PART I · SHORT RESPONSE第一部分 · 简答题SAT-style MCQ + ON/BC short answer · 18 marksSAT 风格选择题 + ON/BC 简答题 · 18 分
Section A · Short ResponseA 节 · 简答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show enough work in the margin that a marker could verify. For short-answer items, label angles consistently and state units (m, cm, degrees) in every answer. Round trigonometric answers to the nearest tenth unless a problem requests an exact form. Calculator permitted throughout Part I unless a part explicitly says otherwise.题型混合：选择题与简答题。选择题请圈出字母，并在空白处保留可验证的过程。简答题需统一标注角，每个答案都要写明单位（m、cm、度）。三角函数答案除非题目要求精确值，否则保留到小数点后一位。第一部分如无特别说明可使用计算器。
Q1EASY🇺🇸 USSAT-style MCQ§1 SOH CAH TOA · HSG-SRT.C.6[3 marks]
In right triangle $ABC$ the right angle is at $C$, and the side lengths satisfy $AB = 13$, $BC = 5$, $CA = 12$. What is the value of $\sin A$?直角三角形 $ABC$ 中直角在 $C$，边长满足 $AB = 13$、$BC = 5$、$CA = 12$。求 $\sin A$ 的值。
In a 30-60-90 triangle, the side opposite the $30^{\circ}$ angle has length $6$. What is the length of the side opposite the $60^{\circ}$ angle?在 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。
(A) $3\sqrt{2}$
(B) $6\sqrt{2}$
(C) $6\sqrt{3}$
(D) $12$
Q4MEDIUM🇨🇦 ONON Provincial-style§3 Missing Side · MPM2D Trigonometry[4 marks]
In right triangle $PQR$, the right angle is at $Q$. The acute angle at $P$ measures $37^{\circ}$ and the hypotenuse $PR$ has length $20$ cm.直角三角形 $PQR$ 中直角在 $Q$。$P$ 处锐角为 $37^{\circ}$，斜边 $PR$ 长 $20$ cm。
(a)Write a trigonometric equation that relates the side $QR$ (opposite $\angle P$) to the hypotenuse and the angle $P$.写出一个三角方程，将 $QR$ 边（与 $\angle P$ 相对）与斜边及角 $P$ 联系起来。[1]
(b)Solve for $QR$ to the nearest tenth of a centimetre.求 $QR$，结果保留到 0.1 cm。[2]
(c)Compute the third side $PQ$ either by another trigonometric ratio or by the Pythagorean theorem, to the nearest tenth.用另一三角比或勾股定理求第三边 $PQ$，结果保留到 0.1。[1]
A right triangle has legs of length $8$ and $15$, with the right angle between them. Let $\theta$ be the acute angle opposite the side of length $15$.直角三角形两直角边长分别为 $8$ 与 $15$，直角夹在两边之间。设 $\theta$ 为与长 $15$ 的边相对的锐角。
(a)State which primary trigonometric ratio (sine, cosine, or tangent) involves the two known legs as the opposite-and-adjacent pair.指出哪一基本三角比（正弦、余弦或正切）以这两条已知直角边作为对边与邻边。[1]
(b)Write the equation $\tan\theta = \tfrac{\text{opp}}{\text{adj}}$ for this triangle, with the numbers substituted.代入数值写出本三角形的方程 $\tan\theta = \tfrac{\text{对边}}{\text{邻边}}$。[1]
(c)Solve for $\theta$ to the nearest tenth of a degree using the inverse-tangent function.用反正切函数求 $\theta$，结果保留到 0.1 度。[2]
(d)State the other acute angle of the triangle, using the complementary-angle relationship $A + B = 90^{\circ}$.利用余角关系 $A + B = 90^{\circ}$ 写出另一个锐角。[1]
PART II · EXTENDED RESPONSE第二部分 · 长答题AP-feeder FRQ + honors · 35 marksAP 衔接长答题 + 荣誉级 · 35 分
Section B · Extended ResponseB 节 · 长答题
