Pythagoras' theorem and trigonometry in 3D

Finding a flat triangle inside a solid Extended

Every 3D question is solved in 2D. The skill is spotting a triangle that lies in one flat plane inside the solid, and that contains the length or angle you want.

A good routine:

1. Mark on the diagram the line or angle you are asked for.
2. Look for a triangle that contains it. Prefer a right-angled triangle: vertical edges meet horizontal lines at right angles, and so do the edges at each corner of a cuboid.
3. If you do not know enough sides of that triangle, find one first from another triangle, often one lying flat in the base.
4. Draw the triangle again on its own, flat, with the right angle marked and the lengths you know written on. Perspective drawings distort angles: a right angle on a 3D sketch rarely looks like one.
5. Use Pythagoras or trigonometry on the flat triangle.

Keep full calculator values between steps. In 3D problems you often use one answer to get the next, and rounding early makes the final answer wrong.

The diagonal of a cuboid Extended

In the cuboid $ABCDEFGH$ below, $G$ is vertically above $C$. The space diagonal $AG$ goes through the inside of the cuboid from one corner to the opposite corner.

It takes Pythagoras twice:

• in the base, triangle $ABC$ has a right angle at $B$, so $A{C}^2=A{B}^2+B{C}^2$;
• then triangle $ACG$ has a right angle at $C$, because the edge $CG$ is vertical and $AC$ is horizontal, so $A{G}^2=A{C}^2+C{G}^2$.

Putting the two together gives a formula for any cuboid with length $l$, width $w$ and height $h$:

$$d^2=l^2+w^2+h^2$$

You do not need to find $AC$ itself first unless the question asks for it: $AG=\sqrt{6^2+4^2+{2.5}^2}$ in one line. The space diagonal is the longest straight rod that fits inside the box.

The angle between a line and a plane Extended

A line that meets a flat surface (a plane) at a slant makes an angle with it. To find that angle:

1. From the top end of the line, drop a perpendicular onto the plane. Where it lands is the foot of the perpendicular.
2. Join the point where the line meets the plane to the foot. This is the projection of the line on the plane: its shadow if the light were directly overhead.
3. The angle between the line and the plane is the angle between the line and its projection.

So the angle is always inside a right-angled triangle: the line is the hypotenuse, the projection lies in the plane, and the perpendicular is the third side.

For the space diagonal $AG$ and the base $ABCD$: $G$ is directly above $C$, so the foot of the perpendicular from $G$ is $C$, and the projection of $AG$ on the base is $AC$. The angle between $AG$ and the base is angle $GAC$, labelled $\theta$ in the diagrams.

Name the angle with three letters, the middle letter being where the line meets the plane. A common mistake is to use an angle between the line and some other line in the plane, such as angle $GAB$; that is not the angle with the plane, because $AB$ is not the projection of $AG$.

Pyramids and cones Extended

In a right pyramid the vertex $V$ is directly above the centre $M$ of the base. In a cone the vertex is directly above the centre of the circular base. So $VM$, the vertical height, is perpendicular to every line in the base through $M$, and you get right-angled triangles such as $VMA$ for free.

For a pyramid on a square or rectangular base:

• $M$ is where the diagonals cross, so $AM$ is half the diagonal $AC$;
• triangle $VMA$ has the right angle at $M$, linking the height $VM$, the half diagonal $AM$ and the sloping edge $VA$;
• the angle between the edge $VA$ and the base is angle $VAM$, because $M$ is the foot of the perpendicular from $V$;
• the distance from $M$ to the middle of a side, $N$, is half the length of the other side, and angle $VNM$ is the angle between the line $VN$ (up the middle of a face) and the base.

For a cone with radius $r$, vertical height $h$ and slant height $l$, the triangle is $r$, $h$, $l$ with the right angle at the centre: $l^2=r^2+h^2$, and the angle between the slant edge and the base has $\cos =\frac{r}{l}$.

Triangles that are not right-angled Extended

Sometimes the triangle you need inside a solid has no right angle: for example the triangle formed by three face diagonals of a cuboid. Then:

1. Find each of its sides with Pythagoras, using the faces they lie on.
2. Use the cosine rule (three sides, want an angle) or the sine rule, as in any other triangle.
3. The area of the triangle comes from $\frac{1}{2}ab\sin C$.

The planning is the same: draw the triangle flat with its three sides written on before you start.