📚 Trigonometry in 3D Shapes | IGCSE数学：三维立体中的三角函数应用
In IGCSE Mathematics, trigonometry is not limited to flat triangles. Many exam questions involve finding lengths, angles, or distances inside three-dimensional solids such as cuboids, pyramids, cones, and cylinders. The key is to identify right-angled triangles within the 3D figure and apply sine, cosine, or tangent correctly.
在 IGCSE 数学中，三角函数并不局限于平面三角形。许多考试题目涉及在长方体、棱锥、圆锥和圆柱等三维立体图形中求长度、角度或距离。关键在于找出立体图形内部的直角三角形，并正确运用正弦、余弦和正切。
1. The Core Idea: Reducing 3D to 2D | 核心思路：将三维转化为二维
Every 3D trigonometry problem can be solved by drawing a suitable 2D triangle. You may need to use Pythagoras’ Theorem first to find a missing side, then use trigonometry to find an angle, or vice versa.
每一个三维三角函数问题都可以通过画出合适的二维三角形来解决。你可能需要先用勾股定理求出一条缺失的边，再用三角函数求角，或者反过来进行。
The most common right-angled triangles in a cuboid are formed by:
长方体中最常见的直角三角形由以下元素构成：
- Edges, face diagonals, and space diagonals.
- 棱、面对角线和体对角线。
- A vertical edge, a horizontal line on the base, and the line from the top vertex to a base corner.
- 一条竖直棱、底面上的一条水平线，以及从顶点到底面角落的连线。
Always make a sketch of the triangle you are using, even if the original diagram is 3D.
务必画出你所使用的三角形草图，即使原图是三维图形也要这样。
2. Review of Key Formulas | 关键公式回顾
In a right-angled triangle with angle θ:
在含有角 θ 的直角三角形中：
sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent
Pythagoras’ Theorem: a² + b² = c², where c is the hypotenuse.
勾股定理：a² + b² = c²，其中 c 是斜边。
For non-right-angled triangles, the sine rule and cosine rule may be needed, but in most IGCSE 3D questions the triangles can be chosen to be right-angled.
对于非直角三角形，可能需要正弦定理和余弦定理，但在大多数 IGCSE 三维问题中，所选三角形可以是直角三角形。
3. Angle Between a Line and a Plane | 直线与平面的夹角
The angle between a line and a plane is defined as the angle between the line and its projection onto the plane. This is always the smallest angle between the line and any line in the plane.
直线与平面的夹角定义为该直线与其在平面上的投影之间的夹角。这总是直线与平面内任意直线所成的最小角。
Example: In a cuboid with length 6 cm, width 4 cm, height 3 cm, find the angle between the diagonal AG and the base ABCD.
例如：在长为 6 cm、宽为 4 cm、高为 3 cm 的长方体中，求体对角线 AG 与底面 ABCD 的夹角。
First identify the projection of AG onto the base. If A is at the base and G is at the top, the projection is the face diagonal AC or AD depending on the notation. Assume G is directly above C, then the projection is AC.
首先确定 AG 在底面上的投影。如果 A 在底面，G 在顶部，投影是面对角线 AC 或 AD，具体取决于标注。假设 G 在 C 的正上方，则投影为 AC。
AC = √(6² + 4²) = √52 = 2√13 cm.
AC = √(6² + 4²) = √52 = 2√13 cm。
Now in the right-angled triangle ACG, the angle at A is the required angle. tan θ = CG / AC = 3 / (2√13).
现在在直角三角形 ACG 中，A 处的角就是所求角。tan θ = CG / AC = 3 / (2√13)。
θ = tan⁻¹(3 / (2√13)) ≈ 22.6°.
Remember: the projection is always on the plane you are measuring from.
记住：投影总是在你用来测量的那个平面上。
4. Angle Between Two Planes | 两个平面的夹角（二面角）
The angle between two planes is found by drawing a line in each plane perpendicular to the line of intersection of the planes, at the same point. The angle between these two perpendicular lines is the required dihedral angle.
两个平面的夹角可以通过在两个平面内分别作垂直于交线的直线，且这两条垂线交于同一点来求得。这两条垂线之间的夹角就是所求的二面角。
For example, in a cube, find the angle between the vertical plane ABGH and the base ABCD.
例如，在正方体中，求竖直平面 ABGH 与底面 ABCD 的夹角。
Here the line of intersection is AB. In the base, draw a line from the midpoint of AB perpendicular to AB, say to the midpoint of DC. In the vertical plane, draw a line from the same midpoint to the top midpoint. The triangle formed is right-angled.