Show every step. Label every angle and side on a diagram before substituting into a trig equation. For exact-value items (Part II Q7), do not switch to a calculator decimal; for non-exact items, state both the unrounded calculator value and the rounded final answer. The honors items Q8 and Q9 require the Law of Sines and the Law of Cosines respectively; both are on syllabus for ON Grade 10 (MPM2D), BC PC 11, and AB Math 20-1.每一步都要写出。代入三角方程前先在图上标好每个角和每条边。精确值题（第二部分 Q7）不得改用小数；非精确题须同时给出未取整的计算器值与最终舍入值。荣誉级 Q8 与 Q9 分别需要正弦定理与余弦定理；二者均属 ON 10 年级（MPM2D）、BC PC 11 及 AB Math 20-1 的考纲。
Q6MEDIUM🇺🇸 USAP-feeder FRQ§3 + §4 Solve a Right Triangle · HSG-SRT.C.8[8 marks]
In right triangle $ABC$, the right angle is at $C$. You are given $AB = 25$ (the hypotenuse) and $BC = 7$ (one leg).直角三角形 $ABC$ 中直角在 $C$。已知 $AB = 25$（斜边）、$BC = 7$（一条直角边）。
(a)Find the length of the third side $CA$. Show the use of the Pythagorean theorem and give an exact answer.求第三边 $CA$ 的长。请使用勾股定理并给出精确答案。[2]
(b)Find $\sin A$, $\cos A$, and $\tan A$ as exact fractions.以精确分数形式写出 $\sin A$、$\cos A$、$\tan A$。[3]
(c)Find $\angle A$ to the nearest tenth of a degree, then state $\angle B$ using the complementary-angle relationship from HSG-SRT.C.7.求 $\angle A$（保留到 0.1 度），再用 HSG-SRT.C.7 的余角关系写出 $\angle B$。[2]
(d)State $\sin B$ as an exact fraction and verify that $\sin B = \cos A$.以精确分数形式写出 $\sin B$，并验证 $\sin B = \cos A$。[1]
Q7MEDIUM🇨🇦 ABAB Diploma-style§2 Special Triangles · Math 10C 4.3[8 marks]
Use only the special-triangle reference values, no calculator decimals, for this entire question. Give every answer in exact (radical) form.整题仅使用特殊三角形参考值，不得使用计算器小数。所有答案均以精确（含根号）形式给出。
(a)State the exact values of $\sin 30^{\circ}$, $\cos 30^{\circ}$, and $\tan 30^{\circ}$.写出 $\sin 30^{\circ}$、$\cos 30^{\circ}$、$\tan 30^{\circ}$ 的精确值。[2]
(b)State the exact values of $\sin 45^{\circ}$, $\cos 45^{\circ}$, and $\tan 45^{\circ}$.写出 $\sin 45^{\circ}$、$\cos 45^{\circ}$、$\tan 45^{\circ}$ 的精确值。[2]
(c)An isosceles right triangle has hypotenuse $10$. Find the exact length of each leg by setting up $\cos 45^{\circ} = \tfrac{\text{leg}}{10}$, and rationalise the denominator.等腰直角三角形斜边为 $10$。利用 $\cos 45^{\circ} = \tfrac{\text{直角边}}{10}$ 求每条直角边的精确长度，并对分母进行有理化。[2]
(d)A 30-60-90 triangle has hypotenuse $14$. Find the exact lengths of both legs (the side opposite $30^{\circ}$ and the side opposite $60^{\circ}$).30-60-90 三角形斜边为 $14$。求两条直角边的精确长度（与 $30^{\circ}$ 相对的边与 $60^{\circ}$ 相对的边）。[2]
Q8HARDHonors🇨🇦 BCBC Provincial-style§6 Law of Sines · PC 11 sine / cosine laws[9 marks]
In triangle $ABC$ (not assumed right), the angle at $A$ measures $52^{\circ}$, the angle at $B$ measures $74^{\circ}$, and the side $a$ (opposite $\angle A$) has length $18$ cm.三角形 $ABC$（不一定为直角）中，$A$ 角为 $52^{\circ}$，$B$ 角为 $74^{\circ}$，边 $a$（与 $\angle A$ 相对）长 $18$ cm。
