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Right-Triangle Trigonometry · Solutions直角三角形三角学 · 解析

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC / AB styles练习题配套解析 · 逐分讲解 · SAT / AP 衔接 / ON / BC / AB 卷型

EASY MEDIUM HARD 🇺🇸 US 🇨🇦 ON 🇨🇦 BC 🇨🇦 AB SAT-style MCQ AP-feeder FRQ ON Provincial-style BC Provincial-style AB Diploma-style Honors
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  简答题 · 解析SAT MCQ + ON/BC short answer · 18 marksSAT 选择题 + ON/BC 简答题 · 18 分

Section A · Worked SolutionsA 节 · 详解

Q1EASY 🇺🇸 US SAT-style MCQ §1 SOH CAH TOA · HSG-SRT.C.6 [3 marks]

Right triangle $ABC$, right angle at $C$, $AB = 13$, $BC = 5$, $CA = 12$. Find $\sin A$.直角三角形 $ABC$，直角在 $C$，$AB = 13$、$BC = 5$、$CA = 12$。求 $\sin A$。

Answer:答案：  (B)  $\dfrac{5}{13}$

(a) Identify opposite and hypotenuse at vertex $A$在顶点 $A$ 处辨认对边与斜边 M1·A1·A1

The right angle is at $C$, so the hypotenuse is the side opposite $C$, namely $AB = 13$. At vertex $A$, the "opposite" side is the one that does not touch $A$, which is $BC = 5$. By SOH:直角在 $C$，所以斜边是 $C$ 的对边 $AB = 13$。在顶点 $A$ 处，"对边"是不与 $A$ 相接的边，即 $BC = 5$。由 SOH：

$$ \sin A \;=\; \frac{\text{opposite}}{\text{hypotenuse}} \;=\; \frac{BC}{AB} \;=\; \frac{5}{13}. $$

This matches option (B). Sanity-check the Pythagorean triple: $5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}$. $\checkmark$与选项 (B) 相符。用勾股数对照核验：$5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}$。$\checkmark$

Why the wrong choices fail.错项分析。
  • (A) $\tfrac{5}{12}$, this is $\tan A$, not $\sin A$; uses the two legs instead of opposite-over-hypotenuse.这是 $\tan A$，不是 $\sin A$；用了两条直角边而非"对/斜"。
  • (C) $\tfrac{12}{13}$, this is $\cos A$ ($CA/AB$), the adjacent-over-hypotenuse pair, a classic sin / cos swap.这是 $\cos A$（$CA/AB$），"邻/斜"对，典型的 sin 与 cos 互换错误。
  • (D) $\tfrac{13}{5}$, the reciprocal $\tfrac{1}{\sin A}$ (this is $\csc A$); cannot exceed $1$, so disqualified at sight for any acute-angle sine.取了倒数 $\tfrac{1}{\sin A}$（即 $\csc A$）；锐角正弦不能大于 $1$，一眼即可排除。
Every primary trig ratio at an acute angle of a right triangle is at most $1$.直角三角形锐角的任一基本三角比都不超过 $1$。 If your candidate answer for $\sin\theta$ or $\cos\theta$ is greater than $1$, you have either swapped a ratio for its reciprocal or written the inverse upside down. The SAT seeds option (D) precisely to catch this. The fast triage on any acute-angle MCQ: which choices are $> 1$? Eliminate them. Then among the remaining, recall SOH CAH TOA and read off the requested ratio.若你给出的 $\sin\theta$ 或 $\cos\theta$ 候选值大于 $1$，要么把三角比写成了它的倒数，要么把反函数写反了。SAT 设置选项 (D) 正是为了引诱这种错误。锐角选择题的快速筛选法：先看哪些选项 $> 1$，直接排除；剩下的再回忆 SOH CAH TOA 读出题目所求的比。
Q2EASY 🇺🇸 US SAT-style MCQ §2 45-45-90 Triangle · HSG-SRT.C.6 [3 marks]

Isosceles right triangle with both legs $7$. Find the hypotenuse, exact form.等腰直角三角形两直角边均为 $7$。求斜边的精确长度。

Answer:答案：  (B)  $7\sqrt{2}$

(a) Apply the 45-45-90 ratio套用 45-45-90 比例 M1·A1·A1

A 45-45-90 triangle has side ratios $1 : 1 : \sqrt{2}$ (legs : hypotenuse). With each leg equal to $7$, the hypotenuse is $7 \cdot \sqrt{2} = 7\sqrt{2}$. Verify by the Pythagorean theorem:45-45-90 三角形的边比为 $1 : 1 : \sqrt{2}$（两直角边 : 斜边）。每条直角边为 $7$ 时，斜边为 $7 \cdot \sqrt{2} = 7\sqrt{2}$。用勾股定理验证：

$$ \text{hyp}^{2} \;=\; 7^{2} + 7^{2} \;=\; 98 \;\Longrightarrow\; \text{hyp} \;=\; \sqrt{98} \;=\; \sqrt{49 \cdot 2} \;=\; 7\sqrt{2}. \;\checkmark $$

This matches option (B).与选项 (B) 相符。

Why the wrong choices fail.错项分析。
  • (A) $7$, copies one leg as the hypotenuse; the hypotenuse of a non-degenerate right triangle is strictly longer than either leg.把一条直角边当作斜边；非退化直角三角形的斜边严格长于任一直角边。
  • (C) $7\sqrt{3}$, uses the 30-60-90 ratio (where the long leg is $\sqrt{3}$ times the short leg) instead of 45-45-90.误用 30-60-90 的比例（长直角边是短直角边的 $\sqrt{3}$ 倍），而非 45-45-90。
  • (D) $14$, doubles a leg; correct for the wrong reason (the hypotenuse of a 30-60-90 with short leg $7$), and incorrect for this isosceles right triangle.把一条直角边乘以 $2$；这是 30-60-90 中短直角边为 $7$ 时的斜边，不适用于本题等腰直角三角形。
Memorise the two side-ratio strips and you bypass the calculator on every special-triangle item.把两条边比例条记牢，所有特殊三角形题都无需计算器。 45-45-90 is $1 : 1 : \sqrt{2}$ and 30-60-90 is $1 : \sqrt{3} : 2$ (short leg : long leg : hypotenuse, with the short leg opposite $30^{\circ}$). The SAT cannot, by design, ask for special-triangle decimals; every right-answer choice is built from these two strips. Train the eye to scan for $\sqrt{2}$ vs $\sqrt{3}$ as the discriminator.45-45-90 比例为 $1 : 1 : \sqrt{2}$，30-60-90 比例为 $1 : \sqrt{3} : 2$（短直角边 : 长直角边 : 斜边，短直角边与 $30^{\circ}$ 相对）。SAT 设计上不会让你在特殊三角形上写小数，正确选项必由这两条比例构成。训练眼力分辨 $\sqrt{2}$ 与 $\sqrt{3}$，迅速锁定答案。
Q3MEDIUM 🇺🇸 US SAT-style MCQ §2 30-60-90 Triangle · HSG-SRT.C.6 [3 marks]

