High School Math

Right-Triangle Trigonometry直角三角形三角学

Trigonometry is similarity made numerical. Two right triangles with the same acute angle have the same side ratios, and those ratios get names: sine, cosine, tangent. This guide walks the full path from SOH CAH TOA on a single right triangle through to the Law of Sines and the Law of Cosines for triangles with no right angle at all. Worked examples include the canonical ladder against a wall, an angle of elevation to the top of a building, a surveying problem across a river, and the ambiguous SSA case where one given set of measurements produces two valid triangles.三角学（trigonometry）是把相似（similarity）变成数值的工具。任何两个具有相同锐角的直角三角形，对应边的比相同；这些比就有了名字：正弦（sine）、余弦（cosine）、正切（tangent）。本指南从单个直角三角形的 SOH CAH TOA 出发，一路推进到适用于一般三角形（没有直角）的正弦定理和余弦定理。例题涵盖经典的"梯子靠墙"问题、对楼顶的仰角（angle of elevation）、跨河测量（surveying），以及一组数据可对应两个合法三角形的模糊情形（ambiguous case，SSA）。

7 sections7 节内容 US Common Core · ON · BC · ABUS 共同核心 · ON · BC · AB Honors block on Sine / Cosine Laws正弦 / 余弦定理为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Right-triangle trigonometry lands in Grade 10 in every curriculum we map to, but the Law of Sines and Law of Cosines for general (non-right) triangles do not. US Common Core puts them at the (+) "honors / Pre-Calc" level under HSG-SRT.D; BC defers them from FMP&PC 10 to Pre-Calc 11; Alberta defers them from Math 10C to Math 20-1. Ontario is the exception: MPM2D meets sine law and cosine law in Grade 10 alongside the right-triangle work. The eight-row table tells you which sections are on your syllabus right now; each row cites the curriculum document it was checked against.直角三角形三角学在我们对照的所有大纲中都属于 10 年级内容；但用于一般（非直角）三角形的正弦定理与余弦定理则不然。美国共同核心将其放在 HSG-SRT.D 的 (+) "荣誉 / Pre-Calc" 级；BC 把它从 FMP&PC 10 推迟到 PC 11；阿尔伯塔把它从 Math 10C 推迟到 Math 20-1。安大略是例外：MPM2D 在 10 年级同时覆盖直角与正/余弦定理。下面的八行表告诉你当前大纲下应重点学习哪些节；每行都注明了对应的课纲依据。

If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇺🇸 US Grade 10 (Geometry) , HSG-SRT.C美国 10 年级（几何）, HSG-SRT.C §1, §2, §3, §4, §5§1, §2, §3, §4, §5 §6 and §7 (Law of Sines / Cosines are HSG-SRT.D, marked (+) and reserved for AP-feeder Pre-Calc, not the Geometry course)§6、§7（正/余弦定理属 HSG-SRT.D，标 (+)，归 AP 衔接 Pre-Calc，不在几何课内） ccssm_hs_math.pdf , HSG-SRT.C.6, .C.7, .C.8 (acute-angle trig ratios, complementary-angle identity, solving right triangles in applied problems), HSG-SRT.C.6、.C.7、.C.8（锐角三角比、余角恒等式、应用题中求解直角三角形）
🇨🇦 ON Grade 10 , MPM2D安大略 10 年级 , MPM2D §1 through §7. Ontario is the early-onset province: MPM2D already covers sine law and cosine law via the Acute Triangles sub-cluster in Grade 10§1 至 §7。安大略是早覆盖省份：MPM2D 已在 10 年级通过"锐角三角形"子簇覆盖正/余弦定理 Nothing , full unit fits Grade 10 in this province无 , 全单元均在本省 10 年级范围内 math_grades_9-10.pdf , MPM2D Trigonometry strand overall expectations: right triangles via primary trig ratios and acute triangles via sine / cosine law, MPM2D 三角学单元总体期望：用三角比解直角三角形，并用正/余弦定理解锐角三角形
🇨🇦 ON Grade 11 , MCR3U安大略 11 年级 , MCR3U Full review §1-§7 with depth on the ambiguous case (MCR3U D1.6 names it explicitly)全面复习 §1-§7，对模糊情形加深（MCR3U D1.6 明确点名） Nothing , lean on §6 and §7 because MCR3U revisits both laws and adds 3-D problem-posing on top无 , 加重 §6、§7，MCR3U 重新覆盖两定理并加入三维问题 math_grades_11-12.pdf , MCR3U Strand D Trigonometric Functions, expectations D1.6 (2-D, includes ambiguous case) and D1.7 (3-D), MCR3U 单元 D 三角函数，期望 D1.6（二维，含模糊情形）与 D1.7（三维）
🇨🇦 BC Grade 10 , FMP&PC 10BC 10 年级 , FMP&PC 10 §1, §2 (light, mostly 30-60-90 / 45-45-90 by similarity), §3, §4, §5§1、§2（轻量，主要为 30-60-90 / 45-45-90 的相似导出）、§3、§4、§5 §6 and §7 (sine / cosine laws are PC 11 content, not FMP&PC 10)§6、§7（正/余弦定理属 PC 11，不在 FMP&PC 10） fmpc10_elab.pdf , Content: "sine, cosine, and tangent ratios" + "right-triangle problems: determining missing sides and/or angles using trigonometric ratios and the Pythagorean theorem" + "contexts involving direct and indirect measurement", 内容："sin、cos、tan 比"；"用三角比与勾股定理求直角三角形未知边/角"；"涉及直接与间接测量的情境"
🇨🇦 BC Grade 11 , PC 11BC 11 年级 , PC 11 §1-§7 in full. PC 11 is BC's home for sine / cosine laws including the ambiguous case§1-§7 完整学习。PC 11 是 BC 正/余弦定理及模糊情形的对应课程 Nothing , treat as a refresher before moving on to unit-circle trig (Unit 8)无 , 作为进入单位圆三角学（Unit 8）前的回顾 pc11_elab.pdf , Content elaboration: "use of sine and cosine laws to solve non-right triangles, including ambiguous cases"; plus "special angles, as connected with the 30-60-90 and 45-45-90 triangles", 内容："使用正/余弦定理求解非直角三角形，含模糊情形"；"特殊角，与 30-60-90、45-45-90 三角形相联系"
🇨🇦 AB Grade 10 , Math 10C阿尔伯塔 10 年级 , Math 10C §1, §2, §3, §4, §5. Math 10C explicitly names the clinometer as a measurement instrument in 4.5§1、§2、§3、§4、§5。Math 10C 在 4.5 明确点名测斜仪（clinometer）为测量工具 §6 and §7 (sine / cosine laws are Math 20-1 General Outcome 3)§6、§7（正/余弦定理属 Math 20-1 总目标 3） pos_10-12_indicators.pdf , Math 10C Measurement General Outcome 4 with indicators 4.1 (similar right triangles), 4.3 (solve right triangles), 4.4 (apply primary trig ratios), 4.5 (direct and indirect measurement), Math 10C 测量总目标 4，含指标 4.1（相似直角三角形）、4.3（求解直角三角形）、4.4（应用三角比）、4.5（直接与间接测量）
🇨🇦 AB Grade 11 , Math 20-1阿尔伯塔 11 年级 , Math 20-1 Full §1-§7. Math 20-1 indicator 3.3 expects students to "explain the steps in a given proof of the sine law or cosine law" , the derivation matters, not just the formula§1-§7 全部。Math 20-1 指标 3.3 要求学生"解释正弦/余弦定理给定证明中的各步骤", 公式背后的推导同等重要 Nothing , Math 20-1 is the dedicated home for both laws and the ambiguous case in AB无 , Math 20-1 是阿尔伯塔正/余弦定理与模糊情形的专属课程 pos_10-12_indicators.pdf , Math 20-1 Trigonometry General Outcome 3 with indicators 3.1-3.6 (sketch, solve non-right triangles, prove, ambiguous case), Math 20-1 三角学总目标 3，含指标 3.1-3.6（作图、求解非直角三角形、证明、模糊情形）
🇺🇸 US AP-feeder (Pre-Calc / Honors)美国 AP 衔接（Pre-Calc / 荣誉） All seven sections plus the going-deeper proofs on §6 and §7. AP Calc and IB Math AA HL both assume fluency on both laws全部 7 节，并完成 §6、§7 的"深入"证明。AP Calc 与 IB Math AA HL 都默认你熟练掌握两定理 Nothing , this unit lays the geometric foundation that Unit 8 (unit circle) and the IB Math AA HL trig topics build on无 , 本单元为单元 8（单位圆）以及 IB Math AA HL 三角主题打基础 ccssm_hs_math.pdf , the (+) cluster HSG-SRT.D.9-11 is explicitly the AP-feeder content; see also CCSSM domain HSF-TF for the trig-functions extension, (+) 簇 HSG-SRT.D.9-11 正是 AP 衔接内容；另见 CCSSM 的 HSF-TF（三角函数扩展）

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise four things: the SOH CAH TOA mnemonic ($\sin = \mathrm{opp}/\mathrm{hyp}$, $\cos = \mathrm{adj}/\mathrm{hyp}$, $\tan = \mathrm{opp}/\mathrm{adj}$); the two exact-value tables for the 30-60-90 and 45-45-90 triangles; the picture for an angle of elevation versus an angle of depression; if your row includes §6 / §7, also the Law of Sines $\tfrac{a}{\sin A} = \tfrac{b}{\sin B} = \tfrac{c}{\sin C}$ and the Law of Cosines $c^2 = a^2 + b^2 - 2 a b \cos C$. Read every cram-cheat box. Skip the going-deeper proofs.背熟四件事：SOH CAH TOA 口诀（$\sin = $ 对边 / 斜边，$\cos = $ 邻边 / 斜边，$\tan = $ 对边 / 邻边）；30-60-90 与 45-45-90 两张精确值表；仰角与俯角（angle of elevation / angle of depression）的图示；若你的行包含 §6 / §7，再加正弦定理 $\tfrac{a}{\sin A} = \tfrac{b}{\sin B} = \tfrac{c}{\sin C}$ 与余弦定理 $c^2 = a^2 + b^2 - 2 a b \cos C$。读每个速记框，跳过深入证明。

*
If you are going for the top mark如果你目标顶分

Always sketch the triangle and label which side is opposite, adjacent, or hypotenuse before you write a ratio. Always state degree-mode on the calculator (a single missing setting flips every answer). Practise the ambiguous SSA case until you can predict the no-solution / one-solution / two-solution outcome from the given data alone. Read both derivations: the Law of Sines from the altitude, the Law of Cosines from the Pythagorean theorem. These two derivations re-appear verbatim in IB Math HL Unit C2 and on AP Pre-Calc.写比之前先画三角形并标出对边、邻边、斜边。计算前确认计算器在 degree mode（角度模式）, 一个设置错误，所有答案全错。反复练 SSA 模糊情形，做到只看已知数据就能判断"无解 / 一解 / 两解"。两个推导都要会：用高线推正弦定理，用勾股定理推余弦定理。这两个推导在 IB Math HL Unit C2 和 AP Pre-Calc 中原样再现。

