Pythagoras and trigonometry in 3D · 三维中的勾股定理与三角学
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| space diagonal/speɪs daɪˈæɡənl/ | 空间对角线 | kōng jiān duì jiǎo xiàn |
| base diagonal/beɪs daɪˈæɡənl/ | 底面对角线 | dǐ miàn duì jiǎo xiàn |
The diagonal through a room
- How long is the longest pole that fits inside a box? It's the space diagonal 空间对角线 — running from one corner to the far opposite corner.
- 3D Pythagoras and trig reduce this to a flat triangle hidden inside the solid.
穿过一个房间的对角线
- 装进一个盒子里的最长的杆有多长?它是空间对角线(space diagonal)——从一个角到对面最远的角。
- 三维勾股定理和三角学把这个简化为隐藏在立体内部的一个平面三角形。
The strategy: find a triangle inside
- Every 3D problem becomes a 2D problem: spot a right-angled triangle inside the solid, then use Pythagoras or trig on it.
- Often you find a base diagonal 底面对角线 first, then bring in the height to form a second triangle.
策略:在内部找一个三角形
- 每个三维问题都变成一个二维问题:发现一个立体内部的直角三角形,然后对它使用勾股定理或三角学。
- 你常常先找一个底对角线(base diagonal),然后引入高来形成第二个三角形。
Pythagoras' theorem · 勾股定理
In a right-angled triangle a² + b² = c². The same idea, applied twice, gives lengths inside 3-D solids. · 在一个直角三角形中 a² + b² = c²。同样的想法,应用两次,给出三维立体内部的长度。
A box has a base 6 cm by 8 cm. Find the length of the base diagonal (cm). · 一个盒子有底 6 cm 乘 8 cm。求底对角线的长度(cm)。
√(6² + 8²) = √100 = 10 cm. · √(6² + 8²) = √100 = 10 cm。
In a 3-D problem you look for a right-angled triangle inside the solid to work with. · 在一个三维问题中,你寻找立体内部的一个直角三角形来处理。
The whole 3-D method is to reduce it to a flat right-angled triangle. · 整个三维方法是把它简化为一个平面直角三角形。
In 3D problems, you often find the ______ diagonal first, then use it with the height. · 在三维问题中,你常常先找______对角线,然后用它和高。
The base diagonal (found using 2D Pythagoras) becomes one side of a second right triangle with the height. · 底对角线(用二维勾股定理求出)成为与高组成的第二个直角三角形的一条边。
Worked example — box diagonal
- A box has base $6\text{ cm} \times 8\text{ cm}$, height $5\text{ cm}$.
- Step 1 — base diagonal: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}$.
- Step 2 — space diagonal: $\sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\text{ cm}$.
例题——盒子对角线
- 一个盒子有底 $6\text{ cm} \times 8\text{ cm}$,高 $5\text{ cm}$。
- 第 1 步——底对角线:$\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}$。
- 第 2 步——空间对角线:$\sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\text{ cm}$。
A box has dimensions 6, 8 and 5 cm. The space diagonal is √(6² + 8² + 5²). Find it (1 dp). · 一个盒子有尺寸 6、8 和 5 cm。空间对角线是 √(6² + 8² + 5²)。求它(1 位小数)。
√(36 + 64 + 25) = √125 ≈ 11.2 cm. · √(36 + 64 + 25) = √125 ≈ 11.2 cm。
A cube has side 3 cm. Its space diagonal is √(3² + 3² + 3²) = √27. Find it (2 dp). · 一个立方体有边 3 cm。它的空间对角线是 √(3² + 3² + 3²) = √27。求它(2 位小数)。
√27 = 3√3 ≈ 5.20 cm. · √27 = 3√3 ≈ 5.20 cm。
Worked example — angle to the base
- The base diagonal ($10$) and height ($5$) form a right triangle.
- $\tan\theta = \dfrac{5}{10} = 0.5$, so $\theta = \tan^{-1}(0.5) = 26.6^{\circ}$.
3D Pythagoras shortcut. The space diagonal of a box $l \times w \times h$ is $\sqrt{l^2 + w^2 + h^2}$. For our box: $\sqrt{36 + 64 + 25} = \sqrt{125} = 11.2$ cm.
Two steps, not one. For the angle, you need the base diagonal AND the height to form the second triangle. Using the space diagonal as a side gives the wrong angle.
例题——到底的角
- 底对角线($10$)和高($5$)形成一个直角三角形。
- $\tan\theta = \dfrac{5}{10} = 0.5$,所以 $\theta = \tan^{-1}(0.5) = 26.6^{\circ}$。
三维勾股定理捷径。 一个 $l \times w \times h$ 的盒子的空间对角线是 $\sqrt{l^2 + w^2 + h^2}$。对我们的盒子:$\sqrt{36 + 64 + 25} = \sqrt{125} = 11.2$ cm。
两步,不是一步。 对于角,你需要底对角线和高来形成第二个三角形。把空间对角线用作一条边会给出错误的角。
The base diagonal is 10 cm and the box height is 5 cm. The angle to the base has tan θ = 5/10. Find θ (degrees, 1 dp). · 底对角线是 10 cm,盒子高是 5 cm。到底的角有 tan θ = 5/10。求 θ(度,1 位小数)。
θ = tan⁻¹(0.5) = 26.6°. · θ = tan⁻¹(0.5) = 26.6°。
Where 3D trig appears
- Architecture: the angle of a roof rafter, the diagonal of a room.
- Engineering: the length of a support strut, the angle of a crane arm.
3D Pythagoras is just 2D Pythagoras applied twice — first on the base, then with the height.
三维三角学出现在哪里
- 建筑:一个屋顶椽的角,一个房间的对角线。
- 工程:一个支撑撑杆的长度,一个起重机臂的角。
三维勾股定理只是二维勾股定理应用两次——先在底上,然后用高。
You've got it
- in 3D, spot a right-angled triangle inside the solid, then use Pythagoras/trig
- often find a base diagonal first, then bring in the height
- box $6 \times 8 \times 5$: base diagonal $10$, space diagonal $\sqrt{125} \approx 11.2$, angle $26.6^{\circ}$
你掌握了
- 在三维中,发现立体内部的直角三角形,然后使用勾股定理/三角学
- 常常先找一个底对角线,然后引入高
- 盒子 $6 \times 8 \times 5$:底对角线 $10$,空间对角线 $\sqrt{125} \approx 11.2$,角 $26.6^{\circ}$