# 二维与三维空间中的应用

> 爱德思国际GCSE进阶纯数学 · 4PM1
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s10-applications-in-two-and-three/

本指南将教你运用三角学规则求解爱德思IGCSE进阶纯数学（4PM1）考试中的二维和三维几何问题。你将学习如何通过从三维图形中提取对应的二维三角形来计算线面角和面面角。

**先修:** [三角形的正弦定理与余弦定理](https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s10-sine-cosine-rules/); [30°、45°、60°的精确三角函数值](https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s10-exact-trig-values/)

## 学习目标

- 从二维和三维几何图形中提取二维直角三角形以求解三角学问题
- 计算三维图形中直线与平面的夹角
- 计算三维图形中两个相交平面之间的二面角

## 二维三角学应用：基础复习

在处理三维问题之前，我们首先回顾如何将三角学规则应用于标准二维问题，例如求解三角形中的未知边长/角度，或是解决导航与测量类问题。这里的核心技能是识别正确的三角形，标注已知数值，并根据情况选用正弦、余弦或正切规则。请记住，考试中你需要自行熟记正弦定理以及三角形面积公式（$\frac{1}{2}ab\sin C$），而余弦定理会在公式表中提供。

> **tip**
>
> 请务必先检查你所处理的三角形是否为直角三角形：如果是，使用SOHCAHTOA规则；如果不是，则使用正弦定理或余弦定理。

**例题:** 一名测量员站在距离竖直塔基120米的位置，他观测塔顶的仰角为32°。计算该塔的高度，结果保留三位有效数字。

1. 画出由测量员、塔基和塔顶构成的直角三角形，标注邻边长度为120m，角度为32°，对边为h（即塔的高度）。
2. $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$
3. 代入已知数值：$\tan(32^\circ) = \frac{h}{120}$
4. $$h = 120 \times \tan(32^\circ) = 74.98 \approx 75.0\ \text{m}$$

*计算器:* allowed

## 计算直线与平面的夹角

对于三维问题，第一步始终是找出与你需要计算的夹角对应的直角三角形。线面角是直线与其在平面上的投影之间的最小夹角。求解该角度的步骤为：1. 找到直线与平面的交点；2. 在直线上任选另一个点，从该点向平面作垂线，得到投影点；3. 由原直线、直线的投影以及所作的垂线构成直角三角形，原直线与其投影之间的夹角即为所求的线面角。

**线面角** — 直线与其在平面上的正交投影之间的最小夹角，取值范围始终在0°到90°之间。

**例题:** 已知一个长方体的长为8cm，宽为6cm，高为4cm。求从长方体前左下角指向后右上角的体对角线与长方体底面平面的夹角，结果保留一位小数。

1. 首先计算该体对角线在底面上的投影长度：该长度即为长方体的底面对角线长度。
2. $$\text{Base diagonal length} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{cm}$$
3. 对应夹角所在的直角三角形中，对边为长方体的高度（4cm），邻边为底面对角线长度（10cm）。
4. $$\tan(\theta) = \frac{4}{10} = 0.4$$
5. $$\theta = \arctan(0.4) \approx 21.8^\circ$$

> **考试提示:** 请务必明确说明你所计算的夹角是由哪两条直线/哪两个平面构成的，避免因作答表述模糊而失分。

*计算器:* allowed

## 计算两个平面之间的夹角（二面角）

两个相交平面之间的夹角（称为二面角）是分别位于两个平面上、且同时垂直于两平面交线的两条直线之间的最小角。计算该角度的步骤如下：1. 确定两个平面的交线。2. 在该交线上选取一点，分别在两个平面内作一条垂直于交线的直线。3. 这两条垂线之间的夹角就是二面角，你可以通过所得直角三角形的三角函数计算出该角度。