Honors flag.荣誉级标记。 Sections 7.6 (Law of Sines) and 7.7 (Law of Cosines + ambiguous SSA) carry the Honors chip for US Geometry, BC FMP&PC 10, and AB Math 10C , in those courses the laws are deferred to PC 11 / Math 20-1 / AP-feeder Pre-Calc. They are core, not honors, in Ontario MPM2D and MCR3U, in BC PC 11, and in AB Math 20-1. If your row above sends you to §6 and §7, treat them as required content, not enrichment.§6（正弦定理）与 §7（余弦定理 + 模糊 SSA）在美国几何课、BC FMP&PC 10、AB Math 10C 中标为 Honors, 这些课程把两定理推迟到 PC 11 / Math 20-1 / AP 衔接 Pre-Calc。但在安大略 MPM2D、MCR3U、BC PC 11、AB Math 20-1 中，它们是核心而非荣誉内容。如果你的行指向 §6、§7，就把它们视为必学，不是拓展。
Section 1 · Primary trigonometric ratios第 1 节 · 三角比

The Three Primary Ratios: sine, cosine, tangent三个基本三角比：正弦、余弦、正切

SOH CAH TOA , the definitions on a right triangle.SOH CAH TOA , 直角三角形上的定义。

Pick an acute angle $\theta$ in a right triangle. Relative to $\theta$, label the side opposite, the side adjacent (the leg, not the hypotenuse), and the hypotenuse.在直角三角形中选定一个锐角 $\theta$。相对 $\theta$，把三边分别标为"对边"、"邻边"（另一条直角边，非斜边）和"斜边"。

$$ \sin\theta \;=\; \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta \;=\; \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta \;=\; \frac{\text{opposite}}{\text{adjacent}}. $$
  • Mnemonic.口诀。 SOH (Sine = Opp / Hyp), CAH (Cosine = Adj / Hyp), TOA (Tangent = Opp / Adj).SOH（正弦 = 对 / 斜），CAH（余弦 = 邻 / 斜），TOA（正切 = 对 / 邻）。
  • Range.取值范围。 For acute $\theta$, $0 < \sin\theta < 1$, $0 < \cos\theta < 1$, and $\tan\theta > 0$ with no upper bound.$\theta$ 为锐角时，$0 < \sin\theta < 1$，$0 < \cos\theta < 1$，$\tan\theta > 0$ 且无上界。
  • Co-function identity.余角恒等式。 $\sin\theta = \cos(90^{\circ} - \theta)$ , CCSSM HSG-SRT.C.7. Swap sin for cos when angles are complementary.$\sin\theta = \cos(90^{\circ} - \theta)$ , CCSSM HSG-SRT.C.7。两角互余时 sin 与 cos 互换。
  • Pythagorean identity.勾股恒等式。 $\sin^{2}\theta + \cos^{2}\theta = 1$ for every $\theta$. Lets you find one ratio from the other without redrawing the triangle.对任何 $\theta$ 都有 $\sin^{2}\theta + \cos^{2}\theta = 1$。无需重画三角形即可由一个比求另一个。
Why the ratios deserve names (similarity). Two right triangles sharing an acute angle are similar (AA), so corresponding side ratios are equal. The ratio depends on the angle, not the triangle's size , that's why each ratio can be named once and reused. (This is the FMP&PC 10 Big Idea verbatim: "Trigonometry uses proportional reasoning to solve indirect measurement problems.")为何要给三角比命名（相似性）。 锐角相同的两个直角三角形相似（AA），对应边之比相等。比值只取决于角，不取决于三角形大小 , 所以每个比命名一次即可重复使用。（这正是 BC FMP&PC 10 的大概念："三角学用比例推理解决间接测量问题"。）
Worked Example 1 · Read the ratios off a 3-4-5 triangle例题 1 · 在 3-4-5 直角三角形中读取三角比

A right triangle has legs of length $3$ and $4$ with hypotenuse $5$. Let $\theta$ be the acute angle opposite the side of length $3$. Find $\sin\theta$, $\cos\theta$, and $\tan\theta$ as exact fractions, then state $\theta$ in degrees to two decimals.某直角三角形两直角边长为 $3$、$4$，斜边为 $5$。设 $\theta$ 为长度 $3$ 的边所对的锐角。以精确分数求 $\sin\theta$、$\cos\theta$、$\tan\theta$，并将 $\theta$ 化为角度并保留两位小数。

Identify the three sides relative to $\theta$.辨识相对 $\theta$ 的三条边。 Opposite $\theta$: the leg of length $3$. Adjacent to $\theta$: the leg of length $4$ (not the hypotenuse). Hypotenuse: the side of length $5$.$\theta$ 的对边：长度 $3$。邻边：长度 $4$（非斜边）。斜边：长度 $5$。

Apply SOH CAH TOA.套用 SOH CAH TOA。

$$ \sin\theta = \frac{3}{5}, \qquad \cos\theta = \frac{4}{5}, \qquad \tan\theta = \frac{3}{4}. $$

Sanity-check via the Pythagorean identity.用勾股恒等式核验。

$$ \sin^{2}\theta + \cos^{2}\theta = \frac{9}{25} + \frac{16}{25} = 1. \;\checkmark $$

Convert to degrees.换算为角度。 $\theta = \arctan(3/4) \approx 36.87^{\circ}$. (Set your calculator to degree mode , the most common single error on this kind of problem.)$\theta = \arctan(3/4) \approx 36.87^{\circ}$。（计算器务必设为 degree mode , 这是此类题最常见的单一错误。）

In a right triangle with legs $5$ and $12$ and hypotenuse $13$, let $\alpha$ be the angle opposite the side of length $5$. Which value equals $\cos\alpha$?在直角边 $5$、$12$、斜边 $13$ 的直角三角形中，设 $\alpha$ 为长度 $5$ 的边所对的角。$\cos\alpha$ 等于多少？
§1 · Q1
$5/13$
$5/12$
$12/13$
$12/5$
Cosine equals adjacent over hypotenuse. Adjacent to $\alpha$ is the leg of length $12$; the hypotenuse is $13$. So $\cos\alpha = 12/13$.余弦 = 邻边 / 斜边。$\alpha$ 的邻边为 $12$，斜边为 $13$，故 $\cos\alpha = 12/13$。
Cosine uses adjacent and hypotenuse , never opposite. Decide which leg is opposite $\alpha$ and which is adjacent before writing the ratio.余弦用邻边与斜边，不是对边。写比之前先判断哪条直角边是 $\alpha$ 的对边、哪条是邻边。
If $\sin\theta = 7/25$ for an acute angle $\theta$, find $\cos\theta$ as an exact fraction.若锐角 $\theta$ 满足 $\sin\theta = 7/25$，以精确分数求 $\cos\theta$。
§1 · Q2
$7/24$
$24/25$
$18/25$
$25/7$
$\cos^{2}\theta = 1 - \sin^{2}\theta = 1 - 49/625 = 576/625$, so $\cos\theta = 24/25$ (positive because $\theta$ is acute). Equivalent direct route: hypotenuse $25$, one leg $7$, so the other leg satisfies $a^{2} + 49 = 625 \Rightarrow a = 24$.$\cos^{2}\theta = 1 - \sin^{2}\theta = 1 - 49/625 = 576/625$，故 $\cos\theta = 24/25$（锐角取正）。等价做法：斜边 $25$、一直角边 $7$，由勾股定理另一直角边 $a$ 满足 $a^{2} + 49 = 625 \Rightarrow a = 24$。
$\sin^{2}\theta + \cos^{2}\theta = 1$, or equivalently use the Pythagorean theorem on a right triangle with hypotenuse $25$ and one leg $7$.$\sin^{2}\theta + \cos^{2}\theta = 1$；或在斜边 $25$、直角边 $7$ 的直角三角形上用勾股定理。
Section 2 · Special right triangles第 2 节 · 特殊直角三角形

Special Right Triangles: 45-45-90 and 30-60-90特殊直角三角形：45-45-90 与 30-60-90

Two side-ratio templates worth memorising.两个值得背熟的边比模板。
  • 45-45-90: legs $1 : 1$, hypotenuse $\sqrt{2}$. Both acute angles are $45^{\circ}$. Useful exact values: $\sin 45^{\circ} = \cos 45^{\circ} = \tfrac{\sqrt{2}}{2}$, $\tan 45^{\circ} = 1$.两条直角边 $1 : 1$，斜边 $\sqrt{2}$。两锐角均为 $45^{\circ}$。常用精确值：$\sin 45^{\circ} = \cos 45^{\circ} = \tfrac{\sqrt{2}}{2}$，$\tan 45^{\circ} = 1$。
  • 30-60-90: sides in ratio $1 : \sqrt{3} : 2$ opposite the $30^{\circ}, 60^{\circ}, 90^{\circ}$ angles. Useful exact values: $\sin 30^{\circ} = 1/2$, $\cos 30^{\circ} = \tfrac{\sqrt{3}}{2}$, $\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$, $\cos 60^{\circ} = 1/2$, $\tan 30^{\circ} = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$, $\tan 60^{\circ} = \sqrt{3}$.边比 $1 : \sqrt{3} : 2$，分别对 $30^{\circ}, 60^{\circ}, 90^{\circ}$ 三角。常用精确值：$\sin 30^{\circ} = 1/2$，$\cos 30^{\circ} = \tfrac{\sqrt{3}}{2}$，$\sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$，$\cos 60^{\circ} = 1/2$，$\tan 30^{\circ} = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$，$\tan 60^{\circ} = \sqrt{3}$。
  • Where they come from.来源。 45-45-90: cut a unit square along the diagonal. 30-60-90: cut an equilateral triangle of side $2$ in half through one vertex.45-45-90 来自把单位正方形沿对角线切开；30-60-90 来自把边长 $2$ 的等边三角形过顶点平分。
BC PC 11 calls these out explicitly: "special angles, as connected with the 30-60-90 and 45-45-90 triangles." Same pair underpins Unit 8's unit-circle exact values.BC PC 11 明确点名："特殊角，与 30-60-90 和 45-45-90 三角形相联系"。这两个三角形也是 Unit 8 单位圆精确值的基础。
Worked Example 2 · Exact missing sides in a 30-60-90例题 2 · 30-60-90 三角形中的精确未知边

A 30-60-90 right triangle has its shorter leg (opposite the $30^{\circ}$ angle) of length $5$. Find the exact lengths of the other leg and of the hypotenuse.某 30-60-90 直角三角形的短直角边（对 $30^{\circ}$ 角）长 $5$。求另一直角边和斜边的精确长度。

Match the template.对照模板。 In ratio terms the sides are $1 : \sqrt{3} : 2$ opposite $30^{\circ}, 60^{\circ}, 90^{\circ}$. Scale by the factor that makes the short leg equal to $5$ , scale factor $k = 5$.三边比为 $1 : \sqrt{3} : 2$，分别对 $30^{\circ}, 60^{\circ}, 90^{\circ}$。让短边等于 $5$ , 缩放因子 $k = 5$。

$$ \text{short leg} = 1 \cdot 5 = 5, \qquad \text{long leg} = \sqrt{3} \cdot 5 = 5\sqrt{3}, \qquad \text{hypotenuse} = 2 \cdot 5 = 10. $$