30-60-90 triangle, side opposite $30^{\circ}$ has length $6$. Find side opposite $60^{\circ}$.30-60-90 三角形中，与 $30^{\circ}$ 角相对的边长为 $6$。求与 $60^{\circ}$ 角相对的边长。

Answer:答案：  (C)  $6\sqrt{3}$

(a) Apply the 30-60-90 ratio套用 30-60-90 比例 M1·A1·A1

A 30-60-90 triangle has side ratios $1 : \sqrt{3} : 2$ for (short leg opposite $30^{\circ}$) : (long leg opposite $60^{\circ}$) : (hypotenuse). With the short leg equal to $6$, the long leg is $6 \sqrt{3}$ and the hypotenuse is $2 \cdot 6 = 12$. The side opposite $60^{\circ}$ is the long leg $= 6\sqrt{3}$. This matches option (C).30-60-90 三角形的边比为 $1 : \sqrt{3} : 2$，对应（与 $30^{\circ}$ 相对的短直角边）：（与 $60^{\circ}$ 相对的长直角边）：斜边。短直角边为 $6$ 时，长直角边为 $6 \sqrt{3}$，斜边为 $2 \cdot 6 = 12$。与 $60^{\circ}$ 相对的边即长直角边 $= 6\sqrt{3}$。与选项 (C) 相符。

Why the wrong choices fail.错项分析。
  • (A) $3\sqrt{2}$, mixes the 45-45-90 hypotenuse pattern with a halved leg; double misstep.把 45-45-90 的斜边规则与减半的边混用，双重错误。
  • (B) $6\sqrt{2}$, uses the 45-45-90 hypotenuse rule on the short leg of a 30-60-90; classic special-triangle confusion ($\sqrt{2}$ vs $\sqrt{3}$).把 45-45-90 的斜边规则用在 30-60-90 短直角边上；典型 $\sqrt{2}$ 与 $\sqrt{3}$ 混淆。
  • (D) $12$, gives the hypotenuse instead of the side opposite $60^{\circ}$; reads the wrong slot of the ratio strip.给的是斜边而非与 $60^{\circ}$ 相对的边；读错比例条中的位置。
The side opposite the bigger angle is longer, always.大角对长边，恒成立。 In any triangle the side-angle ordering matches: the longest side is opposite the largest angle. Inside a 30-60-90, that orders the three sides as short ($30^{\circ}$) $<$ long ($60^{\circ}$) $<$ hypotenuse ($90^{\circ}$), which forces $6 < \text{long leg} < 12$. Only $6\sqrt{3} \approx 10.4$ from the choices fits; (D) $12$ is the hypotenuse and (A), (B) are smaller than $6$ or use the wrong radical. The angle-side ordering rule eliminates three of four choices in one sweep.任意三角形中，边与角的排序一致：最大角对最长边。30-60-90 中三边顺序为短（$30^{\circ}$）$<$ 长（$60^{\circ}$）$<$ 斜边（$90^{\circ}$），由此 $6 < \text{长直角边} < 12$。选项中只有 $6\sqrt{3} \approx 10.4$ 落入此区间；(D) $12$ 是斜边，(A)、(B) 要么小于 $6$ 要么用错了根号。角边排序一条规则即可一扫排除三个选项。
Q4MEDIUM 🇨🇦 ON ON Provincial-style §3 Missing Side · MPM2D Trigonometry [4 marks]

Right triangle $PQR$, right angle at $Q$, $\angle P = 37^{\circ}$, hypotenuse $PR = 20$ cm. (a) Equation. (b) Solve for $QR$. (c) Solve for $PQ$.直角三角形 $PQR$，直角在 $Q$，$\angle P = 37^{\circ}$，斜边 $PR = 20$ cm。(a) 列方程。(b) 求 $QR$。(c) 求 $PQ$。

Answer:答案：  (a) $\sin 37^{\circ} = \tfrac{QR}{20}$  ·  (b) $QR \approx 12.0$ cm  ·  (c) $PQ \approx 16.0$ cm

(a) Trigonometric equation linking $QR$ to the hypotenuse把 $QR$ 与斜边联系的三角方程 A1

At vertex $P$, side $QR$ is the opposite side (it is across from $\angle P$) and $PR = 20$ is the hypotenuse. The ratio that uses opposite and hypotenuse is sine: $\sin 37^{\circ} = \dfrac{QR}{20}.$在顶点 $P$ 处，$QR$ 是对边（与 $\angle P$ 相对），$PR = 20$ 是斜边。使用对边和斜边的比是正弦：$\sin 37^{\circ} = \dfrac{QR}{20}.$

(b) Solve for $QR$求 $QR$ M1·A1

$$ QR \;=\; 20 \sin 37^{\circ} \;\approx\; 20 \cdot 0.6018 \;\approx\; 12.04 \;\approx\; 12.0 \;\text{cm}. $$

(c) Find $PQ$ via cosine (or Pythagorean)用余弦（或勾股定理）求 $PQ$ A1

Use $\cos 37^{\circ} = \dfrac{PQ}{20}$: $PQ = 20 \cos 37^{\circ} \approx 20 \cdot 0.7986 \approx 15.97 \approx 16.0$ cm. (Pythagorean cross-check: $\sqrt{20^{2} - 12.04^{2}} = \sqrt{400 - 144.96} = \sqrt{255.04} \approx 15.97$ cm. $\checkmark$)用 $\cos 37^{\circ} = \dfrac{PQ}{20}$：$PQ = 20 \cos 37^{\circ} \approx 20 \cdot 0.7986 \approx 15.97 \approx 16.0$ cm。（勾股核验：$\sqrt{20^{2} - 12.04^{2}} = \sqrt{400 - 144.96} = \sqrt{255.04} \approx 15.97$ cm。$\checkmark$）