(a)State the Law of Sines in the form $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$, and explain in one sentence why an AAS configuration (two angles plus one side) admits a unique triangle.写出正弦定理 $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$，并用一句话说明 AAS 情形（两角加一边）为何对应唯一三角形。[2]
(b)Find $\angle C$ using the fact that the interior angles of a triangle sum to $180^{\circ}$.利用三角形内角和为 $180^{\circ}$ 求 $\angle C$。[1]
(c)Find side $b$ (opposite $\angle B$) using the Law of Sines, to the nearest tenth of a centimetre.用正弦定理求边 $b$（与 $\angle B$ 相对），保留到 0.1 cm。[3]
(d)Find side $c$ (opposite $\angle C$) using the Law of Sines, to the nearest tenth of a centimetre.用正弦定理求边 $c$（与 $\angle C$ 相对），保留到 0.1 cm。[2]
(e)Compute the area of triangle $ABC$ using $A = \tfrac{1}{2} a b \sin C$ (the formula from HSG-SRT.D.9), to the nearest tenth of a square centimetre.用面积公式 $A = \tfrac{1}{2} a b \sin C$（出自 HSG-SRT.D.9）计算三角形 $ABC$ 的面积，保留到 0.1 cm$^{2}$。[1]
Q9HARDHonors🇨🇦 ONON Provincial-style§7 Law of Cosines + Ambiguous SSA · MCR3U D1.6[10 marks]
Part A. In triangle $XYZ$, side $x$ (opposite $\angle X$) has length $9$, side $y$ has length $12$, and the angle $\angle Z$ between sides $x$ and $y$ measures $108^{\circ}$.A 部分。三角形 $XYZ$ 中，边 $x$（与 $\angle X$ 相对）长 $9$，边 $y$ 长 $12$，$x$ 与 $y$ 之间的夹角 $\angle Z = 108^{\circ}$。
(a)State the Law of Cosines in the form $c^{2} = a^{2} + b^{2} - 2 a b \cos C$, and explain in one sentence why an SAS configuration (two sides and the included angle) admits a unique triangle.写出余弦定理 $c^{2} = a^{2} + b^{2} - 2 a b \cos C$，并用一句话说明 SAS 情形（两边加夹角）为何对应唯一三角形。[2]
(b)Use the Law of Cosines to find side $z$ (opposite $\angle Z$), to the nearest tenth.用余弦定理求边 $z$（与 $\angle Z$ 相对），保留到 0.1。[3]
Part B. A different triangle $LMN$ has side $\ell = 8$ (opposite $\angle L$), side $m = 11$ (opposite $\angle M$), and $\angle L = 35^{\circ}$. This is an SSA configuration, the ambiguous case.B 部分。另一三角形 $LMN$ 中边 $\ell = 8$（与 $\angle L$ 相对）、边 $m = 11$（与 $\angle M$ 相对）、$\angle L = 35^{\circ}$。这是 SSA 情形，即模糊情形。
(c)Use the Law of Sines to write $\sin M = \tfrac{m \sin L}{\ell}$, substitute, and solve for the principal-value angle $M_{1}$ (the one returned by $\sin^{-1}$), to the nearest tenth of a degree.用正弦定理写出 $\sin M = \tfrac{m \sin L}{\ell}$，代入并求主值角 $M_{1}$（即 $\sin^{-1}$ 直接返回的角），保留到 0.1 度。[2]
(d)State the second candidate angle $M_{2} = 180^{\circ} - M_{1}$, and decide whether both, one, or neither candidate gives a valid triangle, by checking that $\angle L + M$ stays strictly below $180^{\circ}$.写出第二个候选角 $M_{2} = 180^{\circ} - M_{1}$，并通过检验 $\angle L + M$ 是否严格小于 $180^{\circ}$，判断两个、一个还是都不构成合法三角形。[2]
(e)Confirm in one sentence why this configuration is "ambiguous" with reference to the SSA naming.用一句话结合 SSA 命名说明此情形为何称作"模糊"。[1]
PART III · MODELING / APPLIED第三部分 · 建模 / 应用Universal · 28 marks通用题 · 28 分
Section C · Modeling and ApplicationsC 节 · 建模与应用