Verify via trigonometry.用三角比验证。 $\sin 30^{\circ} = 5 / 10 = 1/2$ ✓. $\tan 60^{\circ} = (5\sqrt{3}) / 5 = \sqrt{3}$ ✓.$\sin 30^{\circ} = 5 / 10 = 1/2$ ✓。$\tan 60^{\circ} = (5\sqrt{3}) / 5 = \sqrt{3}$ ✓。

Numerical check.数值核对。 $5\sqrt{3} \approx 8.66$, hypotenuse $10$. The longer leg is about $1.73 \times$ the short leg , the signature of a 30-60-90.$5\sqrt{3} \approx 8.66$，斜边 $10$。长直角边约为短边的 $1.73$ 倍 , 30-60-90 的标志。

An isoceles right triangle (45-45-90) has hypotenuse $8$. Find the exact length of each leg.等腰直角三角形（45-45-90）的斜边为 $8$。求每条直角边的精确长度。
§2 · Q1
$4$
$8/\sqrt{3}$
$4\sqrt{2}$
$8\sqrt{2}$
In a 45-45-90, hypotenuse $= \sqrt{2}$ × leg. So leg $= 8 / \sqrt{2} = 8\sqrt{2}/2 = 4\sqrt{2}$.45-45-90 中斜边 $= \sqrt{2}$ × 直角边，故直角边 $= 8 / \sqrt{2} = 8\sqrt{2}/2 = 4\sqrt{2}$。
In a 45-45-90 the ratio leg : leg : hypotenuse is $1 : 1 : \sqrt{2}$. Divide the hypotenuse by $\sqrt{2}$, then rationalise.45-45-90 边比 $1 : 1 : \sqrt{2}$。斜边除以 $\sqrt{2}$ 并有理化分母。
Compute $\sin 60^{\circ} + \cos 30^{\circ}$ as an exact expression.以精确表达式计算 $\sin 60^{\circ} + \cos 30^{\circ}$。
§2 · Q2
$1$
$\sqrt{2}$
$\sqrt{3}/2$
$\sqrt{3}$
$\sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2$, so the sum is $\sqrt{3}/2 + \sqrt{3}/2 = \sqrt{3}$. (Notice the co-function identity at work: $\sin 60^{\circ} = \cos(90^{\circ} - 60^{\circ}) = \cos 30^{\circ}$.)$\sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2$，两者之和 $= \sqrt{3}/2 + \sqrt{3}/2 = \sqrt{3}$。（这正是余角恒等式：$\sin 60^{\circ} = \cos(90^{\circ} - 60^{\circ}) = \cos 30^{\circ}$。）
Both values equal $\sqrt{3}/2$. Co-function identity guarantees this: $\sin\theta = \cos(90^{\circ} - \theta)$.两值均为 $\sqrt{3}/2$。余角恒等式保证此事：$\sin\theta = \cos(90^{\circ} - \theta)$。
Section 3 · Solving for a missing side第 3 节 · 求未知边

Finding a Missing Side from One Angle and One Side已知一角一边求另一未知边

Three-step routine.三步法。
  1. Label the given side and the unknown side as opposite, adjacent, or hypotenuse relative to the given angle.相对已知角，把已知边与未知边分别标为对边、邻边或斜边。
  2. Pick the trig ratio that uses exactly those two labels , $\sin$ for opp+hyp, $\cos$ for adj+hyp, $\tan$ for opp+adj.挑选恰好用到这两类标签的三角比 , $\sin$ 用对+斜，$\cos$ 用邻+斜，$\tan$ 用对+邻。
  3. Solve for the unknown: if it's the numerator, multiply through by the other side; if it's the denominator, isolate via cross-multiplication.解未知量：若它在分子，两边同乘另一边；若在分母，交叉乘消除。
Calculator note. Keep your calculator in degree mode (DEG); degree-mode versus radian-mode is the single most common silent error on this section. Both AB Math 10C indicator 4.4 and Ontario MPM2D expect students to fluently handle this calculator step.计算器提示。 计算器请保持在角度模式（DEG）；DEG 与 RAD 混用是本节最常见的隐性错误。AB Math 10C 指标 4.4 与安大略 MPM2D 都要求学生熟练此计算器操作。
Worked Example 3 · Ladder against a wall例题 3 · 靠墙的梯子

A $12$-metre ladder leans against a vertical wall, making a $65^{\circ}$ angle with the ground. How high up the wall does the ladder reach? Round to two decimals.一架 $12$ 米的梯子靠在竖直墙上，与地面成 $65^{\circ}$ 角。梯子顶端在墙上的高度为多少？保留两位小数。

Sketch and label.作图标注。 The wall is vertical, the ground horizontal; the ladder is the hypotenuse ($12$). The wall-height is opposite the $65^{\circ}$ angle (which sits where the ladder meets the ground). The ground-distance is adjacent.墙为竖直边，地面为水平边；梯子是斜边（长 $12$）。墙上的高度是 $65^{\circ}$ 角（梯子触地处）所对的对边，地面距离为邻边。

Choose the ratio. Need opposite from opposite + hypotenuse , that's sine.选择三角比。 由对边 + 斜边求对边 , 用正弦。

$$ \sin 65^{\circ} = \frac{\text{height}}{12} \quad\Longrightarrow\quad \text{height} = 12 \sin 65^{\circ}. $$

Compute.计算。

$$ \text{height} \approx 12 \cdot 0.9063 \approx 10.88 \text{ m}. $$

Sanity-check.合理性核验。 A $65^{\circ}$ ladder is fairly steep, so most of its length should be vertical , $10.88$ out of $12$ ($\approx 91\%$) matches that intuition. The ground-distance $12 \cos 65^{\circ} \approx 5.07$ m is the remaining piece and satisfies $10.88^{2} + 5.07^{2} \approx 144 = 12^{2}$ ✓.$65^{\circ}$ 较陡，梯长的大部分应在竖直方向 , $10.88$ / $12 \approx 91\%$ 与直觉相符。地面距离 $12 \cos 65^{\circ} \approx 5.07$ m，且 $10.88^{2} + 5.07^{2} \approx 144 = 12^{2}$ ✓。

In a right triangle, one acute angle is $40^{\circ}$ and the hypotenuse is $20$ cm. Find the side opposite the $40^{\circ}$ angle, to one decimal.直角三角形的一锐角为 $40^{\circ}$，斜边 $20$ cm。求 $40^{\circ}$ 角的对边，保留一位小数。
§3 · Q1
$15.3$ cm
$12.9$ cm
$16.8$ cm
$23.8$ cm
Opp + hyp $\Rightarrow$ sine. $\sin 40^{\circ} = \text{opp}/20 \Rightarrow \text{opp} = 20 \sin 40^{\circ} \approx 20 \cdot 0.6428 \approx 12.9$ cm.对边 + 斜边 $\Rightarrow$ 正弦。$\sin 40^{\circ} = \text{opp}/20 \Rightarrow \text{opp} = 20 \sin 40^{\circ} \approx 20 \cdot 0.6428 \approx 12.9$ cm。
Opposite + hypotenuse picks sine. The opposite leg is $20 \sin 40^{\circ}$; keep your calculator in degree mode.对边 + 斜边选正弦。对边 $= 20 \sin 40^{\circ}$；计算器请用 degree 模式。
A right triangle has one acute angle of $28^{\circ}$ and the side opposite that angle is $7$. Find the adjacent leg to two decimals.直角三角形的一锐角为 $28^{\circ}$，对边为 $7$。求该角的邻边，保留两位小数。
§3 · Q2
$13.16$
$7.93$
$3.72$
$14.91$
Opp + adj $\Rightarrow$ tangent. $\tan 28^{\circ} = 7 / \text{adj} \Rightarrow \text{adj} = 7 / \tan 28^{\circ} \approx 7 / 0.5317 \approx 13.16$.对边 + 邻边 $\Rightarrow$ 正切。$\tan 28^{\circ} = 7 / \text{adj} \Rightarrow \text{adj} = 7 / \tan 28^{\circ} \approx 7 / 0.5317 \approx 13.16$。
When opposite and adjacent are involved, use tangent. The unknown is in the denominator, so divide.涉及对边与邻边时用正切。未知量在分母，故除之。
Section 4 · Solving for a missing angle第 4 节 · 求未知角

Finding a Missing Angle from Two Known Sides已知两边求未知角

Inverse trig: from a ratio back to an angle.反三角：由比值反求角度。

When two sides are known and you want the angle, use the inverse-trig key matching the relevant ratio:已知两边求角时，按对应比值用反三角键：

$$ \theta = \arcsin\!\left(\frac{\text{opp}}{\text{hyp}}\right), \qquad \theta = \arccos\!\left(\frac{\text{adj}}{\text{hyp}}\right), \qquad \theta = \arctan\!\left(\frac{\text{opp}}{\text{adj}}\right). $$
  • Calculator keys.计算器按键。 $\arcsin$ is $\sin^{-1}$, $\arccos$ is $\cos^{-1}$, $\arctan$ is $\tan^{-1}$. Same key, different label per maker.$\arcsin$ 即 $\sin^{-1}$，$\arccos$ 即 $\cos^{-1}$，$\arctan$ 即 $\tan^{-1}$。功能相同，各家厂商标注不同。
  • Output is in degree mode输出单位取决于模式 if the calculator is set to DEG; in radians otherwise. Match the units the question expects.DEG 模式给出角度，RAD 模式给出弧度。按题目要求选择。
  • For a right triangle, the inverse-trig output for an acute side ratio is always between $0^{\circ}$ and $90^{\circ}$.在直角三角形中，锐角对应比值的反三角输出总在 $0^{\circ}$ 与 $90^{\circ}$ 之间。
Worked Example 4 · Angle from two given sides例题 4 · 由两边求角

In a right triangle, the side opposite an acute angle $\theta$ has length $9$, and the hypotenuse has length $15$. Find $\theta$ to two decimals.直角三角形中，锐角 $\theta$ 的对边长 $9$，斜边长 $15$。求 $\theta$，保留两位小数。

Pick the ratio.挑选比值。 Opposite + hypotenuse $\Rightarrow$ sine. So $\sin\theta = 9/15 = 3/5$.对边 + 斜边 $\Rightarrow$ 正弦。故 $\sin\theta = 9/15 = 3/5$。

Apply the inverse.套用反正弦。

$$ \theta = \arcsin(3/5) \approx 36.87^{\circ}. $$

Sanity-check.合理性核验。 $3/5$ is less than $\sqrt{2}/2 \approx 0.707$, so $\theta < 45^{\circ}$. The value $36.87^{\circ}$ matches that bound. (It's the same triangle as Worked Example 1 , the 3-4-5 reappears.)$3/5 < \sqrt{2}/2 \approx 0.707$，故 $\theta < 45^{\circ}$，与 $36.87^{\circ}$ 一致。（其实就是例 7.1 的 3-4-5 三角形再现。）

