3D Pythagoras & TrigonometryEdexcel IGCSE Maths: Revision notes

A cuboid with length $l$, width $w$ and height $h$ has a space diagonal running from one corner through the solid to the opposite corner (not across a face).

Label the cuboid $ABCDEFGH$, with base $ABCD$ and top $EFGH$ directly above ($E$ above $A$, $F$ above $B$, etc.). The space diagonal $AG$ (from base corner $A$ to the top corner $G$ diagonally opposite) is found in two steps:

1. Find the diagonal of the base rectangle first: $AC^2 = l^2 + w^2$
2. Then treat $AC$, the vertical edge $CG = h$, and $AG$ as a right-angled triangle: $AG^2 = AC^2 + h^2$

Combining gives the direct formula: $AG = \sqrt{l^2 + w^2 + h^2}$

This works because the space diagonal, the base diagonal and a vertical edge always form a right-angled triangle, with the right angle at the base corner directly below the top vertex.

The base diagonal step gives $AC^2 = l^2 + w^2$. Substituting this into the second right-angled triangle $AG^2 = AC^2 + h^2$ gives $AG^2 = l^2 + w^2 + h^2$, so: $AG = \sqrt{l^2 + w^2 + h^2}$

This single formula works for any cuboid space diagonal, since $l$, $w$ and $h$ are interchangeable (it doesn't matter which edge you call length, width or height). You can use the shortcut directly in an exam, but you must still be able to show the two-triangle working if a question asks you to 'show that' or asks for a specific intermediate length such as a base or face diagonal.

The angle between a line and a plane is measured between the line and its projection onto the plane — this is the smallest possible angle between the line and any line in the plane.

Method for a cuboid, e.g. the angle between the diagonal $AG$ and the base plane $ABCD$:

1. Identify where the line meets the plane: $A$ is in the base plane, $G$ is above the plane.
2. Drop a perpendicular from $G$ straight down to the base — this lands exactly on $C$ (the corner diagonally opposite $A$ on the base), since $CG$ is a vertical edge.
3. The projection of $AG$ onto the base plane is $AC$.
4. The required angle is $\angle GAC$, found in the right-angled triangle $ACG$ (right angle at $C$) using: $\tan(\angle GAC) = \frac{GC}{AC} = \frac{\text{height}}{\text{base diagonal}}$

Once you've identified the projection, you have a right-angled triangle made of: the original line (hypotenuse), the projection (adjacent to the required angle), and a vertical edge (opposite the required angle, perpendicular to the plane).

Because the vertical edge is always opposite the angle and the projection is always adjacent: $\tan(\theta) = \frac{\text{opposite (vertical height)}}{\text{adjacent (projection length)}}$

You can also use sine or cosine with the original 3D line as the hypotenuse: $\sin(\theta) = \frac{\text{height}}{\text{diagonal}}, \quad \cos(\theta) = \frac{\text{projection}}{\text{diagonal}}$

All three give the same angle — tangent is usually quickest since it avoids calculating the full diagonal first.

Full marks require clearly labelled working, not just a final number:

1. Sketch or describe the cuboid with labelled vertices ($ABCDEFGH$) and given lengths.
2. State which triangle you are using and why it is right-angled (justify using the properties of a cuboid — all edges meet at 90 degrees).
3. Calculate the base (or face) diagonal first, leaving it in surd form (e.g. $\sqrt{52}$) rather than rounding early.
4. Use the surd form in the second calculation (space diagonal length, or the trig ratio for the angle) to avoid rounding errors.
5. Round only your final answer, to the accuracy stated in the question (usually 1 or 3 significant figures).