**二面角** — 两个相交平面之间的最小角，由两条垂直于两平面交线的直线测得，取值范围始终在0°到90°之间。

> **warning**
>
> 除非你已确认三维图形的两条边都垂直于两个平面的交线，否则切勿直接计算这两条边的夹角。

**例题:** 一个底面为正方形的棱锥，底面边长为10cm，垂直高度为12cm。计算其中一个三角形侧面与底面平面之间的二面角，结果保留1位小数。

1. 三角形侧面与底面的交线是棱锥的底边（长10cm）。该底边的中点距离底边两端各5cm。
2. 在底面上，从底边中点指向底面中心的垂线长度为5cm（因为底面是正方形，该长度为边长的一半）。
3. 三角形侧面上的垂线是该侧面的斜高，它与棱锥的垂直高度（12cm）以及底面上长5cm的线段共同构成一个直角三角形。
4. $$\tan(\theta) = \frac{\text{perpendicular height of pyramid}}{5} = \frac{12}{5} = 2.4$$
5. $$\theta = \arctan(2.4) \approx 67.4^\circ$$

*计算器:* allowed

## 常见错误

- **错误做法:** Calculating the angle between a line and an edge of the plane, instead of the line's projection onto the plane.
  - 原因: This gives an incorrect angle that is larger than the true smallest line-plane angle, leading to lost marks.
  - 正确做法: Always find the orthogonal projection of the line onto the plane first, then calculate the angle between the original line and its projection.
- **错误做法:** Failing to draw and label the extracted 2D triangle separately in your working.
  - 原因: Examiners cannot follow your reasoning for 3D problems without clear working, and you will lose partial credit even if your final answer is correct.
  - 正确做法: Draw the 2D triangle relevant to the angle you are calculating, label all known sides and angles, and reference it in your steps.
- **错误做法:** Calculating the obtuse angle between two planes instead of the acute angle.
  - 原因: The specification requires the smallest possible angle between two planes, which is always between 0° and 90°.
  - 正确做法: If your calculation gives an angle greater than 90°, subtract it from 180° to get the correct acute dihedral angle.
- **错误做法:** Using the wrong side lengths when setting up trigonometric ratios for 3D problems.
  - 原因: It is easy to mix up the height of a 3D shape with the slant height, leading to incorrect ratio calculations.
  - 正确做法: Label each side of your extracted 2D triangle with its role (opposite/adjacent/hypotenuse) relative to the angle you are calculating before substituting values.
- **错误做法:** Rounding intermediate values too early in multi-step 3D problems.
  - 原因: Early rounding introduces error that can make your final answer fall outside the acceptable mark range.
  - 正确做法: Keep at least 4 significant figures for all intermediate calculations, and only round your final answer to the required number of significant figures.

## 速查表

| Problem Type | Key Steps | Formula to Use |
| --- | --- | --- |
| 2D trigonometry problem | 1. Label right/non-right triangle 2. Identify known sides/angles | SOHCAHTOA (right triangles); sine/cosine rule (non-right) |
| Angle between line and plane | 1. Find line's projection onto plane 2. Extract right triangle with height/projection | $\tan(\theta) = \frac{\text{perpendicular height}}{\text{projection length}}$ |
| Angle between two planes (dihedral) | 1. Find line of intersection of planes 2. Draw perpendiculars on each plane to intersection line | $\sin/\cos/\tan(\theta)$ on right triangle formed by perpendiculars |

## 下一步

Now that you have mastered 2D and 3D trigonometry applications, you are ready to move on to more advanced trigonometric topics in the Edexcel IGCSE Further Pure Maths syllabus. The next core topic is sine and cosine rules, which you will use to solve more complex non-right triangle problems in both 2D and 3D contexts. You will also build on these 3D geometry skills when solving problems involving mensuration of 3D shapes, including calculating surface areas and volumes of pyramids, prisms, and cones. Remember that these 3D trigonometry questions are almost always worth 4-6 marks in the exam, so practice extracting 2D triangles from different 3D figures (cuboids, pyramids, prisms) to build speed and accuracy. Make sure you memorize the steps for calculating line-plane and dihedral angles, as these are frequently tested question types.

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来自 [OwlsPrep](https://www.owlsprep.com) —— A-Level / IB / AP / IGCSE 免费学习指南，依据官方考纲编写。原页面：https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s10-applications-in-two-and-three/