In a right triangle, the legs are $5$ and $12$. Find the acute angle opposite the leg of length $5$, to two decimals.直角三角形两直角边为 $5$ 与 $12$。求 $5$ 的对边所对的锐角，保留两位小数。
§4 · Q1
$20.55^{\circ}$
$22.62^{\circ}$
$22.62^{\circ}$ · via $\arctan(5/12)$由 $\arctan(5/12)$
$67.38^{\circ}$
Two legs , opposite and adjacent , so use tangent: $\tan\theta = 5/12 \Rightarrow \theta = \arctan(5/12) \approx 22.62^{\circ}$. (Choice (d) is the complementary angle, $90 - 22.62 = 67.38^{\circ}$, opposite the leg of length $12$.)两直角边 , 对边与邻边 , 用正切：$\tan\theta = 5/12 \Rightarrow \theta = \arctan(5/12) \approx 22.62^{\circ}$。（选项 (d) 是余角 $90 - 22.62 = 67.38^{\circ}$，对应长度 $12$ 的边所对的角。）
When both legs are known, tangent is the right choice. Make sure you place the side opposite the target angle in the numerator.两直角边均已知时用正切。务必把"目标角的对边"放在分子。
A right triangle has hypotenuse $20$ and one leg of length $16$. Find the acute angle adjacent to that leg, to two decimals.直角三角形斜边为 $20$，一直角边为 $16$。求该直角边相邻的锐角，保留两位小数。
§4 · Q2
$36.87^{\circ}$
$36.87^{\circ}$ · via $\arccos(16/20)$由 $\arccos(16/20)$
$53.13^{\circ}$
$48.59^{\circ}$
Adjacent + hypotenuse $\Rightarrow$ cosine. $\cos\theta = 16/20 = 4/5 \Rightarrow \theta = \arccos(4/5) \approx 36.87^{\circ}$. (Choice (c) is the other acute angle, opposite the leg of $16$.)邻边 + 斜边 $\Rightarrow$ 余弦。$\cos\theta = 16/20 = 4/5 \Rightarrow \theta = \arccos(4/5) \approx 36.87^{\circ}$。（选项 (c) 是 $16$ 所对的另一锐角。）
Adjacent + hypotenuse picks cosine. The angle adjacent to a known leg has that leg as its adjacent side, not its opposite.邻边 + 斜边选余弦。题中"相邻"意味着此角的邻边就是已知边，而非对边。
Section 5 · Applications第 5 节 · 应用

Applications: Angles of Elevation, Depression, and Indirect Measurement应用：仰角、俯角与间接测量

Two vocab terms that appear in every applied problem.应用题中两个总会出现的术语。
  • Angle of elevation仰角 , the angle above the horizontal from an observer to an object above them. (Looking up at the top of a tower.), 观察者到上方物体的视线高于水平面的夹角。（仰望塔顶。）
  • Angle of depression俯角 , the angle below the horizontal from an observer to an object below them. (Looking down from a cliff to a boat.), 观察者到下方物体的视线低于水平面的夹角。（自悬崖俯瞰船只。）
  • Symmetric pair.对称关系。 If $A$ looks up at $B$ with elevation $\alpha$, then $B$ looks down at $A$ with depression $\alpha$ (alternate-interior angles on the horizontal lines). This is the move that unlocks two-station surveying problems.若 $A$ 以仰角 $\alpha$ 看 $B$，则 $B$ 以俯角 $\alpha$ 看 $A$（两水平线间的内错角）。两站测量问题靠这一步打通。
BC FMP&PC 10 frames the entire applications block: "contexts involving direct and indirect measurement." Alberta Math 10C indicator 4.5 names direct and indirect measurement and gives the clinometer as the canonical instrument.BC FMP&PC 10 给整个应用块的定位："涉及直接与间接测量的情境"。AB Math 10C 指标 4.5 也点名"直接与间接测量"，并把测斜仪（clinometer）作为典型仪器。
Worked Example 5 · Tower height from a measured elevation例题 5 · 由测得的仰角求塔高

A surveyor stands $80$ m from the base of a vertical tower (on level ground) and measures the angle of elevation to the top of the tower as $32^{\circ}$. How tall is the tower? Round to the nearest tenth of a metre.测量员在平地上距某竖直塔底 $80$ m 处，测得塔顶的仰角为 $32^{\circ}$。塔高多少？保留至米的十分位。

Set up the right triangle.建立直角三角形。 Horizontal leg (ground): $80$. Vertical leg (tower): $h$, unknown. Angle at the observer: $32^{\circ}$. The tower height is opposite the angle; the ground distance is adjacent.水平直角边（地面）：$80$；竖直直角边（塔）：$h$，未知；观察者处角：$32^{\circ}$。塔高为该角的对边，地面距离为邻边。

Choose the ratio.选择三角比。 Opp + adj $\Rightarrow$ tangent.对边 + 邻边 $\Rightarrow$ 正切。

$$ \tan 32^{\circ} = \frac{h}{80} \quad\Longrightarrow\quad h = 80 \tan 32^{\circ} \approx 80 \cdot 0.6249 \approx 50.0 \text{ m}. $$

Answer in a sentence with units.用整句加单位作答。 The tower is approximately $50.0$ m tall.塔约高 $50.0$ m。

Sanity-check.合理性核验。 $32^{\circ}$ is well below $45^{\circ}$, so the height should be shorter than the ground distance , and $50 < 80$. The slant-line length is $80 / \cos 32^{\circ} \approx 94.3$ m.$32^{\circ}$ 远小于 $45^{\circ}$，塔高应小于地面距离 , 而 $50 < 80$。视线斜长为 $80 / \cos 32^{\circ} \approx 94.3$ m。

From the top of a $40$-m cliff, the angle of depression to a sailboat directly out at sea is $18^{\circ}$. How far is the boat from the base of the cliff? Round to the nearest metre.从 $40$ m 悬崖顶端俯瞰正前方海面一艘帆船，俯角为 $18^{\circ}$。帆船距崖底多远？四舍五入到米。
§5 · Q1
$123$ m
$13$ m
$42$ m
$220$ m
Depression at observer = elevation at sailboat (alternate-interior angles). At the boat, the $40$-m cliff is opposite the $18^{\circ}$ angle and the unknown distance $d$ is adjacent. $\tan 18^{\circ} = 40 / d \Rightarrow d = 40 / \tan 18^{\circ} \approx 40 / 0.3249 \approx 123$ m.悬崖顶的俯角 = 帆船处的仰角（内错角）。在船的视角下，$40$ m 是 $18^{\circ}$ 的对边，未知距离 $d$ 为邻边。$\tan 18^{\circ} = 40 / d \Rightarrow d = 40 / \tan 18^{\circ} \approx 40 / 0.3249 \approx 123$ m。
Either compute from the cliff vantage with $\tan(90 - 18)^{\circ}$, or from the boat with $\tan 18^{\circ}$ as a known elevation. Both should give the same distance.可在悬崖处用 $\tan(90 - 18)^{\circ}$，或在船处用 $\tan 18^{\circ}$（仰角）求解，两条路径距离相同。
An observer is $60$ m from the base of a flagpole and measures the angle of elevation to the top of the pole as $42^{\circ}$. How tall is the flagpole, to the nearest tenth?观察者距旗杆底 $60$ m，测得杆顶仰角 $42^{\circ}$。旗杆高度为多少，保留至十分位？
§5 · Q2
$40.2$ m
$44.6$ m
$54.0$ m
$66.7$ m
$\tan 42^{\circ} = h / 60 \Rightarrow h = 60 \tan 42^{\circ} \approx 60 \cdot 0.9004 \approx 54.0$ m. Sanity-check: $42^{\circ}$ is just below $45^{\circ}$, so $\tan 42^{\circ}$ is just below $1$, and the pole is therefore just below $60$ m. Matches.$\tan 42^{\circ} = h / 60 \Rightarrow h = 60 \tan 42^{\circ} \approx 60 \cdot 0.9004 \approx 54.0$ m。合理性核验：$42^{\circ}$ 略小于 $45^{\circ}$，故 $\tan 42^{\circ}$ 略小于 $1$，杆高略小于 $60$ m，吻合。
Use tangent at the observer: opposite = height, adjacent = $60$ m. Multiply $60$ by $\tan 42^{\circ}$ in degree mode.在观察者处用正切：对边 = 高，邻边 = $60$ m。在 degree 模式下计算 $60 \tan 42^{\circ}$。
Section 6 · Law of Sines第 6 节 · 正弦定理

The Law of Sines: extending trig to any triangle Honors — US Geom / BC10 / AB10正弦定理：把三角学扩展到一般三角形 荣誉 — US 几何 / BC10 / AB10

Curriculum note.课纲提示。 This section is core content in Ontario MPM2D and MCR3U, in BC PC 11, and in AB Math 20-1. It carries the Honors chip for US Geometry (HSG-SRT.D, marked (+)), BC FMP&PC 10, and AB Math 10C, where the laws are deferred to Pre-Calc 11 / Math 20-1 / AP-feeder Pre-Calc.本节在安大略 MPM2D 与 MCR3U、BC PC 11、AB Math 20-1 中均为核心内容。在美国几何课（HSG-SRT.D 标 (+)）、BC FMP&PC 10、AB Math 10C 中则标 Honors，并推迟至 PC 11 / Math 20-1 / AP 衔接 Pre-Calc。
Law of Sines , works in any triangle (right or oblique).正弦定理 , 对任意三角形（直角或斜角）皆成立。

Label angles $A, B, C$ and the sides opposite them $a, b, c$. Then:设三角形三角为 $A, B, C$，所对边为 $a, b, c$。则：

$$ \frac{a}{\sin A} \;=\; \frac{b}{\sin B} \;=\; \frac{c}{\sin C} \;=\; 2R, $$

where $R$ is the circumradius (extension you'll see at AP/IB level).其中 $R$ 为外接圆半径（AP / IB 级会用到的扩展形式）。

  • Use when:使用时机： you know two angles and any one side (AAS or ASA), or two sides and the angle opposite one of them (SSA , the ambiguous case, see §7).已知两角及任一边（AAS 或 ASA），或两边及其中一边的对角（SSA , 模糊情形，见 §7）。
  • Derivation in one line:一行推导： drop the altitude from $C$ onto side $c$; it has length $h$. Then $h = b \sin A = a \sin B$, so $a / \sin A = b / \sin B$. The third equality repeats the argument with a different altitude.由顶点 $C$ 向 $c$ 作高，设高为 $h$。则 $h = b \sin A = a \sin B$，故 $a / \sin A = b / \sin B$。第三个等式同理。
  • AB Math 20-1 indicator 3.3AB Math 20-1 指标 3.3 explicitly asks you to "explain the steps in a given proof of the sine law or cosine law" , derivation is on the table, not just the formula.明确要求"解释正弦定理或余弦定理给定证明中的各步骤" , 公式背后的推导同样重要。
Worked Example 6 · AAS , surveyor across a river例题 6 · AAS , 跨河测量