这里交线是 AB。在底面中，从 AB 的中点作垂直于 AB 的线，例如到 DC 的中点。在竖直平面中，从同一点作到顶部中点的线。所形成的三角形是直角三角形。
If the cube has side s, the horizontal perpendicular has length s, and the vertical perpendicular has length s. The angle between them is 45°.
若正方体边长为 s，水平垂线长为 s，竖直垂线长也为 s，它们之间的夹角为 45°。
For a cuboid with height h and base length l, the angle θ between the plane and the base satisfies tan θ = h / l, where l is the perpendicular distance in the base.
对于高为 h、底面长度为 l 的长方体，平面与底面的夹角 θ 满足 tan θ = h / l，其中 l 是底面内的垂直距离。
5. Trigonometry in a Pyramid | 棱锥中的三角函数
A square-based pyramid has a vertical height from the apex to the centre of the base. This creates several right-angled triangles involving the slant height, the base edge, and the vertical height.
正四棱锥从顶点到底面中心有一条竖直的高。这形成了多个直角三角形，涉及斜高、底边和竖直高。
Common questions: find the slant height, the angle between a slant edge and the base, or the angle between two adjacent triangular faces.
常见问题：求斜高、侧棱与底面的夹角，或两个相邻三角形面之间的夹角。
Example: A square pyramid has base side 8 cm and vertical height 10 cm. Find the angle between a slant edge and the base.
例如：一个正四棱锥底面边长为 8 cm，竖直高为 10 cm。求侧棱与底面的夹角。
The distance from the centre of the base to a vertex is half the diagonal of the square: (8√2)/2 = 4√2 cm.
从底面中心到一个顶点的距离是正方形对角线的一半：(8√2)/2 = 4√2 cm。
In the right-angled triangle formed by the vertical height, this distance, and the slant edge, we have tan θ = 10 / (4√2).
在由竖直高、该距离和侧棱形成的直角三角形中，tan θ = 10 / (4√2)。
θ = tan⁻¹(10 / (4√2)) ≈ 60.5°.
Always locate the right angle: the vertical height is perpendicular to every line in the base that passes through the centre.
务必找到直角：竖直高垂直于底面内任何经过中心的直线。
6. Cone and Cylinder Problems | 圆锥和圆柱问题
In a cone, the slant height, vertical height, and radius form a right-angled triangle. In a cylinder, the longest straight line inside is the space diagonal, found using the diameter and the height.
在圆锥中，斜高、竖直高和底面半径构成直角三角形。在圆柱中，内部最长的直线是体对角线，由直径和高求出。
For a cone with radius r, vertical height h, and slant height l:
对于半径为 r、竖直高 h、斜高 l 的圆锥：
l² = r² + h²
sin θ = h / l, cos θ = r / l, tan θ = h / r
where θ is the angle between the slant side and the base.
其中 θ 是母线（斜边）与底面之间的夹角。
For a cylinder, if you need the angle between a space diagonal and the base, first calculate the base diameter, then use tan θ = height / diameter.
对于圆柱，如果需要求体对角线与底面的夹角，先计算底面直径，然后使用 tan θ = 高 / 直径。
7. Using the Sine and Cosine Rules in 3D | 在三维中运用正弦定理和余弦定理
Sometimes the desired triangle is not right-angled. In such cases, use the sine rule or cosine rule.
有时所需三角形不是直角三角形。在这种情况下，使用正弦定理或余弦定理。
Sine rule: a / sin A = b / sin B = c / sin C.
正弦定理：a / sin A = b / sin B = c / sin C。
Cosine rule: a² = b² + c² − 2bc cos A, or cos A = (b² + c² − a²) / (2bc).
余弦定理：a² = b² + c² − 2bc cos A，或 cos A = (b² + c² − a²) / (2bc)。
For example, to find the angle between two sloping edges of a pyramid, you may need to find all three sides of that triangular face first using Pythagoras in other triangles.
例如，要求棱锥两条斜棱之间的夹角，你可能需要先用勾股定理在其他三角形中求出该三角形的三条边。
Then apply the cosine rule directly. This is often faster than trying to construct a right angle.
然后直接应用余弦定理。这通常比尝试构造直角更快。
8. Step-by-Step Problem-Solving Strategy | 分步解题策略
Follow these steps for any 3D trigonometry question:
对于任何三维三角函数问题，遵循以下步骤：
- Read the question carefully and identify the required length or angle.
- 仔细阅读题目，确定所求的长度或角度。
- Draw a clear 3D sketch, then isolate the triangle that contains the unknown.
- 画清晰的三维草图，然后分离出包含未知量的三角形。
- Decide which theorem or ratio to use: Pythagoras, sin, cos, tan, sine rule, or cosine rule.