Hypotenuse is given $\Rightarrow$ reach for $\sin$ or $\cos$, not $\tan$.题中给斜边 $\Rightarrow$ 用 $\sin$ 或 $\cos$，而非 $\tan$。 The diagnostic for "which ratio?" is: which two sides am I working with? Opposite + hypotenuse $\Rightarrow$ sine. Adjacent + hypotenuse $\Rightarrow$ cosine. Opposite + adjacent (no hypotenuse) $\Rightarrow$ tangent. Whenever the hypotenuse is one of the two sides involved, tangent is never the right tool. ON markers explicitly award the equation-setup mark in (a) before any arithmetic, so write the substituted equation before pressing any keys."用哪个比"的判断法是：当前涉及哪两条边？ 对边 + 斜边 $\Rightarrow$ 正弦；邻边 + 斜边 $\Rightarrow$ 余弦；对边 + 邻边（无斜边）$\Rightarrow$ 正切。只要斜边是涉及的两边之一，正切就一定不是正确工具。ON 评分先给方程列出分（a），再看运算，因此按计算器前务必先写出代入式。
Q5MEDIUM 🇨🇦 BC BC Provincial-style §4 Missing Angle · FMP&PC 10 right-triangle trig [5 marks]

Right triangle, legs $8$ and $15$, right angle between them. $\theta$ opposite the leg of length $15$. (a) Which ratio? (b) Equation. (c) Solve $\theta$. (d) Other acute angle.直角三角形两直角边为 $8$ 与 $15$，直角夹在两边之间。$\theta$ 与长 $15$ 的边相对。(a) 用哪种比？(b) 列方程。(c) 求 $\theta$。(d) 另一锐角。

Answer:答案：  (a) tangent正切  ·  (b) $\tan\theta = \tfrac{15}{8}$  ·  (c) $\theta \approx 61.9^{\circ}$  ·  (d) $\approx 28.1^{\circ}$

(a) Choose the ratio选择三角比 R1

The two known sides are the two legs, the opposite side (length $15$) and the adjacent side (length $8$). Neither is the hypotenuse, so the only primary ratio that uses both is tangent (TOA).两条已知边即两条直角边：对边（长 $15$）与邻边（长 $8$）。两者都不是斜边，所以唯一同时用到二者的基本比是 正切（TOA）。

(b) Substituted equation代入式 A1

$$ \tan \theta \;=\; \frac{\text{opp}}{\text{adj}} \;=\; \frac{15}{8}. $$

(c) Solve via inverse tangent用反正切求解 M1·A1

$$ \theta \;=\; \tan^{-1}\!\bigl(\tfrac{15}{8}\bigr) \;=\; \tan^{-1}(1.875) \;\approx\; 61.93^{\circ} \;\approx\; 61.9^{\circ}. $$

(d) Complementary partner互余角 A1

Acute angles in a right triangle sum to $90^{\circ}$, so the other acute angle is $90^{\circ} - 61.9^{\circ} = 28.1^{\circ}$.直角三角形两锐角之和为 $90^{\circ}$，故另一锐角为 $90^{\circ} - 61.9^{\circ} = 28.1^{\circ}$。

The complementary-angle rule ($A + B = 90^{\circ}$) makes the second angle free.余角法则（$A + B = 90^{\circ}$）让第二个角"白送"。 In every right triangle, once you have one acute angle you have both, no second computation needed. This is HSG-SRT.C.7 in action ($\sin\theta = \cos(90^{\circ} - \theta)$) and BC PC 10 lists it under right-triangle trig. The marker awards A1 in (d) explicitly for citing the complementary relationship rather than re-running $\tan^{-1}(8/15)$. The structural move: always state $\theta_{1}$ and $\theta_{2} = 90^{\circ} - \theta_{1}$ together as a pair.直角三角形里，一旦求出一个锐角，另一个就同时确定，无需再算第二次。这正是 HSG-SRT.C.7 的内容（$\sin\theta = \cos(90^{\circ} - \theta)$），BC PC 10 也将其列入直角三角形三角学。评分明确：在 (d) 写余角关系即可得 A1，比再算一次 $\tan^{-1}(8/15)$ 更稳。结构性做法：每次同时写出 $\theta_{1}$ 与 $\theta_{2} = 90^{\circ} - \theta_{1}$。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  长答题 · 解析AP-feeder FRQ + honors · 35 marksAP 衔接长答 + 荣誉级 · 35 分

Section B · Worked SolutionsB 节 · 详解

Q6MEDIUM 🇺🇸 US AP-feeder FRQ §3 + §4 Solve a Right Triangle · HSG-SRT.C.8 [8 marks]

Right triangle $ABC$, right angle at $C$, $AB = 25$, $BC = 7$. (a) $CA$. (b) $\sin A, \cos A, \tan A$. (c) $\angle A$ and $\angle B$. (d) $\sin B$ and check $\sin B = \cos A$.直角三角形 $ABC$，直角在 $C$，$AB = 25$、$BC = 7$。(a) $CA$。(b) $\sin A, \cos A, \tan A$。(c) $\angle A$ 与 $\angle B$。(d) $\sin B$ 并验证 $\sin B = \cos A$。

Answer:答案：  (a) $CA = 24$  ·  (b) $\sin A = \tfrac{7}{25}, \cos A = \tfrac{24}{25}, \tan A = \tfrac{7}{24}$  ·  (c) $\angle A \approx 16.3^{\circ}$, $\angle B \approx 73.7^{\circ}$  ·  (d) $\sin B = \tfrac{24}{25} = \cos A$ ✓

(a) Find the missing leg by the Pythagorean theorem用勾股定理求未知直角边 M1·A1

With hypotenuse $AB = 25$ and one leg $BC = 7$:斜边 $AB = 25$，一直角边 $BC = 7$：

$$ CA^{2} \;=\; AB^{2} - BC^{2} \;=\; 625 - 49 \;=\; 576 \;\Longrightarrow\; CA \;=\; 24. $$

Recognise the Pythagorean triple $(7, 24, 25)$, no decimal needed.识别勾股数 $(7, 24, 25)$，无需小数。

(b) Read the three primary ratios at vertex $A$在顶点 $A$ 处读出三大三角比 A1·A1·A1

At $A$, the opposite side is $BC = 7$, the adjacent side is $CA = 24$, the hypotenuse is $AB = 25$.在 $A$ 处，对边 $BC = 7$，邻边 $CA = 24$，斜边 $AB = 25$。

$$ \sin A \;=\; \frac{7}{25}, \qquad \cos A \;=\; \frac{24}{25}, \qquad \tan A \;=\; \frac{7}{24}. $$