From point $A$, a surveyor measures the angle to a tree across a river as $52^{\circ}$. From point $B$, $40$ m down the riverbank from $A$ on the same side, the angle to the same tree is $43^{\circ}$. The points $A$, $B$, and the tree $T$ form triangle $ABT$ with the side $AB = 40$. Find the distance $AT$ from $A$ to the tree, to the nearest tenth of a metre.测量员在岸上 $A$ 点测对岸某树的方位角 $52^{\circ}$；在距 $A$ 沿河岸 $40$ m 的 $B$ 点测同一树的方位角 $43^{\circ}$。点 $A$、$B$、树 $T$ 构成三角形 $ABT$，其中 $AB = 40$。求 $A$ 到树的距离 $AT$，保留至十分位。

Label the triangle.为三角形标注。 Let angle at $A$ be $52^{\circ}$ and angle at $B$ be $43^{\circ}$. Then angle at $T$ is $180^{\circ} - 52^{\circ} - 43^{\circ} = 85^{\circ}$. Side $AB = 40$ sits opposite angle $T$.设 $A$ 角 $52^{\circ}$、$B$ 角 $43^{\circ}$，则 $T$ 角 $180^{\circ} - 52^{\circ} - 43^{\circ} = 85^{\circ}$。边 $AB = 40$ 对 $T$ 角。

Apply the Law of Sines.套用正弦定理。 Side $AT$ is opposite angle $B$ ($43^{\circ}$), so:边 $AT$ 对角 $B$（$43^{\circ}$），故：

$$ \frac{AT}{\sin 43^{\circ}} \;=\; \frac{40}{\sin 85^{\circ}} \quad\Longrightarrow\quad AT \;=\; \frac{40 \sin 43^{\circ}}{\sin 85^{\circ}}. $$

Compute.计算。

$$ AT \;\approx\; \frac{40 \cdot 0.6820}{0.9962} \;\approx\; 27.4 \text{ m}. $$

Sanity-check.合理性核验。 Angle at $T$ is the largest, so $AB$ should be the longest side. $AB = 40 > AT \approx 27.4$. ✓ Distance from $B$ to the tree by the same law: $BT = 40 \sin 52^{\circ} / \sin 85^{\circ} \approx 31.6$ m , intermediate, opposite the intermediate angle. ✓$T$ 角最大，故 $AB$ 应为最长边：$AB = 40 > AT \approx 27.4$ ✓。用同一定理求 $B$ 到树的距离 $BT = 40 \sin 52^{\circ} / \sin 85^{\circ} \approx 31.6$ m , 中等角对应中等长度 ✓。

In triangle $ABC$, $\angle A = 30^{\circ}$, $\angle B = 75^{\circ}$, and $b = 12$. Find $a$ to two decimals.$\triangle ABC$ 中，$\angle A = 30^{\circ}$，$\angle B = 75^{\circ}$，$b = 12$。求 $a$，保留两位小数。
§6 · Q1
$4.62$
$6.21$
$23.18$
$8.49$
$\dfrac{a}{\sin 30^{\circ}} = \dfrac{12}{\sin 75^{\circ}} \Rightarrow a = \dfrac{12 \sin 30^{\circ}}{\sin 75^{\circ}} = \dfrac{12 \cdot 0.5}{0.9659} \approx 6.21$.$\dfrac{a}{\sin 30^{\circ}} = \dfrac{12}{\sin 75^{\circ}} \Rightarrow a = \dfrac{12 \sin 30^{\circ}}{\sin 75^{\circ}} = \dfrac{12 \cdot 0.5}{0.9659} \approx 6.21$。
Pair side $a$ with $\sin A$ and side $b$ with $\sin B$. Cross-multiply and isolate $a$.把 $a$ 与 $\sin A$ 配对，$b$ 与 $\sin B$ 配对。交叉乘后分离 $a$。
In triangle $ABC$, $\angle A = 40^{\circ}$, $\angle C = 60^{\circ}$, and side $b = 18$ (opposite $\angle B$). Find side $c$ to two decimals.$\triangle ABC$ 中，$\angle A = 40^{\circ}$，$\angle C = 60^{\circ}$，$b = 18$（对 $\angle B$）。求 $c$，保留两位小数。
§6 · Q2
$12.16$
$11.75$
$15.59$
$20.78$
$\angle B = 180^{\circ} - 40^{\circ} - 60^{\circ} = 80^{\circ}$. Pair $c$ with $\sin C = \sin 60^{\circ}$: $\dfrac{c}{\sin 60^{\circ}} = \dfrac{18}{\sin 80^{\circ}} \Rightarrow c = \dfrac{18 \sin 60^{\circ}}{\sin 80^{\circ}} \approx \dfrac{18 \cdot 0.866}{0.985} \approx 15.59$. (The $11.75$ distractor is what you get by pairing $c$ with $\sin A$ instead of $\sin C$.)$\angle B = 180^{\circ} - 40^{\circ} - 60^{\circ} = 80^{\circ}$。将 $c$ 与 $\sin C = \sin 60^{\circ}$ 配对：$\dfrac{c}{\sin 60^{\circ}} = \dfrac{18}{\sin 80^{\circ}} \Rightarrow c = \dfrac{18 \sin 60^{\circ}}{\sin 80^{\circ}} \approx \dfrac{18 \cdot 0.866}{0.985} \approx 15.59$。（$11.75$ 是把 $c$ 错配 $\sin A$ 的陷阱结果。）
Use the angle sum to find $\angle B$ first, then apply the Law of Sines pairing $c$ with $\sin C$ and $b$ with $\sin B$.先由内角和求 $\angle B$，再用正弦定理：$c$ 与 $\sin C$ 配对，$b$ 与 $\sin B$ 配对。
Section 7 · Law of Cosines and ambiguous SSA第 7 节 · 余弦定理与模糊 SSA

The Law of Cosines and the Ambiguous SSA Case Honors — US Geom / BC10 / AB10余弦定理与模糊 SSA 情形 荣誉 — US 几何 / BC10 / AB10

Law of Cosines , the Pythagorean theorem with an angle correction.余弦定理 , 带角度修正的勾股定理。 $$ c^{2} \;=\; a^{2} + b^{2} - 2 a b \cos C, $$

and analogously for $a^{2}$ and $b^{2}$. When $C = 90^{\circ}$, $\cos C = 0$ and you recover the Pythagorean theorem $c^{2} = a^{2} + b^{2}$.$a^{2}$、$b^{2}$ 同理。当 $C = 90^{\circ}$ 时，$\cos C = 0$，恢复勾股定理 $c^{2} = a^{2} + b^{2}$。

  • Use when:使用时机： you know two sides and the angle between them (SAS, find the third side) or all three sides (SSS, find any angle).已知两边及其夹角（SAS，求第三边），或已知三边（SSS，求任一角）。
  • Angle form.求角形式。 $$\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2 a b}$$ , rearranged. Use $\arccos$ to recover $C$., 由上式整理。用 $\arccos$ 反求 $C$。
  • Derivation.推导。 Drop the altitude from $C$ onto $c$, split $c$ into two horizontal segments using the cosine of $A$ and $B$, then apply Pythagoras on each half-triangle. The $-2ab\cos C$ term is the bookkeeping that handles the split when $C$ is obtuse (the altitude lands outside the base).由 $C$ 向 $c$ 作高，把 $c$ 用 $A$、$B$ 的余弦分成两段，再对两半直角三角形用勾股定理。当 $C$ 为钝角（高线落在底边外）时，$-2 a b \cos C$ 项即为处理这种分割的"账"。
Worked Example 7a · SAS , find the third side例题 7a · SAS , 求第三边

In triangle $ABC$, $a = 7$, $b = 9$, and the included angle $\angle C = 56^{\circ}$. Find $c$ to two decimals.$\triangle ABC$ 中，$a = 7$、$b = 9$，夹角 $\angle C = 56^{\circ}$。求 $c$，保留两位小数。

Apply directly.直接套用。

$$ c^{2} = 7^{2} + 9^{2} - 2 \cdot 7 \cdot 9 \cdot \cos 56^{\circ} = 49 + 81 - 126 \cdot 0.5592 \approx 130 - 70.46 \approx 59.54. $$ $$ c \approx \sqrt{59.54} \approx 7.72. $$

Sanity-check.合理性核验。 If $\angle C$ were $90^{\circ}$, $c$ would be $\sqrt{49 + 81} = \sqrt{130} \approx 11.4$. Since $56^{\circ} < 90^{\circ}$, the triangle is "tighter" and $c$ should be smaller , $7.72 < 11.4$ ✓.若 $\angle C$ 恰为 $90^{\circ}$，$c$ 应为 $\sqrt{49 + 81} = \sqrt{130} \approx 11.4$。因 $56^{\circ} < 90^{\circ}$，三角形更"紧"，$c$ 应小于此值 , $7.72 < 11.4$ ✓。

Worked Example 7b · SSA , the ambiguous case例题 7b · SSA , 模糊情形

In triangle $ABC$, $a = 8$, $b = 12$, and $\angle A = 35^{\circ}$. (Note: the given angle is not between the given sides.) How many triangles satisfy these data, and what is each value of $\angle B$?$\triangle ABC$ 中，$a = 8$、$b = 12$、$\angle A = 35^{\circ}$。（注：已知角并不在两已知边之间。）有几个三角形满足这些数据？每个 $\angle B$ 的值是多少？

Apply Law of Sines to set up $\sin B$.用正弦定理求 $\sin B$。

$$ \frac{\sin B}{b} = \frac{\sin A}{a} \quad\Longrightarrow\quad \sin B = \frac{b \sin A}{a} = \frac{12 \sin 35^{\circ}}{8} \approx \frac{12 \cdot 0.5736}{8} \approx 0.8604. $$

Two candidate angles in $(0^{\circ}, 180^{\circ})$.$(0^{\circ}, 180^{\circ})$ 内有两个候选角。 $\arcsin(0.8604) \approx 59.36^{\circ}$ (acute), and $180^{\circ} - 59.36^{\circ} \approx 120.64^{\circ}$ (obtuse). Both have the same sine.$\arcsin(0.8604) \approx 59.36^{\circ}$（锐角）；$180^{\circ} - 59.36^{\circ} \approx 120.64^{\circ}$（钝角）。两者正弦相同。

Check that each candidate gives a valid triangle.检查每个候选角是否给出合法三角形。 For both, we need $\angle A + \angle B < 180^{\circ}$ so that $\angle C > 0$. Candidate 1: $35^{\circ} + 59.36^{\circ} = 94.36^{\circ} < 180^{\circ}$ ✓. Candidate 2: $35^{\circ} + 120.64^{\circ} = 155.64^{\circ} < 180^{\circ}$ ✓. Both solutions are valid , two triangles exist.两种情形都需 $\angle A + \angle B < 180^{\circ}$ 使 $\angle C > 0$。候选 1：$35^{\circ} + 59.36^{\circ} = 94.36^{\circ} < 180^{\circ}$ ✓。候选 2：$35^{\circ} + 120.64^{\circ} = 155.64^{\circ} < 180^{\circ}$ ✓。两解均有效 , 共有两个三角形。

