📚 3D Trigonometry | 三维三角学
3D trigonometry extends the familiar tools of right‑angled triangle trigonometry, Pythagoras’ theorem, and the sine and cosine rules into three‑dimensional shapes. In the Edexcel IGCSE Mathematics specification, this topic requires you to work with cuboids, pyramids, cones and other solids, finding unknown lengths and angles that cannot be seen on a single flat face. Mastering 3D trigonometry sharpens your spatial reasoning and provides a powerful link between algebra and geometry.
三维三角学将我们熟悉的直角三角学、毕达哥拉斯定理以及正弦余弦定理扩展到三维图形中。在爱德思 IGCSE 数学大纲里，这个主题要求你处理长方体、棱锥、圆锥以及其他立体，求出无法在单一平面上直接看到的未知边长和角度。掌握三维三角学不仅能强化你的空间想象力，还在代数与几何之间架起了一座坚固的桥梁。
1. Understanding 3D Space | 理解三维空间
In three‑dimensional geometry we describe points with three coordinates (x, y, z), and solids are built from faces, edges and vertices. When you look at a 2D diagram of a 3D object, some lengths and angles are hidden or distorted. The key skill is to 'see' right‑angled triangles that are lying in different planes.
在三维几何中，我们用三个坐标 (x, y, z) 来描述点，立体由面、棱和顶点构成。当你观看三维物体的平面示意图时，一些长度和角度会被遮挡或变形。核心技巧是“看”到位于不同平面内的直角三角形。
Imagine a cuboid with a vertical edge, a horizontal edge on the base, and a diagonal running across the base. The space diagonal then connects the top of the vertical edge to the opposite corner on the base. Recognising these right‑angled configurations is the foundation of all 3D trigonometry calculations.
想象一个长方体，有一条竖直棱、底面的一条水平棱以及底面的一条对角线。然后空间对角线将竖直棱的顶端与底面的对角连接起来。识别这些直角三角形构成是所有三维三角学计算的基础。
2. Pythagoras’ Theorem in 3D | 三维毕达哥拉斯定理
You already know that in a right‑ dimensions we can apply this twice to find a space diagonal. First, find the diagonal of a rectangular face using the two edges of that face. Then treat that face diagonal and the third perpendicular edge as the legs of a new right‑angled triangle whose hypotenuse is the space diagonal.
你已经知道，在直角三角形中，a² + b² = c²。在三维中我们可以两次应用该定理来求空间对角线。首先，用长方形的两条棱求出该面的对角线。然后将这个面对角线和与之垂直的第三条棱作为一个新直角三角形的两腰，斜边就是空间对角线。
d = √l² + w² + h²
This formula gives the length of the longest diagonal inside a cuboid with length l, width w and height h. Every step relies on identifying a right angle: the face diagonal forms a right angle with the third edge because the edge is perpendicular to the whole face.
该公式给出了长、宽、高分别为 l、w、h 的长方体中最长对角线的长度。每一步都依赖于识别直角：面对角线与第三条棱构成直角，因为该棱垂直于整个面。
3. Identifying Right-Angled Triangles in 3D | 识别三维中的直角三角形
Before you can use any trigonometric ratio, you must locate a right‑angled triangle that contains the unknown side or angle. In a solid such as a pyramid with a rectangular base, the height is perpendicular to the base, creating right angles with every line in the base that passes through the foot of the height.
在使用任何三角比之前，你必须找到一个包含未知边或角的直角三角形。在底面为矩形的棱锥等立体中，高是垂直于底面的，因此与底面内所有通过垂足的直线都构成直角。
Look for lines that are perpendicular to a whole plane: an edge meeting a face at 90°, a slant height drawn to the midpoint of a base edge, or a line from the apex directly down to the centre of the base. Drawing these triangles separately outside the main diagram can make the problem much clearer.
寻找垂直于整个平面的直线：与面成 90° 相交的棱、画到底边中点的斜高、或者从顶点直接垂直落到底面中心的线段。将这些三角形单独抽出来画在主图之外，能让问题清晰很多。
4. Diagonal of a Cuboid | 长方体的对角线
Consider a cuboid with dimensions 3 cm, 4 cm, and 12 cm. The base diagonal is √3² + 4² = √25 = 5 cm. Then the space diagonal becomes the hypotenuse of a right‑angled triangle with legs 5 cm and 12 cm, so its length is √5² + 12² = √169 = 13 cm.
