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This page sets out the method for 3D Pythagoras & trigonometry in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the 3D Pythagoras & trigonometry topic page.
When a length or angle is asked for inside a solid, find and sketch the flat right-angled triangle that contains it.
When you need the long diagonal through the inside of a cuboid, from one corner to the opposite corner.
The face diagonal of a 3 by 4 face is 5, because 3² + 4² = 9 + 16 = 25 and √25 = 5.
The space diagonal of a cuboid with edges a, b, c is d = √(a² + b² + c²). For 3, 4, 12, d = 13.
The space diagonal, height, and base diagonal form a right-angled triangle inside the cuboid.
tan θ = height / base diagonal. For height 12 and base diagonal 5, tan θ = 12/5.
sin θ = height / space diagonal. For height 12 and space diagonal 13, sin θ = 12/13.
The angle is made with the base plane, so the base diagonal is the adjacent side, not a single edge.
Before you write a formula, name the triangle. On a cuboid, the edges meet at right angles, so any face is a right-angled triangle once you draw its diagonal. A 3 cm by 4 cm face has diagonal 5 cm, because 3^2 + 4^2 = 5^2. If the third edge is 12 cm, stand that 5 cm diagonal next to the 12 cm edge. Then 5^2 + 12^2 = 13^2, so the space diagonal is 13 cm. Sketch this second triangle flat on the page. A 3D picture hides the right angle.
Label the three sides of that flat triangle: opposite, adjacent and hypotenuse if you need an angle, or the two legs and the hypotenuse if you need a length. The right angle is never at the corner you are measuring an angle from, unless the question really is a right angle there. For the angle between the space diagonal and the base, the right angle is where the vertical edge meets the floor, at the far end of the floor diagonal.
Method: 1. Draw the face diagonal first, then a second right-angled triangle for the space diagonal. 2. On a 3cm by 4cm face, the diagonal is 5cm. 3. Next to a 12cm edge, 5^2+12^2=13^2, so the space diagonal is 13cm.
You may do the Pythagoras in one line. Square each edge, add, and square-root: √(l^2 + w^2 + h^2). For edges 6 cm, 8 cm and 5 cm, the floor of 6 by 8 has diagonal 10 cm, and the space diagonal is √(10^2 + 5^2) = √(125). Leave a square root exact if the question asks, or round at the end. √(125) = 5√(5), which is useful as a check even when you also want a decimal.
Do not square-root each edge and then add. √(6) + √(8) + √(5) is not a diagonal. Do not use only two edges if the line goes through the interior. A face diagonal uses two edges. A space diagonal uses all three. If your length is shorter than the longest edge, it cannot be a space diagonal, because the straight line through the box is longer than every edge.
Method: 1. Square the three edges, add, then square-root: sqrtl^2+w^2+h^2. 2. Edges 6cm, 8cm and 5cm give sqrt10^2+5^2=√(125). 3. Simplify √(125)=5√(5) when an exact length is useful.
The shadow of the space diagonal on the base is the floor diagonal. Call the angle with the base θ. Then tan θ = fracheightfloor diagonal. For a floor diagonal of 10 cm and a height of 5 cm, tan θ = 5/10 = 0.5, so θ = tan^-1(0.5) = 26.6^° to 1 decimal place. If you already have the space diagonal, sin θ = fracheightspace diagonal gives the same angle. Use one ratio, not a mixture.
Set the calculator to degrees before you press inverse tan. Round the angle once, at the end, usually to 1 decimal place. A short cuboid gives a small angle with the base. A tall cuboid gives a larger one. If your angle is over 45^°, the height is greater than the floor diagonal, so check that this matches the numbers. The angle with the vertical edge is the other angle in the triangle, 90^° minus the angle with the base. Read which one the question asks for.
Method: 1. Use tanθ=dfracheightfloor diagonal for the angle with the base. 2. Height 5cm and floor diagonal 10cm give tanθ=0.5. 3. The angle is tan^-1(0.5)=26.6^° to 1 decimal place.
On a square-based pyramid the apex sits above the centre of the base. There are two different sloping lengths. The slant height of a face runs to the midpoint of a base edge, so the horizontal distance is half the base side. The slant edge runs to a base corner, so the horizontal distance is half the diagonal of the base.