- 决定使用哪个定理或比值：勾股定理、正弦、余弦、正切、正弦定理或余弦定理。
- Calculate intermediate lengths first if they are needed.
- 如果需要，先计算中间长度。
- Give your final answer to a sensible degree of accuracy, usually 1 decimal place for angles.
- 最终答案保留合理的精度，角度通常保留一位小数。
Always check whether the triangle you are using is truly right-angled.
始终检查你所用的三角形是否真的是直角三角形。
9. Common Mistake: Wrong Projection | 常见错误：投影找错
The most frequent error in line-plane angle problems is using the wrong line as the projection. The projection of a point onto a plane is the foot of the perpendicular from that point to the plane.
在线面角问题中最常见的错误是将错误的线当作投影。一个点在平面上的投影是从该点到平面的垂线的垂足。
For a line segment between two points, project both endpoints onto the plane, then connect the two projections.
对于连接两点的线段，将两个端点分别投影到平面上，然后连接这两个投影点。
Example: In a cuboid, the projection of diagonal BH onto the base ABCD is BD, not AC. Verify which vertices are on the base.
例如：在长方体中，体对角线 BH 在底面 ABCD 上的投影是 BD，不是 AC。请确认哪些顶点在底面上。
Drawing a separate 2D diagram of the base often helps avoid this mistake.
单独画出底面的二维图通常有助于避免这个错误。
10. Worked Example: Cuboid Diagonal | 例题：长方体对角线
A cuboid has dimensions 5 cm by 7 cm by 11 cm. Find the angle between the space diagonal and the longest face diagonal.
一个长方体的尺寸为 5 cm × 7 cm × 11 cm。求体对角线与最长面对角线之间的夹角。
The longest face diagonal is on the face with sides 7 cm and 11 cm.
最长的面对角线位于边长为 7 cm 和 11 cm 的面上。
Face diagonal length = √(7² + 11²) = √170 cm.
面对角线长度 = √(7² + 11²) = √170 cm。
The space diagonal length = √(5² + 7² + 11²) = √195 cm.
体对角线长度 = √(5² + 7² + 11²) = √195 cm。
These two diagonals and one edge (5 cm) form a triangle. The angle between them is opposite the 5 cm edge.
这两条对角线和一条棱（5 cm）构成一个三角形。它们之间的夹角对着 5 cm 的边。
Using the cosine rule: cos θ = (170 + 195 − 25) / (2 × √170 × √195) = 340 / (2√33150) = 170 / √33150.
使用余弦定理：cos θ = (170 + 195 − 25) / (2 × √170 × √195) = 340 / (2√33150) = 170 / √33150。
θ ≈ cos⁻¹(170 / 182.1) ≈ 21.0°.
Notice that this triangle is not right-angled, so the cosine rule is necessary.
注意这个三角形不是直角三角形，因此必须使用余弦定理。
11. Practice Points for Exams | 考试练习要点
These skills are frequently tested in IGCSE Paper 4 (Calculator) questions.
这些技能在 IGCSE Paper 4（计算器）中经常被考查。
- Memorise the definitions of line-plane angle and plane-plane angle.
- 牢记线面角和二面角的定义。
- Always write down the triangle you are using, even if it is not drawn.
- 总是写出你正在使用的三角形，即使图中没有画。
- Use exact values such as √3 until the final calculation to avoid rounding errors.
- 在最终计算前使用精确值如 √3，以避免舍入误差。
- Check if the required answer is a length or an angle; adjust your calculator mode correctly.
- 检查所求的是长度还是角度；正确调整计算器模式。
Practise with past paper questions on cuboids, pyramids, and cones.
用历年真题练习长方体、棱锥和圆锥相关题目。
12. Summary | 总结
To solve 3D trigonometry problems, always extract a 2D triangle from the solid. Use Pythagoras’ Theorem to find missing lengths, then apply sin, cos, or tan to find angles. For non-right-angled triangles, use the sine or cosine rule.
解决三维三角函数问题的关键是始终从立体中提取一个二维三角形。用勾股定理求缺失的边长，然后用正弦、余弦或正切求角。对于非直角三角形，使用正弦定理或余弦定理。
Remember: the angle between a line and a plane is measured with its projection, and the angle between two planes is measured by two perpendiculars to their line of intersection.
记住：直线与平面的夹角是通过其投影来测量的，两个平面的夹角是通过垂直于它们交线的两条线来测量的。
With careful diagrams and systematic steps, you can master this topic.
只要仔细画图并系统地解题，你就能掌握这个专题。
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