(c) Find $\angle A$ and $\angle B$求 $\angle A$ 与 $\angle B$ M1·A1

$\angle A = \sin^{-1}(7/25) = \sin^{-1}(0.28) \approx 16.26^{\circ} \approx 16.3^{\circ}$. By the complementary-angle relationship (HSG-SRT.C.7), $\angle B = 90^{\circ} - 16.3^{\circ} = 73.7^{\circ}$.$\angle A = \sin^{-1}(7/25) = \sin^{-1}(0.28) \approx 16.26^{\circ} \approx 16.3^{\circ}$。由余角关系（HSG-SRT.C.7），$\angle B = 90^{\circ} - 16.3^{\circ} = 73.7^{\circ}$。

(d) $\sin B$ and the complementary identity$\sin B$ 与余角恒等式 A1

At $B$, the opposite side is $CA = 24$, the hypotenuse is $AB = 25$, so $\sin B = \tfrac{24}{25}$. This equals $\cos A = \tfrac{24}{25}$. $\checkmark$ This is the statement of HSG-SRT.C.7: $\sin B = \sin(90^{\circ} - A) = \cos A$.在 $B$ 处，对边 $CA = 24$，斜边 $AB = 25$，所以 $\sin B = \tfrac{24}{25}$。这与 $\cos A = \tfrac{24}{25}$ 相等。$\checkmark$ 此即 HSG-SRT.C.7：$\sin B = \sin(90^{\circ} - A) = \cos A$。

"Opposite at $A$" and "adjacent at $B$" are the same physical side."$A$ 的对边"与"$B$ 的邻边"是同一条边。 The complementary-angle identity is not a coincidence: in a right triangle, the leg opposite $A$ is exactly the leg adjacent to $B$ (and vice versa), because $A$ and $B$ are the two non-right vertices on the two ends of that leg. So $\sin A = \cos B$ and $\cos A = \sin B$ fall out automatically from the labels. AP graders look for this geometric explanation in (d) rather than a second calculator computation; cite the labels and you get the A1 even before stating the numerical value.余角恒等式并非巧合：直角三角形中，$A$ 的对边正是 $B$ 的邻边（反之亦然），因为 $A$ 与 $B$ 正是同一条边两端的两个非直角顶点。所以 $\sin A = \cos B$ 与 $\cos A = \sin B$ 由标号即可读出。AP 评分在 (d) 找的是这种几何说明，而非再按一次计算器；只要说清标号即可获得 A1，先于数值结果。
Q7MEDIUM 🇨🇦 AB AB Diploma-style §2 Special Triangles · Math 10C 4.3 [8 marks]

No-calculator exact values. (a) $\sin 30, \cos 30, \tan 30$. (b) $\sin 45, \cos 45, \tan 45$. (c) 45-45-90 hypotenuse $10$, find legs. (d) 30-60-90 hypotenuse $14$, find both legs.不用计算器的精确值。(a) $\sin 30, \cos 30, \tan 30$。(b) $\sin 45, \cos 45, \tan 45$。(c) 45-45-90 三角形斜边为 $10$，求直角边。(d) 30-60-90 三角形斜边为 $14$，求两条直角边。

Answer:答案：  (a) $\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{3}}{3}$  ·  (b) $\tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{2}}{2}, 1$  ·  (c) $5\sqrt{2}$  ·  (d) $7$ and与 $7\sqrt{3}$

(a) 30-60-90 exact values30-60-90 精确值 A1·A1

From the side strip $1 : \sqrt{3} : 2$ (short : long : hypotenuse):由边比例条 $1 : \sqrt{3} : 2$（短 : 长 : 斜边）：

$$ \sin 30^{\circ} \;=\; \frac{1}{2}, \qquad \cos 30^{\circ} \;=\; \frac{\sqrt{3}}{2}, \qquad \tan 30^{\circ} \;=\; \frac{1}{\sqrt{3}} \;=\; \frac{\sqrt{3}}{3}. $$

(b) 45-45-90 exact values45-45-90 精确值 A1·A1

From the side strip $1 : 1 : \sqrt{2}$:由边比例条 $1 : 1 : \sqrt{2}$：

$$ \sin 45^{\circ} \;=\; \cos 45^{\circ} \;=\; \frac{1}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{2}, \qquad \tan 45^{\circ} \;=\; 1. $$

(c) 45-45-90 with hypotenuse $10$45-45-90，斜边 $10$ M1·A1

$\cos 45^{\circ} = \dfrac{\text{leg}}{10}$ gives $\text{leg} = 10 \cdot \dfrac{\sqrt{2}}{2} = 5\sqrt{2}$. Both legs are equal by symmetry of the isosceles right triangle, so each leg is $5\sqrt{2}$. Pythagorean cross-check: $(5\sqrt{2})^{2} + (5\sqrt{2})^{2} = 50 + 50 = 100 = 10^{2}$. $\checkmark$$\cos 45^{\circ} = \dfrac{\text{直角边}}{10}$ 给出 $\text{直角边} = 10 \cdot \dfrac{\sqrt{2}}{2} = 5\sqrt{2}$。由等腰直角三角形对称性两条直角边相等，故每条均为 $5\sqrt{2}$。勾股核验：$(5\sqrt{2})^{2} + (5\sqrt{2})^{2} = 50 + 50 = 100 = 10^{2}$。$\checkmark$

(d) 30-60-90 with hypotenuse $14$30-60-90，斜边 $14$ A1·A1

With hypotenuse $14$, the short leg (opposite $30^{\circ}$) is half the hypotenuse: $14 / 2 = 7$. The long leg (opposite $60^{\circ}$) is $\sqrt{3}$ times the short leg: $7\sqrt{3}$.斜边为 $14$ 时，短直角边（与 $30^{\circ}$ 相对）为斜边的一半：$14 / 2 = 7$。长直角边（与 $60^{\circ}$ 相对）为短直角边的 $\sqrt{3}$ 倍：$7\sqrt{3}$。