考虑一个尺寸为 3 cm、4 cm 和 12 cm 的长方体。底面对角线为 √3² + 4² = √25 = 5 cm。那么空间对角线成为一个两腰为 5 cm 和 12 cm 的直角三角形的斜边，其长度为 √5² + 12² = √169 = 13 cm。
You can also use the one‑step formula d = √3² + 4² + 12² = √9+16+144 13 cm. In exam questions, always write the intermediate step to show your method clearly.
你也可以直接用一步公式 d = √3² + 4² + 12² = √9+16+144 = √169 = 13 cm。在考试题中，总是写出中间步骤以清晰地展示你的方法。
5. 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 orthogonal projection onto the plane. In a cuboid, if you want the angle between the space diagonal AG and the base ABCD, you first drop a perpendicular from G to the base (which is simply edge CG). The projection of AG onto the base is the line AC.
直线与平面的夹角定义为该直线与其在平面上的正投影之间的夹角。在长方体中，如果你想要求空间对角线 AG 与底面 ABCD 之间的夹角，首先从 G 向底面作一条垂线（就是棱 CG）。AG 在底面上的投影是线段 AC。
Then triangle ACG is right‑ θ = CG/AG. Often you have already found AC and AG using Pythagoras, so either ratio works.
然后三角形 ACG 在点 C 处是直角三角形。你可以使用 tanθ = 对边/邻边 = CG/AC，或者 sinθ = CG/AG。通常你已经用毕达哥拉斯定理求出了 AC 和 AG，所以两个比都可以用。
| Formula | Used when |
| tanθ = height / base diagonal | You know the height and the diagonal of the base |
| sinθ = height / space diagonal | You know the height and the space diagonal |
这一定义在棱锥和圆锥中同样适用，只要你能准确找出直线在平面上的投影即可。
6. Angle Between Two Planes | 两平面的夹角
The angle between two intersecting planes is measured by the angle between two lines, one in each plane, that are both perpendicular to the line of intersection. In exam questions, this often appears as finding the angle between a sloping face and the base of a pyramid.
两相交平面的夹角是通过在每个平面内各取一条与交线垂直的直线，并测量这两条直线之间的夹角来得到的。在考试题目中，这通常表现为计算棱锥的斜面与底面之间的夹角。
For a square‑based pyramid, the line of intersection is one edge of the base. From the midpoint of that edge, draw one line to the apex (the slant height) and one line along the base to the centre. These two lines are both perpendicular to the edge, so the angle between them is the angle between the face and the base.
对于一个底面为正方形的棱锥，其交线是底面的一条边。从该边的中点画一条线段到顶点（即斜高），再沿着底面画一条线段到中心。这两条线段都与底边垂直，所以它们之间的夹角就是侧面与底面的夹角。
le) = vertical height / (half the base side) when the apex lies directly above the centre.
在由斜高、中点到中心的垂线段以及垂直高度构成的三角形中使用三角学。当顶点位于中心正上方时，通常有 tan(角度) = 垂直高度 / (底面边长的一半)。
7. Pyramids: Slant Height and Face Angles | 棱锥：斜高与面角
A pyramid with a regular polygonal base has several key lengths: the vertical height h, the slant height l (the distance from the apex to the midpoint of a base edge), and the lateral edge s (from the apex to a vertex of the base). The right‑angled triangle that links h, s, and the distance from the centre to a vertex lets you find unknown lengths.
底面为正多边形的棱锥有几个关键长度：垂直高度 h、斜高 l（从顶点到底面边中点的距离）以及侧棱 s（从顶点到底面顶点的距离）。连接 h、s 和中心到顶点距离的直角三角形可以帮助你求出未知长度。
/l or tanθ = h/apothem. Identifying which triangle to use is a major part of the question.
若要计算侧面与底面的夹角，可使用由 h、l 和边心距（从中心到边的垂线）构成的三角形。侧面与底面的夹角 θ 满足：sinθ = h/l 或 tanθ = h/边心距。识别该用哪一个三角形是解题的一大关键。
When the base is a square of side a, the apothem is a/2 and the distance from the centre to a vertex is a/√2. Always label these on a separate sketch to avoid confusion.