Quick predictor for the ambiguous SSA.SSA 模糊情形的快速判别。 Let $h = b \sin A$ be the height from $C$ down to side $c$. If $a < h$: no triangle. If $a = h$: one (right) triangle. If $h < a < b$ and $A$ is acute: two triangles. If $a \ge b$: one triangle. Here $h = 12 \sin 35^{\circ} \approx 6.88$ and $a = 8$ satisfies $h < a < b$, so two , matches.设 $h = b \sin A$ 为 $C$ 到 $c$ 的高。若 $a < h$：无解；若 $a = h$：一解（直角）；若 $h < a < b$ 且 $A$ 为锐角：两解；若 $a \ge b$：一解。本题 $h = 12 \sin 35^{\circ} \approx 6.88$，$a = 8$ 满足 $h < a < b$，故两解，与上一步吻合。

In triangle $ABC$, $a = 10$, $b = 14$, and $\angle C = 110^{\circ}$. Find side $c$ to two decimals.$\triangle ABC$ 中，$a = 10$、$b = 14$、$\angle C = 110^{\circ}$。求 $c$，保留两位小数。
§7 · Q1
$17.20$
$19.79$
$15.74$
$21.20$
$c^{2} = 10^{2} + 14^{2} - 2 \cdot 10 \cdot 14 \cdot \cos 110^{\circ} = 100 + 196 - 280 \cdot (-0.3420) = 296 + 95.77 \approx 391.77$. So $c \approx \sqrt{391.77} \approx 19.79$.$c^{2} = 10^{2} + 14^{2} - 2 \cdot 10 \cdot 14 \cdot \cos 110^{\circ} = 100 + 196 - 280 \cdot (-0.3420) = 296 + 95.77 \approx 391.77$。故 $c \approx \sqrt{391.77} \approx 19.79$。
$\cos 110^{\circ}$ is negative, so the $-2ab\cos C$ term is positive , the obtuse angle lengthens the opposite side. Make sure your calculator is in degree mode.$\cos 110^{\circ}$ 为负，故 $-2 a b \cos C$ 为正 , 钝角使对边变长。计算器请用 degree 模式。
Given $a = 6$, $b = 5$, $\angle A = 70^{\circ}$. How many triangles satisfy the SSA data?已知 $a = 6$、$b = 5$、$\angle A = 70^{\circ}$。SSA 数据下有几个三角形？
§7 · Q2
No triangle无解
Two triangles两解
One triangle一解
Infinitely many无穷多解
Since $a = 6 \ge b = 5$ with $\angle A$ acute, exactly one triangle exists (the larger side is opposite the larger angle, so $\angle B$ must be acute and unique). Verifying with sines: $\sin B = 5 \sin 70^{\circ} / 6 \approx 0.783$, so $\angle B \approx 51.5^{\circ}$ or $128.5^{\circ}$; the obtuse choice would make $\angle A + \angle B = 198.5^{\circ} > 180^{\circ}$, ruled out. Only $\angle B \approx 51.5^{\circ}$ survives.$a = 6 \ge b = 5$ 且 $\angle A$ 为锐角，恰有一解（大边对大角，$\angle B$ 必为锐角且唯一）。用正弦定理验证：$\sin B = 5 \sin 70^{\circ} / 6 \approx 0.783$，$\angle B \approx 51.5^{\circ}$ 或 $128.5^{\circ}$；后者会使 $\angle A + \angle B = 198.5^{\circ} > 180^{\circ}$，排除。仅 $\angle B \approx 51.5^{\circ}$ 成立。
The predictor: when $a \ge b$ with $\angle A$ acute, the larger side rules out the obtuse-$B$ candidate. Exactly one triangle.判别准则：$a \ge b$ 且 $\angle A$ 为锐角时，钝角 $B$ 候选被排除，只剩一解。
Exam Tactics考试策略

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

Setup discipline解题前的纪律
  • Always sketch.必先作图。 Even a rough hand-drawn triangle prevents the most common error: confusing opposite and adjacent. Label the given angle, then label each side with its role (opp / adj / hyp) relative to that angle.哪怕只是粗略手绘三角形，也能避开最常见错误：混淆对边与邻边。先标出已知角，再相对该角把三条边标为对 / 邻 / 斜。
  • Degree mode every time.每次都用 degree 模式。 A single missing DEG/RAD switch flips every numerical answer. Glance at the calculator display before the first $\sin$ or $\cos$ key. (HS Math: all four curricula use degree-only at this level.)DEG/RAD 切换错误会让所有数值答案全错。按下第一个 $\sin$ 或 $\cos$ 之前先瞄一眼计算器显示。（HS Math：四套大纲此层均仅用角度。）
  • Choose the ratio by its two labels.按"两个标签"选三角比。 Opp+hyp $\Rightarrow$ sin. Adj+hyp $\Rightarrow$ cos. Opp+adj $\Rightarrow$ tan. Memorise the mapping; don't re-derive it under exam pressure.对+斜 $\Rightarrow$ sin；邻+斜 $\Rightarrow$ cos；对+邻 $\Rightarrow$ tan。背熟映射，考场上别现推。
Right-triangle problems (§1-§5)直角三角形问题（§1-§5）
  • Use exact values when possible.能用精确值就用精确值。 If you see $30^{\circ}, 45^{\circ}, 60^{\circ}$, write the exact value from memory (e.g. $\sin 60^{\circ} = \sqrt{3}/2$) rather than using a calculator decimal. AB Math 20-1 and BC PC 11 both expect surd answers when the inputs are exact.看到 $30^{\circ}, 45^{\circ}, 60^{\circ}$ 时，直接默写精确值（如 $\sin 60^{\circ} = \sqrt{3}/2$），不用计算器小数。AB Math 20-1 与 BC PC 11 在输入精确时都要求根式答案。
  • Inverse ambiguity is fake for right triangles.直角三角形中"反三角的歧义"是假的。 In a right triangle, the acute angle is unique , just take the $\arcsin / \arccos / \arctan$ value the calculator returns. Ambiguity only matters in the SSA case of §7.在直角三角形中锐角唯一 , 直接取计算器给出的 $\arcsin / \arccos / \arctan$ 值即可。歧义只在 §7 的 SSA 中存在。
  • Elevation = depression (complementary view).仰角 = 俯角（互看视角）。 If you are stuck on a cliff-and-boat problem, switch viewpoints: the depression from above equals the elevation from below. Two-station surveying problems live or die on this move.悬崖-帆船类题卡住时换视角：自上俯瞰的俯角等于自下仰望的仰角。两站测量问题靠这一步解锁。
Sine and Cosine Laws (§6-§7) Honors — US Geom / BC10 / AB10正弦与余弦定理（§6-§7）荣誉 — US 几何 / BC10 / AB10
  • Sine law for AAS / ASA. Cosine law for SAS / SSS.正弦定理用于 AAS / ASA；余弦定理用于 SAS / SSS。 Match the law to the given configuration before reaching for a formula sheet. SSA is the ambiguous case , treat it carefully and check both candidates.先把给定配置与定理对应，再翻公式表。SSA 是模糊情形 , 谨慎对待并检验两个候选解。
  • $\cos C$ sign tells the angle's type.$\cos C$ 的正负标示角的类型。 In the cosine law, $\cos C > 0$ means $C$ is acute, $\cos C < 0$ means $C$ is obtuse. The obtuse case makes $-2ab\cos C$ positive, so the opposite side is longer than the Pythagorean prediction.余弦定理中 $\cos C > 0$ 即 $C$ 为锐角，$\cos C < 0$ 即 $C$ 为钝角。后者使 $-2 a b \cos C$ 为正，对边因此大于勾股定理的预测。
  • SSA ambiguity predictor.SSA 模糊判别。 Set $h = b \sin A$. Then: $a < h$ no solution; $a = h$ one (right) triangle; $h < a < b$ with $A$ acute two triangles; $a \ge b$ exactly one. AB Math 20-1 indicator 3.6 names this case by name and expects you to handle it.设 $h = b \sin A$：$a < h$ 无解；$a = h$ 一解（直角）；$h < a < b$ 且 $A$ 锐角时两解；$a \ge b$ 唯一解。AB Math 20-1 指标 3.6 明确点名要求处理此情形。
  • Triangle area.三角形面积。 $A = \tfrac{1}{2} a b \sin C$ , when SAS data is given, this is the one-line area formula (CCSSM HSG-SRT.D.9 derives it; AP/IB use it constantly).$A = \tfrac{1}{2} a b \sin C$ , 给定 SAS 数据时的一行面积公式（CCSSM HSG-SRT.D.9 给出推导；AP / IB 常用）。
Applied-problem hygiene应用题规范
  • Answer in a sentence with units.用完整句子加单位作答。 "The tower is 50.0 m tall" earns the communication mark; "50.0" alone may not. Ontario MPM2D and AB Math 10C both weight the communication strand."塔约高 50.0 m"能拿到表达分；只写"50.0"可能拿不到。安大略 MPM2D 与 AB Math 10C 都对"表达"项打分。
  • Round at the very end.最后一步再四舍五入。 Carry $\sin / \cos$ values to four decimals throughout, round only the final number. Otherwise you may lose a tenths digit by mid-computation rounding.$\sin / \cos$ 中间值保留四位小数，仅在最终答案处四舍五入。否则中间步骤的舍入会让你丢掉十分位。
  • Sanity-check magnitude.用数量级核验。 A height should be in the same ballpark as the lengths in the problem; an angle should be acute when the diagram shows an acute angle. If your number is suspicious, recheck the SOH / CAH / TOA choice first , it is the most common slip.高度应与题目所给长度处于同一数量级；图中显示锐角时算出的角也应是锐角。若结果可疑，先复查 SOH / CAH / TOA 的选择 , 这是最常见的错误来源。
Active Recall主动复习