Exact form means no decimals, and Alberta markers strip a mark for a calculator value where exact is asked."精确值"意味着不可写小数，AB 评分若题中要求精确值而写出计算器值会扣分。 Math 20-1 / Math 10C Diploma rubrics name "exact value" explicitly: the answer must be in radical form, with the radical rationalised (no $\tfrac{1}{\sqrt{2}}$, write $\tfrac{\sqrt{2}}{2}$). The two side-strip ratios $1 : 1 : \sqrt{2}$ and $1 : \sqrt{3} : 2$ encode every special-angle value, and you should be able to write all six exact ratios for $30^{\circ}, 45^{\circ}, 60^{\circ}$ from memory in under a minute. This skill becomes the unit-circle backbone in Pre-Calc 12.Math 20-1 / Math 10C Diploma 评分明文规定"精确值"：答案须为含根号形式，且对分母有理化（不写 $\tfrac{1}{\sqrt{2}}$，写 $\tfrac{\sqrt{2}}{2}$）。两条边比例 $1 : 1 : \sqrt{2}$ 与 $1 : \sqrt{3} : 2$ 编码了所有特殊角的值，你应能在一分钟内默写出 $30^{\circ}, 45^{\circ}, 60^{\circ}$ 的全部六个精确比。该技能将在 Pre-Calc 12 单位圆部分继续成为骨架。
Q8HARDHonors 🇨🇦 BC BC Provincial-style §6 Law of Sines · PC 11 sine / cosine laws [9 marks]

Triangle $ABC$, $\angle A = 52^{\circ}$, $\angle B = 74^{\circ}$, $a = 18$ cm. (a) State law + AAS uniqueness. (b) $\angle C$. (c) $b$. (d) $c$. (e) Area via $\tfrac{1}{2} a b \sin C$.三角形 $ABC$，$\angle A = 52^{\circ}$、$\angle B = 74^{\circ}$、$a = 18$ cm。(a) 写定理 + AAS 唯一性。(b) $\angle C$。(c) $b$。(d) $c$。(e) 用 $\tfrac{1}{2} a b \sin C$ 求面积。

Answer:答案：  (b) $\angle C = 54^{\circ}$  ·  (c) $b \approx 22.0$ cm  ·  (d) $c \approx 18.5$ cm  ·  (e) Area面积 $\approx 159.9$ cm$^{2}$

(a) Law of Sines and AAS uniqueness正弦定理与 AAS 唯一性 R1·A1

Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$. An AAS configuration (two angles plus one side opposite a known angle) determines the triangle uniquely because the two angles fix the third (sum $= 180^{\circ}$), and the AAA-shape is then scaled to a unique size by the one given side, every other side follows from the Law of Sines.正弦定理：$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$。AAS 情形（两角加一已知角对边）唯一确定三角形：两角固定第三角（和为 $180^{\circ}$），AAA 形状再由这一条已知边唯一缩放，其余各边由正弦定理推出。

(b) Third angle第三角 A1

$\angle C = 180^{\circ} - 52^{\circ} - 74^{\circ} = 54^{\circ}$.

(c) Side $b$ by Law of Sines用正弦定理求 $b$ M1·A1·A1

$$ \frac{b}{\sin 74^{\circ}} \;=\; \frac{18}{\sin 52^{\circ}} \;\Longrightarrow\; b \;=\; \frac{18 \sin 74^{\circ}}{\sin 52^{\circ}} \;\approx\; \frac{18 \cdot 0.9613}{0.7880} \;\approx\; \frac{17.303}{0.7880} \;\approx\; 21.96 \;\approx\; 22.0 \;\text{cm}. $$

(d) Side $c$ by Law of Sines用正弦定理求 $c$ M1·A1

$$ c \;=\; \frac{18 \sin 54^{\circ}}{\sin 52^{\circ}} \;\approx\; \frac{18 \cdot 0.8090}{0.7880} \;\approx\; 18.48 \;\approx\; 18.5 \;\text{cm}. $$

(e) Area via $\tfrac{1}{2} a b \sin C$用 $\tfrac{1}{2} a b \sin C$ 求面积 A1

$$ \text{Area} \;=\; \tfrac{1}{2} \cdot 18 \cdot 21.96 \cdot \sin 54^{\circ} \;\approx\; \tfrac{1}{2} \cdot 18 \cdot 21.96 \cdot 0.8090 \;\approx\; 159.9 \;\text{cm}^{2}. $$

(HSG-SRT.D.9 derives this formula from the perpendicular drop $h = b \sin C$ inside the base-times-height area formula.)（HSG-SRT.D.9 由作高 $h = b \sin C$ 代入"底乘高"面积公式推导此式。）

AAS is unique, SSA is the only configuration that can fail uniqueness.AAS 唯一；五大情形中只有 SSA 可能失唯一。 Of the five triangle-configuration acronyms (SSS, SAS, ASA, AAS, SSA), the first four always give a unique triangle (when one exists), and the Law of Sines or Cosines resolves them in one shot. SSA, the "ambiguous case", is the lone outlier, see Q9. Naming the configuration explicitly before reaching for the law tells the marker you know which tool to deploy: Law of Sines for AAS / ASA / SSA, Law of Cosines for SAS / SSS. BC PC 11 markers reward the configuration-naming as a setup mark.五种情形（SSS、SAS、ASA、AAS、SSA）中，前四种只要三角形存在便唯一，正弦或余弦定理一步解出。唯独 SSA "模糊情形"是例外，见 Q9。先说出情形再用定理，向评分员表明你知道工具：AAS / ASA / SSA 用正弦定理，SAS / SSS 用余弦定理。BC PC 11 评分将命名情形列为列式分。
Q9HARDHonors 🇨🇦 ON ON Provincial-style §7 Law of Cosines + Ambiguous SSA · MCR3U D1.6 [10 marks]

Part A: triangle $XYZ$, $x = 9, y = 12, \angle Z = 108^{\circ}$. (a) Law + SAS uniqueness. (b) Side $z$. Part B: triangle $LMN$, $\ell = 8, m = 11, \angle L = 35^{\circ}$ (SSA). (c) $M_{1}$. (d) $M_{2}$ + validity check. (e) Why "ambiguous".A 部分：三角形 $XYZ$，$x = 9, y = 12, \angle Z = 108^{\circ}$。(a) 写定理 + SAS 唯一性。(b) 边 $z$。B 部分：三角形 $LMN$，$\ell = 8, m = 11, \angle L = 35^{\circ}$（SSA）。(c) $M_{1}$。(d) $M_{2}$ + 验证有效性。(e) 为何"模糊"。

Answer:答案：  (b) $z \approx 17.1$  ·  (c) $M_{1} \approx 52.1^{\circ}$  ·  (d) $M_{2} \approx 127.9^{\circ}$, both candidates valid (two triangles)两候选均有效（两个三角形）

