3D Pythagoras and trigonometry

In 3D problems, find a right-angled triangle inside the shape, then use Pythagoras or trigonometry on it. You often need two triangles, one after the other.

For a cuboid of length l, width w and height h, the space diagonal (from one corner to the opposite corner, through the inside) is:

d = √(l² + w² + h²)

Example: a cuboid is 3 cm by 4 cm by 12 cm. Find the length of its space diagonal.

1. Diagonal of the base: √(3² + 4²) = √25 = 5 cm
2. This, the height and the space diagonal make a right-angled triangle
3. √(5² + 12²) = √169 = 13 cm

Check with the formula: √(9 + 16 + 144) = √169 = 13.

A plane is a flat surface, such as the base of a cuboid. To find the angle between a line and a plane:

1. From the top of the line, drop a vertical line down to the plane.
2. Join where it lands to the bottom of the line. This is the line's projection onto the plane.
3. The angle you want is between the line and its projection, in a right-angled triangle.

Example: in the 3 cm by 4 cm by 12 cm cuboid, find the angle between the space diagonal and the base, to 1 d.p.

1. The projection is the base diagonal, 5 cm, and the height is 12 cm
2. tan θ = 12/5
3. θ = tan⁻¹(2.4) = 67.4°

For a pyramid with its top directly above the centre of a square base, use the right-angled triangle made by the height, half the diagonal of the base, and a sloping edge.

Example: a pyramid has a square base 6 cm by 6 cm and a height of 4 cm. Find the length of a sloping edge, and the angle between it and the base, to 1 d.p.

1. Base diagonal: √(6² + 6²) = √72, so half of it is √18 = 4.24… cm
2. Sloping edge: √(18 + 4²) = √34 = 5.83… = 5.8 cm
3. tan θ = 4 ÷ 4.24…
4. θ = 43.3°