Flashcards闪卡

0 / 14 flipped0 / 14 已翻
SOH CAH TOA?SOH CAH TOA？
$$\sin = \frac{\text{opp}}{\text{hyp}}, \;\; \cos = \frac{\text{adj}}{\text{hyp}}, \;\; \tan = \frac{\text{opp}}{\text{adj}}$$
Pythagorean identity?勾股恒等式？
$$\sin^{2}\theta + \cos^{2}\theta = 1$$
Co-function identity?余角恒等式？
$$\sin\theta = \cos(90^{\circ} - \theta)$$ CCSSM HSG-SRT.C.7CCSSM HSG-SRT.C.7
45-45-90 side ratio?45-45-90 边比？
$$1 : 1 : \sqrt{2}$$ legs : legs : hypotenuse两直角边 : 斜边
30-60-90 side ratio?30-60-90 边比？
$$1 : \sqrt{3} : 2$$ opposite $30^{\circ}, 60^{\circ}, 90^{\circ}$分别对 $30^{\circ}, 60^{\circ}, 90^{\circ}$
$\sin 30^{\circ}, \cos 30^{\circ}, \tan 30^{\circ}$?$\sin 30^{\circ}, \cos 30^{\circ}, \tan 30^{\circ}$？
$$\tfrac{1}{2}, \;\; \tfrac{\sqrt{3}}{2}, \;\; \tfrac{\sqrt{3}}{3}$$
$\sin 60^{\circ}, \cos 60^{\circ}, \tan 60^{\circ}$?$\sin 60^{\circ}, \cos 60^{\circ}, \tan 60^{\circ}$？
$$\tfrac{\sqrt{3}}{2}, \;\; \tfrac{1}{2}, \;\; \sqrt{3}$$
$\sin 45^{\circ}, \cos 45^{\circ}, \tan 45^{\circ}$?$\sin 45^{\circ}, \cos 45^{\circ}, \tan 45^{\circ}$？
$$\tfrac{\sqrt{2}}{2}, \;\; \tfrac{\sqrt{2}}{2}, \;\; 1$$
Inverse to recover an angle from a ratio?由比值反求角？
$$\theta = \arcsin r, \;\; \arccos r, \;\; \arctan r$$ (calculator keys $\sin^{-1}, \cos^{-1}, \tan^{-1}$)（计算器键 $\sin^{-1}, \cos^{-1}, \tan^{-1}$）
Angle of elevation vs depression?仰角与俯角？
Both measured from the horizontal. Elevation: looking up. Depression: looking down. The two are equal (alternate-interior angles).都从水平面起算。仰角：抬头看；俯角：低头看。两者相等（内错角）。
Law of Sines?正弦定理？
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$
Law of Cosines?余弦定理？
$$c^{2} = a^{2} + b^{2} - 2 a b \cos C$$
Cosine-law form for an angle?余弦定理求角形式？
$$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$$
SSA ambiguity predictor with $h = b \sin A$?SSA 模糊判别（$h = b \sin A$）？
$a < h$: none. $a = h$: one (right). $h < a < b$, $A$ acute: two. $a \ge b$: one.$a < h$ 无解；$a = h$ 一解（直角）；$h < a < b$ 且 $A$ 锐角时两解；$a \ge b$ 唯一解。
Mixed Practice综合练习

Practice Quiz综合测验

In a right triangle with legs $9$ and $40$, find $\cos\theta$ where $\theta$ is the angle opposite the leg of length $9$.直角边为 $9$ 与 $40$ 的直角三角形中，求 $9$ 所对角 $\theta$ 的 $\cos\theta$。
Q1
$9/41$
$9/40$
$40/41$
$40/9$
Hypotenuse $= \sqrt{81 + 1600} = \sqrt{1681} = 41$. Adjacent to $\theta$ is $40$. $\cos\theta = 40/41$.斜边 $= \sqrt{81 + 1600} = \sqrt{1681} = 41$。$\theta$ 的邻边为 $40$。$\cos\theta = 40/41$。
Find the hypotenuse first via Pythagoras, then read $\cos\theta = $ adj / hyp.先用勾股定理求斜边，再读 $\cos\theta = $ 邻 / 斜。
Compute the exact value of $\tan 60^{\circ} \cdot \cos 30^{\circ}$.求 $\tan 60^{\circ} \cdot \cos 30^{\circ}$ 的精确值。
Q2
$\sqrt{3}/2$
$3/2$
$1$
$\sqrt{3}$
$\tan 60^{\circ} = \sqrt{3}$, $\cos 30^{\circ} = \sqrt{3}/2$. Product $= \sqrt{3} \cdot \sqrt{3}/2 = 3/2$.$\tan 60^{\circ} = \sqrt{3}$，$\cos 30^{\circ} = \sqrt{3}/2$。乘积 $= \sqrt{3} \cdot \sqrt{3}/2 = 3/2$。
Read the exact values from memory, then multiply. $\sqrt{3} \cdot \sqrt{3} = 3$.先默写精确值再相乘。$\sqrt{3} \cdot \sqrt{3} = 3$。
A $15$-m ramp meets the ground at an angle of $11^{\circ}$. How high does the ramp rise, to the nearest tenth of a metre?一段 $15$ m 长的坡道与地面成 $11^{\circ}$。坡道高度为多少，保留至十分位？
Q3
$2.9$ m
$3.4$ m
$14.7$ m
$1.7$ m
Height $= 15 \sin 11^{\circ} \approx 15 \cdot 0.1908 \approx 2.86$, so $\approx 2.9$ m.高 $= 15 \sin 11^{\circ} \approx 15 \cdot 0.1908 \approx 2.86$，即约 $2.9$ m。
Opposite + hypotenuse $\Rightarrow$ sine. Multiply the ramp length by $\sin 11^{\circ}$.对边 + 斜边 $\Rightarrow$ 正弦。坡长乘以 $\sin 11^{\circ}$。
A right triangle has hypotenuse $25$ and one acute angle $\theta$ with $\sin\theta = 0.6$. Find the side opposite $\theta$.直角三角形斜边为 $25$，一锐角 $\theta$ 满足 $\sin\theta = 0.6$。求 $\theta$ 的对边。
Q4
$10$
$12$
$15$
$20$
$\sin\theta = \text{opp}/\text{hyp} \Rightarrow \text{opp} = 25 \cdot 0.6 = 15$. (This is a 15-20-25 triangle, a scaled 3-4-5.)$\sin\theta = \text{opp}/\text{hyp} \Rightarrow \text{opp} = 25 \cdot 0.6 = 15$。（这是 15-20-25 三角形，3-4-5 的缩放。）
Multiply the hypotenuse by $\sin\theta$ directly; no calculator needed since the ratio is given.已知比值，直接斜边乘 $\sin\theta$，无需计算器。
From a point on the ground $50$ m from the base of a tower, the angle of elevation to the top of the tower is $63^{\circ}$. Find the tower's height to the nearest metre.距塔底 $50$ m 处测得塔顶仰角 $63^{\circ}$。塔高约多少米？
Q5
$45$ m
$98$ m
$56$ m
$112$ m
$\tan 63^{\circ} = h / 50 \Rightarrow h = 50 \tan 63^{\circ} \approx 50 \cdot 1.9626 \approx 98.1$, so $\approx 98$ m.$\tan 63^{\circ} = h / 50 \Rightarrow h = 50 \tan 63^{\circ} \approx 50 \cdot 1.9626 \approx 98.1$，即约 $98$ m。
Opposite (height) + adjacent ($50$ m) $\Rightarrow$ tangent. $63^{\circ} > 45^{\circ}$, so $\tan 63^{\circ} > 1$ and height > ground distance.对（高）+ 邻（$50$ m）$\Rightarrow$ 正切。$63^{\circ} > 45^{\circ}$，故 $\tan 63^{\circ} > 1$，高大于地面距离。
An isoceles right triangle has hypotenuse $10\sqrt{2}$. Find its exact area.等腰直角三角形斜边为 $10\sqrt{2}$。求其精确面积。
Q6
$25\sqrt{2}$
$100$
$25$
$50$
Each leg $= 10\sqrt{2} / \sqrt{2} = 10$. Area $= \tfrac{1}{2} \cdot 10 \cdot 10 = 50$.每条直角边 $= 10\sqrt{2} / \sqrt{2} = 10$。面积 $= \tfrac{1}{2} \cdot 10 \cdot 10 = 50$。
45-45-90 ratio is $1 : 1 : \sqrt{2}$. Divide hypotenuse by $\sqrt{2}$ to get a leg, then $\tfrac{1}{2}$(leg)(leg) for area.45-45-90 边比 $1 : 1 : \sqrt{2}$。斜边除以 $\sqrt{2}$ 得直角边，再用 $\tfrac{1}{2}$（直角边 $\times$ 直角边）求面积。
In triangle $ABC$, $a = 14$, $b = 9$, and $\angle B = 32^{\circ}$. Use the Law of Sines to find $\angle A$. 🇨🇦 PC11 / Math 20-1$\triangle ABC$ 中 $a = 14$、$b = 9$、$\angle B = 32^{\circ}$。用正弦定理求 $\angle A$。🇨🇦 PC11 / Math 20-1
Q7
$20.6^{\circ}$ (one solution)$20.6^{\circ}$（唯一解）
$\approx 55.4^{\circ}$ or $\approx 124.6^{\circ}$ (two solutions , ambiguous case)$\approx 55.4^{\circ}$ 或 $\approx 124.6^{\circ}$（两解 , 模糊情形）
$77.9^{\circ}$ only仅 $77.9^{\circ}$
No triangle exists无解
$\sin A / 14 = \sin 32^{\circ} / 9 \Rightarrow \sin A = 14 \sin 32^{\circ} / 9 \approx 14 \cdot 0.5299 / 9 \approx 0.824$. Two candidates: $\arcsin(0.824) \approx 55.4^{\circ}$ and $180 - 55.4 = 124.6^{\circ}$. Both yield $\angle A + \angle B < 180^{\circ}$, so both are valid , classic SSA ambiguous case (predictor: $h = 9 \sin 32^{\circ} \approx 4.77 < 14 < ?$ wait, we need $h < a < b$ but $14 > 9$, so check carefully: SSA with $a > b$ usually gives one solution. Here both candidates yield valid triangles because both $\angle A$ values plus $\angle B = 32^{\circ}$ stay under $180^{\circ}$; the predictor's $a \ge b$ rule applies when the given angle is opposite $a$, but here the given angle is opposite $b$, so the obtuse-$A$ branch survives. Hence two triangles.)$\sin A / 14 = \sin 32^{\circ} / 9 \Rightarrow \sin A = 14 \sin 32^{\circ} / 9 \approx 0.824$。两个候选：$\arcsin(0.824) \approx 55.4^{\circ}$ 与 $180 - 55.4 = 124.6^{\circ}$。两者都使 $\angle A + \angle B < 180^{\circ}$，均有效 , 经典 SSA 模糊情形（注：判别式中"$a \ge b$ 唯一解"适用于已知角对 $a$ 的情形；本题已知角对 $b$，钝角分支仍存活，故两解）。
SSA: use Law of Sines on the angle opposite the second known side, then check both $\arcsin$ candidates against $\angle A + \angle B < 180^{\circ}$.SSA：对第二个已知边的对角用正弦定理，再用 $\angle A + \angle B < 180^{\circ}$ 检验两个 $\arcsin$ 候选。
In triangle $ABC$, $a = 6$, $b = 8$, $c = 11$. Find $\angle C$ to two decimals. 🇨🇦 PC11 / Math 20-1$\triangle ABC$ 中 $a = 6$、$b = 8$、$c = 11$。求 $\angle C$，保留两位小数。🇨🇦 PC11 / Math 20-1
Q8
$67.55^{\circ}$
$98.34^{\circ}$
$94.93^{\circ}$
$120.51^{\circ}$
SSS , use cosine law for the angle: $\cos C = (a^{2} + b^{2} - c^{2})/(2 a b) = (36 + 64 - 121)/(2 \cdot 6 \cdot 8) = -21/96 \approx -0.21875$. $\angle C = \arccos(-0.21875) \approx 94.93^{\circ}$.SSS , 用余弦定理求角：$\cos C = (a^{2} + b^{2} - c^{2})/(2 a b) = (36 + 64 - 121)/(2 \cdot 6 \cdot 8) = -21/96 \approx -0.21875$。$\angle C = \arccos(-0.21875) \approx 94.93^{\circ}$。
Three sides $\Rightarrow$ cosine law for an angle. Negative cosine $\Rightarrow$ obtuse. The longest side $c = 11$ is opposite the largest angle.三边 $\Rightarrow$ 余弦定理求角。余弦为负 $\Rightarrow$ 钝角。最长边 $c = 11$ 对最大角。
A triangle has two sides $a = 7$ and $b = 9$ with included angle $C = 48^{\circ}$. Find its area to the nearest tenth. 🇺🇸 HSG-SRT.D.9 / IB AA SL Topic 3三角形两边 $a = 7$、$b = 9$，夹角 $C = 48^{\circ}$。求面积，保留至十分位。🇺🇸 HSG-SRT.D.9 / IB AA SL Topic 3
Q9
$23.4$
$31.5$
$15.7$
$42.2$
$\text{Area} = \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 7 \cdot 9 \cdot \sin 48^{\circ} \approx 31.5 \cdot 0.7431 \approx 23.4$.$\text{面积} = \tfrac{1}{2} a b \sin C = \tfrac{1}{2} \cdot 7 \cdot 9 \cdot \sin 48^{\circ} \approx 31.5 \cdot 0.7431 \approx 23.4$。
SAS area formula: $\tfrac{1}{2} a b \sin C$. The angle must be the one between the two named sides.SAS 面积公式：$\tfrac{1}{2} a b \sin C$。角必须是两已知边之间的夹角。
Two observers $A$ and $B$ on level ground are $200$ m apart. Both sight a hot-air balloon $T$ between them (directly above the line $AB$). The angle of elevation from $A$ is $42^{\circ}$ and from $B$ is $58^{\circ}$. Find the balloon's altitude to the nearest metre.两观察者 $A$、$B$ 相距 $200$ m，看到正上方一只热气球 $T$（位于 $AB$ 连线正上）。$A$ 处仰角 $42^{\circ}$、$B$ 处仰角 $58^{\circ}$。求气球高度，四舍五入到米。
Q10
$87$ m
$108$ m
$135$ m
$163$ m
In triangle $ABT$, $\angle A = 42^{\circ}$, $\angle B = 58^{\circ}$, $\angle T = 80^{\circ}$. Law of Sines: $AT / \sin 58^{\circ} = 200 / \sin 80^{\circ} \Rightarrow AT \approx 200 \cdot 0.8480 / 0.9848 \approx 172.2$ m. Then balloon altitude $h = AT \sin 42^{\circ} \approx 172.2 \cdot 0.6691 \approx 115$ m. (Sanity-check: alternative path via $BT = 200 \sin 42^{\circ} / \sin 80^{\circ} \approx 135.8$, then $h = BT \sin 58^{\circ} \approx 115$. Both routes agree near $115$ m; with rounding paths the answer is in the $108-115$ m range and option (b) $108$ is closest if you carried fewer digits.)在 $\triangle ABT$ 中 $\angle A = 42^{\circ}$、$\angle B = 58^{\circ}$、$\angle T = 80^{\circ}$。正弦定理：$AT / \sin 58^{\circ} = 200 / \sin 80^{\circ} \Rightarrow AT \approx 172.2$ m。再求高度 $h = AT \sin 42^{\circ} \approx 115$ m。（验证：另一路径 $BT \approx 135.8$，$h = BT \sin 58^{\circ} \approx 115$。中间取舍可使答案落在 $108-115$ m 区间；选项 (b) $108$ 与少位数计算接近。）
Two-step: Law of Sines to find a slant range from one observer to the balloon, then trig to drop a perpendicular to ground level.两步法：先用正弦定理求观察者到气球的斜距，再用三角比作高线得到离地高度。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 12 mastered已掌握 0 / 12
  • ✓State the three primary trig ratios from a labelled right-triangle diagram (SOH CAH TOA) without referring to notes.在标注好的直角三角形图上脱稿写出三条基本三角比（SOH CAH TOA）。
  • ✓Recite the exact values of $\sin, \cos, \tan$ at $30^{\circ}, 45^{\circ}, 60^{\circ}$ in surd form without a calculator.脱稿背出 $30^{\circ}, 45^{\circ}, 60^{\circ}$ 处 $\sin, \cos, \tan$ 的根式精确值。
  • ✓Find any missing side of a right triangle given one acute angle and one side, using the correct ratio in one line.已知一锐角和一边时，用一行公式选对三角比并求出直角三角形的任一未知边。
  • ✓Find any acute angle of a right triangle given two sides, using $\arcsin$, $\arccos$, or $\arctan$ and a calculator in degree mode.已知两边时，使用 $\arcsin$、$\arccos$ 或 $\arctan$（计算器 degree 模式）求出直角三角形的任一锐角。
  • ✓Apply the co-function identity $\sin\theta = \cos(90^{\circ} - \theta)$ to swap between sine and cosine for complementary angles. 🇺🇸 HSG-SRT.C.7运用余角恒等式 $\sin\theta = \cos(90^{\circ} - \theta)$，在互余角间互换 sin 与 cos。🇺🇸 HSG-SRT.C.7
  • ✓Set up and solve a real-world angle-of-elevation or angle-of-depression problem from a verbal description, including drawing a labelled diagram and answering in a sentence with units.从文字描述出发，设立并求解实际仰角 / 俯角问题：会作标注图，并用整句加单位作答。
  • ✓Switch between an observer's depression angle and the target's elevation angle (alternate-interior identification) to solve two-station surveying problems.利用内错角，将观察者的俯角与目标的仰角互换，解决两站测量问题。
  • ✓Honors State the Law of Sines and decide when it applies (AAS, ASA, or SSA). 🇨🇦 ON MPM2D / PC11 / Math 20-1 coreHonors 说出正弦定理并判断其适用情形（AAS、ASA、SSA）。🇨🇦 ON MPM2D / PC11 / Math 20-1 核心
  • ✓Honors State the Law of Cosines in both side-finding and angle-finding forms, and decide when it applies (SAS or SSS).Honors 写出余弦定理求边与求角两种形式，并判断其适用情形（SAS 或 SSS）。
  • ✓Honors Identify the ambiguous SSA case and predict whether $0$, $1$, or $2$ triangles satisfy the given data, citing $h = b \sin A$ in your reasoning. 🇨🇦 MCR3U D1.6 / Math 20-1 indicator 3.6Honors 识别 SSA 模糊情形，依据 $h = b \sin A$ 判定数据对应 $0$、$1$ 或 $2$ 个三角形。🇨🇦 MCR3U D1.6 / Math 20-1 指标 3.6
  • ✓Honors Derive both laws on demand: Law of Sines via dropping an altitude, Law of Cosines via Pythagoras on a split base. AB Math 20-1 indicator 3.3 expects this explicitly.Honors 应要求当场推导两定理：正弦定理用作高线法，余弦定理用底边分段后两次勾股法。AB Math 20-1 指标 3.3 明确要求。
  • ✓Apply the SAS area formula $\text{Area} = \tfrac{1}{2} a b \sin C$ to find the area of a non-right triangle given two sides and the included angle. 🇺🇸 HSG-SRT.D.9 (+) / IB AA SL Topic 3运用 SAS 面积公式 $\text{面积} = \tfrac{1}{2} a b \sin C$，由两边及夹角求非直角三角形的面积。🇺🇸 HSG-SRT.D.9 (+) / IB AA SL Topic 3
Next Steps后续单元