(a) Law of Cosines and SAS uniqueness余弦定理与 SAS 唯一性 R1·A1

Law of Cosines: $c^{2} = a^{2} + b^{2} - 2 a b \cos C$. An SAS configuration (two sides with the included angle) admits a unique triangle because the two sides and the angle between them rigidly fix the third side (and hence the remaining angles), the law turns this rigidity into a closed-form length.余弦定理：$c^{2} = a^{2} + b^{2} - 2 a b \cos C$。SAS 情形（两边加夹角）唯一确定三角形：两边及夹角刚性地固定第三边（并由此固定其余两角），余弦定理将这种刚性转化为闭式长度。

(b) Side $z$边 $z$ M1·A1·A1

With $x = 9, y = 12, \angle Z = 108^{\circ}$:$x = 9, y = 12, \angle Z = 108^{\circ}$：

$$ z^{2} \;=\; x^{2} + y^{2} - 2 x y \cos Z \;=\; 81 + 144 - 2(9)(12) \cos 108^{\circ}. $$

$\cos 108^{\circ} \approx -0.3090$, so$\cos 108^{\circ} \approx -0.3090$，故

$$ z^{2} \;\approx\; 225 - 216 \cdot (-0.3090) \;=\; 225 + 66.75 \;=\; 291.75 \;\Longrightarrow\; z \;\approx\; \sqrt{291.75} \;\approx\; 17.08 \;\approx\; 17.1. $$

(Sanity-check: the obtuse $\angle Z = 108^{\circ}$ forces $z$ to be the longest side, $z > \max(9, 12) = 12$. $\checkmark$)（核验：钝角 $\angle Z = 108^{\circ}$ 使 $z$ 必为最长边，$z > \max(9, 12) = 12$。$\checkmark$）

(c) Principal-value $M_{1}$ via Law of Sines用正弦定理求主值 $M_{1}$ M1·A1

$$ \sin M \;=\; \frac{m \sin L}{\ell} \;=\; \frac{11 \sin 35^{\circ}}{8} \;\approx\; \frac{11 \cdot 0.5736}{8} \;\approx\; \frac{6.310}{8} \;\approx\; 0.7887. $$

$M_{1} = \sin^{-1}(0.7887) \approx 52.07^{\circ} \approx 52.1^{\circ}$.

(d) Second candidate $M_{2}$ and validity第二候选 $M_{2}$ 与有效性 M1·R1

Supplementary candidate: $M_{2} = 180^{\circ} - 52.1^{\circ} = 127.9^{\circ}$. Check the triangle-angle sum constraint $\angle L + M < 180^{\circ}$ for each:补角候选：$M_{2} = 180^{\circ} - 52.1^{\circ} = 127.9^{\circ}$。对每个候选检验三角形角和约束 $\angle L + M < 180^{\circ}$：

  • $M_{1} = 52.1^{\circ}$: $L + M_{1} = 35^{\circ} + 52.1^{\circ} = 87.1^{\circ} < 180^{\circ}$. Valid.有效。 $\checkmark$
  • $M_{2} = 127.9^{\circ}$: $L + M_{2} = 35^{\circ} + 127.9^{\circ} = 162.9^{\circ} < 180^{\circ}$. Valid.有效。 $\checkmark$

Both candidates produce a legitimate triangle, so this SSA data corresponds to two distinct triangles.两候选均给出合法三角形，故此 SSA 数据对应 两个不同三角形。

(e) Why "ambiguous"为何"模糊" R1

The SSA configuration gives the side $m$, the side $\ell$, and the angle $\angle L$ (which is opposite $\ell$, not between the two given sides). Because $\sin$ takes the same value at $M$ and at $180^{\circ} - M$, the Law of Sines cannot, on its own, distinguish the acute from the obtuse candidate, that ambiguity is the "S, S, A" name's signature.SSA 情形给出的是边 $m$、边 $\ell$ 与角 $\angle L$（角 $L$ 与边 $\ell$ 相对，并不夹在两已知边之间）。由于 $\sin$ 在 $M$ 与 $180^{\circ} - M$ 处取同一值，单凭正弦定理无法区分锐角与钝角候选；这正是"S, S, A"命名所标记的模糊。

SSA's ambiguity is rooted in $\sin$ not being one-to-one on $(0^{\circ}, 180^{\circ})$.SSA 的模糊源于 $\sin$ 在 $(0^{\circ}, 180^{\circ})$ 上非单射。 Inside a triangle, $\sin$ takes each value (except $\sin 90^{\circ} = 1$) at exactly two angles, one acute and one obtuse. The Law of Cosines avoids this trap because $\cos$ is one-to-one on $(0^{\circ}, 180^{\circ})$, that is why SAS and SSS are never ambiguous. The MCR3U D1.6 rubric awards the R1 in (e) explicitly for naming the $\sin$-non-injectivity reason rather than just stating "two triangles can fit". For the angle check in (d), the always-required sanity test is $L + M < 180^{\circ}$, when it fails, that candidate triangle is rejected and SSA collapses to a single solution.三角形内，$\sin$ 对每个值（除 $\sin 90^{\circ} = 1$ 外）都恰对应两个角，一锐一钝。余弦定理躲开这一陷阱，因为 $\cos$ 是 $(0^{\circ}, 180^{\circ})$ 上的单射，这就是 SAS、SSS 不会出现模糊的原因。MCR3U D1.6 评分在 (e) 处的 R1 明确奖励"指出 $\sin$ 非单射"，而非只写"两三角形都可拼出"。(d) 处的常规核验为 $L + M < 180^{\circ}$；若不满足，则该候选三角形被淘汰，SSA 退化为单解。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模 / 应用 · 解析Universal · 28 marks通用 · 28 分

Section C · Modeling and ApplicationsC 节 · 建模与应用

Q10MEDIUM 🇺🇸 US AP-feeder FRQ §5 Angle of Elevation · HSG-SRT.C.8 [9 marks]

Surveyor $60$ m from flagpole base, eye level $1.6$ m, angle of elevation $32^{\circ}$. (a) Diagram. (b) Tangent equation $\Rightarrow$ $h$. (c) Total flagpole height. (d) Angle from $40$ m.测量员距旗杆底 $60$ m，眼睛高度 $1.6$ m，仰角 $32^{\circ}$。(a) 图示。(b) 正切方程 $\Rightarrow$ $h$。(c) 旗杆总高。(d) 从 $40$ m 处的仰角。

Answer:答案：  (b) $h \approx 37.5$ m  ·  (c) total总高 $\approx 39.1$ m  ·  (d) $\approx 43.1^{\circ}$, angle grows as distance shrinks距离缩短时仰角增大

