side shorter than the other given side? (iii) does $A + B_{2} < 180^{\circ}$? Skipping any of these costs the R1 reasoning mark.

3D trig: identify the right triangle first setup M1

"Line to plane" $\to$ project line onto plane, work in the right triangle formed by line, projection, and dropped perpendicular. "Plane to plane" $\to$ erect perpendiculars to the common edge inside each plane and measure the angle at the foot. Cosine rule on a face triangle is C2.3 not 3D — do not conflate.

Cross-check identities verification

$\sin^{2}\theta + \cos^{2}\theta = 1$ catches arithmetic errors at Q1; $\tan = \sin / \cos$ confirms the third ratio; $r < R$ via Euler ($R \ge 2r$) sanity-checks Q5; extended sine rule $\tfrac{c}{\sin C} = 2R$ cross-checks any circumradius computation.

"先引定理"纪律 M1 卫生

C2 的每个 M1 都以"代入前先引定理"为前提。"正弦定理：$\tfrac{a}{\sin A} = \tfrac{b}{\sin B}$，故 $\ldots$"得 M1；直接写"$b = \tfrac{8 \sin 60^{\circ}}{\sin 40^{\circ}}$"不得。余弦定理、$\tfrac{1}{2}ab\sin C$、勾股、SOHCAHTOA 同理。

精确 vs 十进制 A1 卫生

Paper 1（不可计算器）保留 $\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$、$\sqrt{39}$、$20\sqrt{41}$ 等根号形式 — 不要十进制。Paper 2（可计算器）按要求精度给十进制（角 1 位小数、长度 3 位有效数字），除非题目要求"精确"。

方位角格式 符号 A1

三位数、前导零、自北顺时针：$060^{\circ}$、$098.7^{\circ}$、$240^{\circ}$ — 绝不写 $60^{\circ}$ 或 $98.7^{\circ}$。每个测站画北箭头可防第二常见错误（反方位角搞混）。

SSA：两个还是一个三角形？ R1 卫生

给两边和非夹角时，求出 $\sin B$ 后必测：(i) $\sin B \le 1$？(ii) 非夹角对边短于另一已知边？(iii) $A + B_{2} < 180^{\circ}$？任一未做即丢 R1 推理分。

3D 三角：先认直角三角形 建图 M1

"线与面" $\to$ 将线投影到面，在线、投影与所作垂线构成的直角三角形里算。"面与面" $\to$ 在两面内分别作公共棱的垂线，在脚处量夹角。一个面三角形内用余弦定理属 C2.3，非 3D — 别混淆。

身份交叉核对 验证

$\sin^{2}\theta + \cos^{2}\theta = 1$ 抓 Q1 算术错误；$\tan = \sin / \cos$ 核对第三个比；$r < R$（Euler 不等式 $R \ge 2r$）核对 Q5；扩展正弦定理 $\tfrac{c}{\sin C} = 2R$ 核对任何外接圆计算。