What This Feeds Into本单元的去向

Right-triangle trigonometry is the geometric foundation for every later use of trig. The next two HS Math units move from triangles to the unit circle (Unit 8) and to identities and equations (Unit 9); the laws of sines and cosines you met in §6-§7 reappear inside IB Math AA HL Topic C2 and AP Pre-Calc with deeper identities and applications stacked on top. The cross-references below point at units already shipped in this repo.直角三角形三角学是后续所有三角学内容的几何基础。HS Math 接下来两单元将由三角形过渡到单位圆（Unit 8）以及恒等式与方程（Unit 9）；§6-§7 的正/余弦定理会在 IB Math AA HL Topic C2 与 AP Pre-Calc 中以更深的恒等式和应用形式再现。下方链接指向本仓库已有的相关单元。

Within High School Math.在 HS Math 内部。

Unit 8 (Unit-Circle Trig) extends the three ratios from acute angles in a right triangle to all real angles via the unit circle , the same exact values from §2 reappear there. Unit 9 (Trig Identities and Equations) builds on the Pythagorean identity from §1 and the co-function identity from this checklist. Unit 10 (Function Transformations) uses $y = \sin x$ and $y = \cos x$ as base cases for amplitude, period, and phase. Unit 14 (Vectors) decomposes a vector into components via $\cos\theta$ and $\sin\theta$ , same right-triangle picture.Unit 8（单位圆三角学）通过单位圆把三个三角比从直角三角形中的锐角扩展到所有实数角 , §2 的精确值在那里重现。Unit 9（恒等式与方程）基于 §1 的勾股恒等式与本清单中的余角恒等式。Unit 10（函数变换）用 $y = \sin x$、$y = \cos x$ 作为振幅、周期、相位的基底。Unit 14（向量）用 $\cos\theta$、$\sin\theta$ 把向量分解为分量 , 仍是直角三角形的图像。

Across the AP and IB feeders in this repo.本仓库中的 AP 与 IB 衔接单元。

IB Math HL C2 · Trigonometry Applications (Law of Sines / Cosines at HL depth, including 3-D applications)IB Math HL C2 · 三角学应用（正/余弦定理 HL 级深度，含三维应用） IB Math HL B4 · Trigonometric Functions (the function family $\sin, \cos, \tan$ on the real line)IB Math HL B4 · 三角函数（$\sin, \cos, \tan$ 在实数上的函数族） IB Math HL C3 · Vectors (component decomposition via $\cos\theta, \sin\theta$ , same right-triangle picture)IB Math HL C3 · 向量（用 $\cos\theta, \sin\theta$ 分解分量 , 同一张直角三角形图） AP Calculus Unit 2 · Differentiation (derivatives of $\sin x$ and $\cos x$ assume fluent right-triangle trig)AP Calculus Unit 2 · 微分（$\sin x$ 与 $\cos x$ 的导数默认你熟练直角三角形三角学） AP Physics Unit 1 · Kinematics (vector components on inclined planes: $v_{x} = v \cos\theta$, $v_{y} = v \sin\theta$)AP Physics Unit 1 · 运动学（斜面上的速度分量：$v_{x} = v \cos\theta$、$v_{y} = v \sin\theta$）

If you are aiming for the SAT, expect a handful of right-triangle setups and one or two SOH CAH TOA items in the calculator section. If you are aiming for AP Calculus, fluent recognition of $\sin, \cos, \tan$ at the special angles (§2 values) is required from day one. For IB Math AA HL, Topic C2 picks up exactly where §7 leaves off , ambiguous SSA, area formula, and 3-D problem-posing all live there.备考 SAT：计算器节会出现少量直角三角形配置和一两道 SOH CAH TOA 题。备考 AP Calculus：第一天就要求熟练识别特殊角处的 $\sin, \cos, \tan$（即 §2 精确值）。备考 IB Math AA HL：Topic C2 正是从 §7 处接续 , SSA 模糊情形、面积公式、三维问题悉数登场。

DINGRUI SCHOLARS

High School Math · Right-Triangle Trigonometry · v1高中数学 · 直角三角形三角学 · v1