Trigonometry in 3D

Every 3D question is a 2D triangle in disguise

A three-dimensional question looks harder than it is because the picture is drawn flat on the page. Nothing new is being asked. Inside every cuboid, prism and pyramid there is a right-angled triangle, and once you name its three vertices and draw it out on its own, the question is ordinary Pythagoras or ordinary SOHCAHTOA.

The method never changes. Find the right-angled triangle that contains the length or the angle you want; check that you know two of its three sides, or one side and one angle; draw that triangle flat, mark what you know on the drawing, and solve it. If the triangle you have chosen does not have a right angle in it, it is the wrong triangle.

Two steps are normal rather than exceptional. A length inside a solid usually needs the diagonal of a face first, and the angle is then found from that diagonal. Keep the full unrounded value in your calculator between the two steps and round only at the end.

• The right angle is almost always at a vertex on the base, or where a perpendicular from the apex meets the base
• Two of the three sides of the triangle you need are usually already given in the question; the third is the answer
• Name the triangle in your working - 'in triangle AGC' - so the examiner can see which right angle you used

A cuboid: the face diagonal, the space diagonal and the angle with the base

In a cuboid of length $l$, width $w$ and height $h$, the diagonal of a face is $\sqrt{l^2+w^2}$ and the space diagonal - corner to opposite corner, through the inside of the solid - is $\sqrt{l^2+w^2+h^2}$. That second formula is the first one used twice, which is why the space diagonal is a two-step calculation and not a new rule to learn.

A cuboid measures 8 cm by 5 cm by 4 cm. The diagonal of the base is $\sqrt{8^2+5^2}=\sqrt{89}=9.43$ cm, and the space diagonal is $\sqrt{89+4^2}=\sqrt{105}=10.2$ cm.

The angle between the space diagonal and the base lives in the right-angled triangle made by the space diagonal, the base diagonal and the vertical edge. The angle is at the corner of the base: $\tan \theta =\frac{4}{9.43}$, so $\theta ={23.0}^{\circ }$. The space diagonal is the hypotenuse, the base diagonal lies along the floor and the vertical edge is the height.

• Use the base diagonal, never a base edge, as the horizontal side: 8 cm would give 26.6 degrees, which is wrong
• The vertical edge is the height of that triangle and is never the hypotenuse

A prism: work in the cross-section

A prism has the same cross-section all the way along, so the length of the prism is a straight measurement quite separate from anything inside the cross-section. A line that runs from one end of the prism to the other combines the two: Pythagoras in the cross-section first, then Pythagoras again with the length.

A prism has a right-angled triangular cross-section with the two shorter sides 6 cm and 8 cm, and a length of 15 cm. The longest line from a vertex at one end to a vertex at the other end is found from the triangle made by the cross-section's hypotenuse and the length of the prism: $\sqrt{(6^2+8^2)+{15}^2}=\sqrt{100+225}=\sqrt{325}=18.0$ cm.

The same two steps give an angle. If the prism stands on one of its triangular faces, the length of the prism is the vertical height, so the angle between that longest line and the base is ${\tan }^{-1}\left(\frac{15}{10}\right)={56.3}^{\circ }$. The 15 cm is the height because the prism is standing on its end, and 10 cm is the hypotenuse of the cross-section.

A pyramid: drop the perpendicular to the centre of the base

The vertical height of a pyramid is measured from the apex straight down to the centre of the base, and that perpendicular is what creates the right-angled triangles. On a square base of side $s$, the distance from the centre to a vertex is $\frac{s\sqrt{2}}{2}$, half the base diagonal, and the distance from the centre to the midpoint of an edge is $\frac{s}{2}$, half the side.

A pyramid has a square base of side 10 cm and a vertical height of 12 cm. From the centre of the base to a vertex is $5\sqrt{2}=7.07$ cm, so a slant edge is $\sqrt{{7.07}^2+{12}^2}=\sqrt{50+144}=\sqrt{194}=13.9$ cm.

The angle between a slant face and the base uses the midpoint of a base edge instead, and that distance is 5 cm, not 7.07 cm: $\tan \theta =\frac{12}{5}$, so $\theta ={67.4}^{\circ }$. Choosing the wrong one of those two distances is the commonest error in pyramid questions, so decide first whether the line you are dealing with ends at a vertex of the base or at the middle of an edge.

• Centre to a vertex: half the base diagonal, $5\sqrt{2}=7.07$ cm
• Centre to the midpoint of an edge: half the side, 5 cm
• The slant height of a face is $\sqrt{{12}^2+5^2}=13$ cm, shorter than the slant edge of 13.9 cm, because the middle of an edge is nearer the centre than a vertex is

The angle between a line and a plane

The angle between a line and a plane is the angle between the line and its projection on the plane. In practice: drop a perpendicular from one end of the line onto the plane, join the point where it lands to the other end of the line, and the angle you want is in the right-angled triangle you have just drawn.

In the 8 cm by 5 cm by 4 cm cuboid, the space diagonal leaves a corner of the base and rises 4 cm. Its projection on the base is the base diagonal, 9.43 cm, so the angle between the space diagonal and the base is ${\tan }^{-1}\left(\frac{4}{9.43}\right)={23.0}^{\circ }$. The right-angled triangle has the space diagonal as its hypotenuse, the base diagonal along the floor and the vertical edge as the height.

The perpendicular from a point to a plane meets it at the foot of the perpendicular, directly below the point. A line that already lies in the plane makes an angle of zero with it, and a line perpendicular to the plane makes an angle of 90 degrees.

The angle between two planes, and drawing the 2D triangle you solve

Two planes meet in a line. The angle between them is the angle between two lines drawn in the planes, each perpendicular to that common line at the same point. In a cuboid the angle between a side face and the base is 90 degrees, so the interesting cases are pyramids, where the planes are a slant face and the base.

Take the pyramid with a 10 cm square base and a vertical height of 12 cm. The base and the slant face through the edge AB meet along AB. In the base, the line from the centre O to the midpoint M of AB is perpendicular to AB and is 5 cm long. In the face, the line from the apex E to M is perpendicular to AB. The angle between the two planes is therefore the angle OME, and $\tan \theta =\frac{12}{5}$, so $\theta ={67.4}^{\circ }$.

Drawing that two-dimensional triangle is the last step of every 3D question, and it is worth doing even when the answer looks obvious. Sketch the triangle, mark the right angle, mark the lengths you know and put the angle you want at the vertex it belongs to. If your sketch has two right angles, or a hypotenuse that is not the longest side, the triangle you chose is the wrong one.

• Mark the right angle first - if there is not one, it is the wrong triangle
• The hypotenuse is the line running along or through the solid, never the vertical height
• Put the angle where the question puts it, so a second part of the question can use it