(a) Diagram setup作图与标注 A1·A1

The right triangle has the horizontal sight line at eye level as one leg ($60$ m), the vertical segment from eye level up to the flagpole's top as the other leg ($h$), and the sloped line of sight as the hypotenuse. The $32^{\circ}$ angle of elevation sits at the surveyor's eye, between the horizontal leg and the hypotenuse. The right angle is at the foot of $h$ on the horizontal sight line.直角三角形：一条直角边为眼睛高度水平视线（$60$ m），另一条为从眼睛高度到旗杆顶的竖直段（$h$），斜视线为斜边。$32^{\circ}$ 仰角位于测量员眼处，介于水平直角边与斜边之间。直角位于 $h$ 与水平视线的交点。

(b) Solve $h$求 $h$ M1·A1·A1

Opposite ($h$) and adjacent ($60$) give tangent:对边（$h$）与邻边（$60$）对应正切：

$$ \tan 32^{\circ} \;=\; \frac{h}{60} \;\Longrightarrow\; h \;=\; 60 \tan 32^{\circ} \;\approx\; 60 \cdot 0.6249 \;\approx\; 37.49 \;\approx\; 37.5 \;\text{m}. $$

(c) Total flagpole height旗杆总高 A1

The $h$ above is measured from eye level, not the ground. Total flagpole height: $37.5 + 1.6 = 39.1$ m.上面的 $h$ 是从眼睛高度起算，不是从地面起算。旗杆总高：$37.5 + 1.6 = 39.1$ m。

(d) New angle from $40$ m从 $40$ m 处的新仰角 M1·A1·R1

The flagpole height above eye level is fixed at $37.5$ m (the pole has not moved). At the new horizontal distance $40$ m:旗杆在眼睛高度以上的高度仍为 $37.5$ m（旗杆未动）。新水平距离 $40$ m 时：

$$ \tan \theta \;=\; \frac{37.5}{40} \;=\; 0.9372 \;\Longrightarrow\; \theta \;=\; \tan^{-1}(0.9372) \;\approx\; 43.14^{\circ} \;\approx\; 43.1^{\circ}. $$

As the horizontal distance decreases (the surveyor walks closer), $\tan\theta = \tfrac{\text{vertical}}{\text{horizontal}}$ grows, so $\theta$ grows. Geometrically, the line of sight tilts up more steeply when you stand closer to a fixed vertical object.水平距离减小（测量员走近）时，$\tan\theta = \tfrac{\text{竖直}}{\text{水平}}$ 增大，故 $\theta$ 增大。几何上，越靠近固定竖直物体，视线越陡。

"Angle of elevation" lives at the observer, "angle of depression" lives at the elevated viewer, and both equal each other by alternate-interior angles on parallel horizontals."仰角"位于观察者处，"俯角"位于高处观察者处，二者由平行水平线的内错角相等。 The right-triangle setup is the same in either framing, what changes is at which vertex you place the labelled angle. AP graders explicitly want the diagram in (a) before the equation in (b), the diagram mark separates students who set up correctly from those who fish for a formula. The eye-level offset ($+1.6$ m) is a classic AP twist: read the prompt carefully to see whether "height of pole" or "height above eye level" is requested.直角三角形结构在两种描述下相同，差别只在角放在哪个顶点。AP 评分明确要求 (a) 中先画图再列方程；图示分把会列式的学生与碰公式的学生区分开。眼睛高度补偿（$+1.6$ m）是经典 AP 陷阱：仔细读题，看清问的是"旗杆高度"还是"眼睛以上的高度"。
Q11MEDIUM 🇨🇦 AB AB Diploma-style §5 Clinometer + Angle of Depression · Math 10C 4.5 [9 marks]

Cliff platform $48$ m above lake. Angle of depression to boat $= 18^{\circ}$. (a) Diagram + alternate-interior. (b) Tangent equation $\Rightarrow$ $d$. (c) Repeat with $26^{\circ}$. (d) Distance travelled + direction.崖顶平台距湖面 $48$ m。到船的俯角 $= 18^{\circ}$。(a) 图示 + 内错角。(b) 正切方程 $\Rightarrow$ $d$。(c) 改用 $26^{\circ}$ 重做。(d) 移动距离 + 方向。

Answer:答案：  (b) $d \approx 148$ m  ·  (c) $\approx 98$ m  ·  (d) $\approx 50$ m toward the cliff向悬崖方向

(a) Diagram and alternate-interior angle图示与内错角 A1·R1

The right triangle: vertical leg $48$ m from the platform down to the lake surface, horizontal leg $d$ from the cliff foot to the boat, hypotenuse from platform down to boat. The platform's horizontal sight line and the lake surface are parallel; the sloped sight line is a transversal. By the alternate-interior-angle theorem, the $18^{\circ}$ angle of depression at the platform equals the angle of elevation $18^{\circ}$ at the boat, which is the angle inside the right triangle at the boat.直角三角形：竖直直角边 $48$ m（平台至湖面），水平直角边 $d$（崖底至船），斜边为平台至船。平台水平视线与湖面平行；斜视线为截线。由内错角定理，平台处 $18^{\circ}$ 俯角等于船处 $18^{\circ}$ 仰角，即三角形在船处的内角。

(b) Solve $d$ for $18^{\circ}$$18^{\circ}$ 时求 $d$ M1·A1·A1

At the boat-vertex, opposite is $48$ m (vertical to platform) and adjacent is $d$ (horizontal):船处顶点上对边为 $48$ m（至平台的竖直距离），邻边为 $d$（水平距离）：

$$ \tan 18^{\circ} \;=\; \frac{48}{d} \;\Longrightarrow\; d \;=\; \frac{48}{\tan 18^{\circ}} \;\approx\; \frac{48}{0.3249} \;\approx\; 147.7 \;\approx\; 148 \;\text{m}. $$

(c) Solve $d$ for $26^{\circ}$$26^{\circ}$ 时求 $d$ M1·A1

$$ d_{\text{new}} \;=\; \frac{48}{\tan 26^{\circ}} \;\approx\; \frac{48}{0.4877} \;\approx\; 98.4 \;\approx\; 98 \;\text{m}. $$

(d) Distance travelled + direction移动距离与方向 A1·A1

The boat moved from $148$ m to $98$ m horizontally: $\Delta = 148 - 98 = 50$ m. Direction: toward the cliff (closer, because the new distance is smaller and the angle of depression increased).船的水平位置从 $148$ m 变为 $98$ m：$\Delta = 148 - 98 = 50$ m。方向：朝悬崖方向（新距离更小且俯角增大，意味着更近）。