Draw the diagram first; mark the right angle, the angle of elevation or depression, and the segments whose lengths you know. State your variable with units. State the answer in a one-sentence sign-off ("the building is about ___ metres tall"). Calculator permitted throughout Part III; round trigonometric answers to the nearest tenth unless asked otherwise.先画图，标出直角、仰角或俯角、以及所有已知边。变量写明单位。结论用一句话作答（"大楼约 ___ 米高"）。第三部分允许使用计算器；三角函数答案除非另作说明，保留到 0.1。
Q10MEDIUM🇺🇸 USAP-feeder FRQ§5 Angle of Elevation · HSG-SRT.C.8[9 marks]
A surveyor stands $60$ metres from the base of a vertical flagpole on level ground. From her line of sight (eye level $1.6$ m above the ground), the angle of elevation to the top of the flagpole is $32^{\circ}$.一位测量员站在水平地面上，距离垂直旗杆底部 $60$ 米。她的视线（眼睛高度距地面 $1.6$ m）到旗杆顶部的仰角为 $32^{\circ}$。
(a)Sketch the right triangle formed by the horizontal sight line at eye level, the vertical segment from eye level to the top of the flagpole, and the sloped sight line. Label the $32^{\circ}$ angle, the horizontal leg $60$ m, and the vertical leg $h$ (the height above eye level).画出由眼睛高度水平视线、眼睛高度至旗杆顶的竖直段、以及斜视线组成的直角三角形。标注 $32^{\circ}$ 角、水平直角边 $60$ m，以及竖直直角边 $h$（眼睛高度以上的高度）。[2]
(b)Set up a tangent equation for $h$ and solve for $h$ to the nearest tenth of a metre.为 $h$ 列出正切方程并求解，结果保留到 0.1 m。[3]
(c)State the total height of the flagpole above the ground, accounting for the $1.6$ m eye level.考虑 $1.6$ m 眼睛高度，写出旗杆距地面的总高度。[1]
(d)The surveyor walks $20$ m closer (now $40$ m from the base, same eye level). State the new angle of elevation, to the nearest tenth of a degree, and explain in one sentence why the angle of elevation increases as she approaches the pole.测量员再向前走 $20$ m（现距底 $40$ m，眼睛高度不变）。写出新的仰角（保留到 0.1 度），并用一句话说明为何走近时仰角增大。[3]
Q11MEDIUM🇨🇦 ABAB Diploma-style§5 Clinometer + Angle of Depression · Math 10C 4.5[9 marks]
A geographer at the edge of a cliff uses a clinometer to measure the angle of depression to a small boat on the lake below. From a viewing platform $48$ m above the lake surface, the angle of depression to the boat is $18^{\circ}$. (The angle of depression is measured below the horizontal sight line.)一位地理学家在悬崖边缘用测角仪测量湖面小船的俯角。在距离湖面 $48$ m 高的观景平台上，到船的俯角为 $18^{\circ}$。（俯角自水平视线向下度量。）
(a)Sketch the right triangle. Mark the horizontal sight line at the viewing platform, the $18^{\circ}$ angle of depression, the $48$ m vertical from platform to lake surface, and the unknown horizontal distance $d$ from the foot of the cliff to the boat. Identify the angle inside the triangle at the boat using the alternate-interior-angle theorem.画出直角三角形。标出观景平台处水平视线、$18^{\circ}$ 俯角、平台到湖面 $48$ m 的竖直距离，以及自崖底到船的未知水平距离 $d$。用内错角定理标出三角形在船处的角。[2]
(b)Set up a tangent equation that uses the $18^{\circ}$ angle at the boat (the angle of elevation back up to the platform), and solve for $d$ to the nearest metre.利用船处的 $18^{\circ}$ 角（即回望平台的仰角）列出正切方程，求 $d$，结果保留到整数 m。[3]