当底面是边长为 a 的正方形时，边心距为 a/2，中心到顶点的距离为 a/√2。务必在单独的草图上标注清楚，避免混淆。
8. Cones: Slant Height and Semi-Vertical Angle | 圆锥：斜高与半顶角
A right circular cone gives a clean right‑angled triangle when you take a cross‑ ius r, the vertical height h, and the slant height l are related by r² + h² = l². The semi‑ ngle α (the angle between the axis and the slant side) satisfies tanα = r/h or sinα = r/l.
对直圆锥做一个过顶点和底面中心的截面，你会得到一个完美的直角三角形。底面半径 r、垂直高度 h 和斜高 l 满足 r² + h² = l²。半顶角 α（轴与斜边之间的夹角）满足 tanα = r/h 或 sinα = r/l。
Many IGCSE problems ask for the angle between the slant side and the base. This is exactly the semi‑vertical angle α, because the base is a plane perpendicular to the axis. If you need the sector angle of the curved surface when flattened, that involves the ratio of the circumference to the slant height, but that is a separate topic.
许多 IGCSE 题目会要求计算斜边与底面的夹角。这正是半顶角 α，因为底面是与轴垂直的平面。如果需要展开后曲面的扇形角，那涉及圆的周长与斜高的比例，不过那属于另一个主题。
9. Using the Sine and Cosine Rules in 3D | 在三维中使用正弦和余弦定理
Not all triangles in 3D solids are right‑angled. When you need to find an angle in a scalene triangle, or when you know two sides and a non‑included angle, the sine and cosine rules become essential. For example, a tetrahedron often contains faces that are general triangles.
并非所有三维立体中的三角形都是直角三角形。当需要求一个不等边三角形中的角，或者已知两条边和一个非夹角时，正弦定理和余弦定理就变得非常重要。例如，四面体常常包含一般的三角形面。
In a triangular pyramid, you might know the lengths of all three edges meeting at the apex and the lengths of the base edges. To find the angle between two sloping faces, you often need to work with a triangle formed inside the solid that requires the cosine rule.
在一个三棱锥中，你可能已知从顶点出发的三条棱的长度以及底面边长。为了求两个侧面之间的夹角，往往需要在立体内部构造一个三角形，此时就需要用到余弦定理。
a² = b² + c² − 2bc·cosA
Always write the rule you are using before substituting numbers. Apply the same disciplined approach: extract the triangle from the 3D diagram, label all known lengths and angles, and then decide whether you need the sine rule, the cosine rule, or standard right‑triangle trigonometry.
在代入数字前，务必先写出所使用的定理。采用同样有条理的方法：从三维图中提取出三角形，标注所有已知的边长和角度，然后决定需要用正弦定理、余弦定理还是常规的直角三角学方法。
10. Practice and Problem-Solving Tips | 练习与解题技巧
Begin every 3D trigonometry question by redrawing the solid and highlighting the triangle you plan to use. Draw that triangle separately, labelling lengths and the unknown angle clearly. Never attempt to work entirely within the 3D sketch — it almost always leads to mistakes.
解每道三维三角学问题时，先重新画出立体并突出你要使用的三角形。把这个三角形单独画出来，清楚地标注边长和未知角。永远不要试图完全在原三维草图中完成计算——这样几乎总会出错。
Check whether a triangle is right‑angled: look for a line perpendicular to a plane, or a tangent to a circle.
检查三角形是否是直角三角形：寻找垂直于平面的直线，或者圆上的切线。
Use exact values where possible, and only round at the final answer to the required degree of accuracy.
尽可能使用精确值，只在最后答案处按要求精确度舍入。
If the question involves angles, decide whether you are being asked for the angle between two lines, a line and a plane, or two planes — the method differs for each.
如果题目涉及角，要判断问的是两直线的夹角、直线与平面的夹角还是两平面的夹角——每种情况的求法都不同。
Finally, practice with a variety of solids: cuboids, square‑based pyramids, tetrahedrons, cones, and even composite shapes such as a pyramid on top of a cuboid. The more shapes you work with, the quicker you will recognise the hidden right‑angled triangles.
最后，要针对各种立体进行练习：长方体、正四棱锥、四面体、圆锥，甚至是组合体，如长方体上叠加一个棱锥。你处理的立体形状越多，就越能快速识别出隐藏的直角三角形。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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