Angle of depression goes up when the target moves closer; the cliff platform is the fixed pivot.目标越近，俯角越大；悬崖平台为定点。 The intuition: stand at the top of a cliff and watch a boat sail away, your eyes have to drop less and less (smaller depression angle) to keep it in view as it recedes; conversely, as it approaches, your eyes drop more steeply. The alternate-interior-angle theorem in (a) is the formal justification for putting the labelled angle at the boat's vertex, where the right-triangle ratio applies most cleanly. Math 10C 4.5 explicitly names the clinometer as the field instrument that measures this depression angle, AB Diploma rubrics reward students who name the instrument and the underlying theorem.直觉：站在悬崖顶看船渐远，眼睛要往下看得越来越浅（俯角变小）；反之走近时俯角加大。(a) 中的内错角定理为把标注角放在 船处顶点提供正式依据，那里三角比最干净。Math 10C 4.5 明确指出测角仪是测量俯角的现场仪器，AB Diploma 评分会奖励指明仪器与定理的学生。
Q12HARDHonors 🇨🇦 ON 🇺🇸 US ON Provincial-style §6 + §7 Surveying in 3-D · MCR3U D1.7 / HSG-SRT.D.11 [10 marks]

Two surveyors $A, B$ are $80$ m apart on near bank; both sight point $P$ on far cliff. $\angle BAP = 58^{\circ}$, $\angle ABP = 72^{\circ}$. (a) $\angle APB$. (b) $AP$. (c) $BP$. (d) Vertical height via elevation $24^{\circ}$ from $A$. (e) 3-D justification.两位测量员 $A, B$ 在近岸相距 $80$ m，均观测远岸崖上点 $P$。$\angle BAP = 58^{\circ}$、$\angle ABP = 72^{\circ}$。(a) $\angle APB$。(b) $AP$。(c) $BP$。(d) 用 $A$ 处仰角 $24^{\circ}$ 求竖直高度。(e) 三维做法说明。

Answer:答案：  (a) $\angle APB = 50^{\circ}$  ·  (b) $AP \approx 99.3$ m  ·  (c) $BP \approx 88.6$ m  ·  (d) vertical竖直高度 $\approx 40.4$ m

(a) Third angle第三角 A1

Inside triangle $ABP$: $\angle APB = 180^{\circ} - 58^{\circ} - 72^{\circ} = 50^{\circ}$.三角形 $ABP$ 内：$\angle APB = 180^{\circ} - 58^{\circ} - 72^{\circ} = 50^{\circ}$。

(b) Slant distance $AP$ by Law of Sines用正弦定理求斜距 $AP$ M1·A1·A1

In triangle $ABP$, $AB = 80$ is opposite $\angle APB = 50^{\circ}$, and $AP$ is opposite $\angle ABP = 72^{\circ}$:三角形 $ABP$ 中，$AB = 80$ 与 $\angle APB = 50^{\circ}$ 相对，$AP$ 与 $\angle ABP = 72^{\circ}$ 相对：

$$ \frac{AP}{\sin 72^{\circ}} \;=\; \frac{80}{\sin 50^{\circ}} \;\Longrightarrow\; AP \;=\; \frac{80 \sin 72^{\circ}}{\sin 50^{\circ}} \;\approx\; \frac{80 \cdot 0.9511}{0.7660} \;\approx\; 99.33 \;\approx\; 99.3 \;\text{m}. $$

(c) Slant distance $BP$ by Law of Sines用正弦定理求斜距 $BP$ M1·A1

$BP$ is opposite $\angle BAP = 58^{\circ}$:$BP$ 与 $\angle BAP = 58^{\circ}$ 相对：

$$ BP \;=\; \frac{80 \sin 58^{\circ}}{\sin 50^{\circ}} \;\approx\; \frac{80 \cdot 0.8480}{0.7660} \;\approx\; 88.56 \;\approx\; 88.6 \;\text{m}. $$

(d) Vertical height via right-triangle trig from $A$从 $A$ 用直角三角形三角比求竖直高度 M1·A1·A1

The right triangle is: hypotenuse $AP \approx 99.3$ m (the slant), vertical leg $= $ height of $P$ above ground at $A$, horizontal leg $=$ ground projection of $AP$. Using the angle of elevation $24^{\circ}$ at $A$:直角三角形：斜边 $AP \approx 99.3$ m（斜距），竖直直角边 $=$ $P$ 相对 $A$ 处地面的高度，水平直角边 $=$ $AP$ 在地面上的投影。在 $A$ 处仰角 $24^{\circ}$：

$$ \sin 24^{\circ} \;=\; \frac{\text{vertical}}{AP} \;\Longrightarrow\; \text{vertical} \;=\; 99.33 \cdot \sin 24^{\circ} \;\approx\; 99.33 \cdot 0.4067 \;\approx\; 40.39 \;\approx\; 40.4 \;\text{m}. $$

(e) 3-D justification via MCR3U D1.7基于 MCR3U D1.7 的三维论证 R1

Splitting the 3-D problem into a horizontal triangle (solved by Law of Sines for the slant ranges) and a vertical right triangle (solved by primary trig for the height) lets a surveyor measure inaccessible heights without ever crossing the river, exactly the 3-D toolkit named in MCR3U D1.7.把三维问题拆为水平三角形（用正弦定理求斜距）与竖直直角三角形（用基本三角比求高），即可让测量员无需过河便测出不可达高度。这正是 MCR3U D1.7 命名的三维工具组。

2-D + 2-D = 3-D: project, solve, then lift.2D + 2D = 3D：投影、求解、再抬升。 Any 3-D measurement problem decomposes into (i) a horizontal-plane triangle, where the Law of Sines or Cosines gives slant or planar distances, and (ii) a vertical right triangle, where primary trig (the angle of elevation at the observer) converts a slant into a vertical lift. This is the same projector-and-lift split that AP Physics uses for projectile motion (horizontal $x$-component, vertical $y$-component) and that AP Calc BC uses for parametric curves. ON markers (MCR3U D1.7) reward the structural sentence in (e) over a numerical refinement, name the decomposition explicitly and the R1 lands.任意三维测量问题都可拆为 (i) 水平面三角形（用正/余弦定理求斜距或平面距离）+ (ii) 竖直直角三角形（用基本三角比，观察者仰角，把斜距转化为竖直高度）。这与 AP Physics 处理抛体运动的横（$x$）纵（$y$）分量分解、AP Calc BC 处理参数曲线的方法一致。ON 评分（MCR3U D1.7）奖励 (e) 中的结构性说明，而非更精的数值；明确写出分解，R1 必到手。