(c)A few minutes later the geographer remeasures: the angle of depression has increased to $26^{\circ}$. Find the new horizontal distance from the cliff foot to the boat.数分钟后再测，俯角增大为 $26^{\circ}$。求自崖底到船的新水平距离。[2]
(d)State how far the boat has travelled between the two measurements, and the direction (toward or away from the cliff).写出两次测量之间船行驶的距离及方向（接近或远离悬崖）。[2]
Two surveyors stand $80$ m apart on level ground on the near bank of a river. They both sight the same fixed point $P$ at the top of a cliff on the far bank. From surveyor $A$, the line of sight to $P$ makes an angle of $58^{\circ}$ with line segment $AB$. From surveyor $B$, the line of sight to $P$ makes an angle of $72^{\circ}$ with line segment $BA$. (The two sight lines and segment $AB$ form triangle $ABP$ in a plane that is tilted up from the ground.)两位测量员在河近岸水平地面上相距 $80$ m。他们同时观测远岸崖顶上的同一固定点 $P$。从测量员 $A$ 看，视线 $AP$ 与线段 $AB$ 成 $58^{\circ}$ 角；从测量员 $B$ 看，视线 $BP$ 与线段 $BA$ 成 $72^{\circ}$ 角。（两条视线与 $AB$ 共同构成三角形 $ABP$，该三角形所在平面相对于地面向上倾斜。）
(a)State the third angle $\angle APB$ inside triangle $ABP$.写出三角形 $ABP$ 中第三个角 $\angle APB$。[1]
(b)Use the Law of Sines on triangle $ABP$ to find the slant distance $AP$ from surveyor $A$ to point $P$, to the nearest tenth of a metre.在三角形 $ABP$ 上用正弦定理求测量员 $A$ 到 $P$ 的斜距 $AP$，保留到 0.1 m。[3]
(c)Use the Law of Sines on triangle $ABP$ to find the slant distance $BP$ from surveyor $B$ to point $P$, to the nearest tenth of a metre.在三角形 $ABP$ 上用正弦定理求测量员 $B$ 到 $P$ 的斜距 $BP$，保留到 0.1 m。[2]
(d)Surveyor $A$ then tilts the theodolite upward and reads an angle of elevation of $24^{\circ}$ from horizontal to point $P$ along the slant $AP$. Use the right-triangle relationship $\sin 24^{\circ} = \tfrac{\text{vertical}}{AP}$ to find the vertical height of $P$ above the ground at $A$, to the nearest tenth of a metre.测量员 $A$ 接着将经纬仪向上倾斜，沿斜线 $AP$ 测得仰角 $24^{\circ}$。利用直角三角形关系 $\sin 24^{\circ} = \tfrac{\text{竖直高度}}{AP}$ 求 $P$ 相对于 $A$ 处地面的竖直高度，保留到 0.1 m。[3]
(e)Confirm in one sentence why combining the Law of Sines (for the horizontal triangle) with right-triangle trig (for the vertical lift) lets a surveyor measure inaccessible heights without crossing the river, citing MCR3U D1.7.用一句话说明：为何将正弦定理（处理水平三角形）与直角三角形三角比（处理竖直高度）结合，可让测量员无需过河即可测得不可达高度，并引用 MCR3U D1.7。[1]
🇺🇸 US Common Core美国共同核心HSG-SRT.C.6 · HSG-SRT.C.7 · HSG-SRT.C.8 · HSG-SRT.D.9 (+) · HSG-SRT.D.10 (+) · HSG-SRT.D.11 (+)
🇨🇦 British Columbia不列颠哥伦比亚FMP&PC 10 right-triangle trig + indirect measurement · PC 11 sine / cosine laws including ambiguous case + special anglesFMP&PC 10 直角三角形三角比 + 间接测量 · PC 11 正/余弦定理（含模糊情形）+ 特殊角
🇨🇦 Alberta阿尔伯塔Math 10C GO 4, indicators 4.1–4.5 (right-triangle + clinometer) · Math 20-1 GO 3, indicators 3.1–3.6 (sine / cosine laws, ambiguous case)Math 10C 总体目标 4，指标 4.1–4.5（直角三角形 + 测角仪）· Math 20-1 总体目标 3，指标 3.1–3.6（正/余弦定理，模糊情形）
Full 4-column Syllabus Map lives in ../Study Guides/Unit_7_Right-Triangle_Trigonometry.html.完整 4 栏大纲对照见 ../Study Guides/Unit_7_Right-Triangle_Trigonometry.html。