For a base of side 10 cm, the base diagonal is √(10^2 + 10^2) = √(200), and half of it is 5√(2). The square of that half-diagonal is 50. With height 12 cm, the slant edge is √(12^2 + 50) = √(194), which is 13.9 cm to 1 decimal place. The angle between that edge and the base has tan θ = frac12√(50). Inverse tan gives 59.5^° to 1 decimal place. Draw the flat right-angled triangle first: height opposite, half-diagonal adjacent, slant edge as hypotenuse.
Method: 1. For a slant edge, use half the base diagonal as the horizontal leg. 2. A base of side 10cm has half-diagonal squared equal to 50. 3. Height 12cm gives slant edge √(144+50)=13.9cm to 1 decimal place.
Most 3D questions give you a cuboid or a pyramid with corners labelled in order. A typical set-up says: ABCDEFGH is a cuboid, ABCD is the base, and E is directly above A. That tells you which edges are vertical and which line is the space diagonal. The diagram is marked "not drawn accurately", so never measure it. Every length you need is either printed or found by calculation.
The command words matter. "Calculate" or "work out" means show the numbers you used. "Show that" means you must reach the printed value from your own working, so write one more line than you normally would. "Give your answer to 3 significant figures" means keep the unrounded value in your calculator until the very last step, then round once.
Method: 1. Write down which edges are vertical, which face is the base, and which line the question names. 2. Sketch the flat right-angled triangle that holds that line, and label every side you know. 3. Use Pythagoras for a length or a trig ratio for an angle, and write the calculation line. 4. Round once at the end and add the unit, such as cm or degrees.
Tip: Redraw the flat triangle on its own. It is much easier to see the right angle in a small clear sketch than in the 3D picture.
A space diagonal needs two steps. First find the diagonal of the base using the two flat sides, then use that answer as one short side of a second right-angled triangle, with the vertical edge as the other. The result is √(l^2 + w^2 + h^2). In an exam you earn the method mark for the first triangle, so do not jump straight to the one-line formula without showing where it came from.
Non-calculator versions are built on friendly numbers. Sides such as 2, 3, 6 give a diagonal of 7, and 3, 4, 12 give 13. Others leave a surd, such as √(50) = 5√(2). When the question says "give your answer as a surd" or "show that the length is exactly", do not use a decimal at all. Keep the square root until the last line and simplify it.
Method: 1. Find the diagonal of the base, either as a number or as a square root. 2. Draw a second triangle with that diagonal and the vertical edge as its legs. 3. Add the squares of the two legs and take the square root. 4. Simplify any surd, or round if a decimal was asked for.
The angle between a line and a flat surface is found by dropping a perpendicular from the top of the line to the surface. For a space diagonal AG of a cuboid, the foot of that perpendicular is C, so the angle is the one at A in triangle ACG. The opposite side is the vertical edge CG, the adjacent side is the base diagonal AC, and the hypotenuse is AG.
Choose the ratio from the two sides you actually know. If you know CG and AC use tangent. If you know CG and AG use sine. If you know AC and AG use cosine. Using the original sides AB or BC by mistake is a frequent slip, because they are not in the right-angled triangle. Questions also work backwards: given the angle and one length, find a missing edge, then a volume.
Method: 1. Identify the foot of the perpendicular on the plane. 2. Mark the right angle and the angle you want in the flat triangle. 3. Label opposite, adjacent and hypotenuse from that angle. 4. Choose sin, cos or tan, rearrange, and use the unrounded value.
Tip: In a worked-backwards question, write the formula first, for example CG = AC × tan θ, and only then substitute.
A square-based pyramid with its apex directly above the centre hides three useful right-angled triangles. The height with half of the base diagonal gives the slant edge. The height with half of the base side gives the slant height of a face. The slant height with half the base side gives the edge again. Always check that you are using half of the diagonal, not the whole diagonal.
The half-diagonal of a square of side s is dfracs√(2)2, which is about 0.707s. Questions often ask for the angle between an edge and the base, the angle between a sloping face and the base, or the total surface area. The face angle uses the half side, and the edge angle uses the half diagonal. Name your triangle in the working, for example "triangle VMN", so the marker can see which one you mean.
Method: 1. Mark the centre of the base and the apex directly above it. 2. Use half the base diagonal for edge questions, or half the base side for face questions. 3. Pick the right-angled triangle that contains the length or angle asked for. 4. Calculate, and check that the hypotenuse is the longest side.
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