3D Pythagoras and Trigonometry
Three dimensional Pythagoras and trigonometry apply the two dimensional rules to right-angled triangles that sit inside a solid. The key skill is finding the triangle, which almost always means identifying a right angle between a vertical edge and a line in the horizontal base. For a cuboid with edges a, b and c, the space diagonal has length equal to the square root of a squared plus b squared plus c squared, which comes from applying Pythagoras twice: once across the base to get the base diagonal, then in the vertical triangle formed by that diagonal, the height and the space diagonal. The angle between a line and a plane is the angle between the line and its projection onto that plane, found by dropping a perpendicular from the top of the line to the plane. In a right pyramid or cone the apex sits directly above the centre of the base, so the vertical height, the slant height and the horizontal distance from the centre to a base point form a right-angled triangle. Where a triangle inside the solid is not right-angled, the sine rule, the cosine rule and the area formula still apply.
Before you start
Make sure you're comfortable with these topics first:
Method
- Sketch the solid and mark every length you are given on it. Then sketch the single triangle you intend to use as a separate two dimensional diagram, away from the solid.
- Find the right angle. It is almost always where a vertical edge meets the horizontal base, or between the height of a pyramid and a line drawn in the base.
- For a length in a cuboid, apply Pythagoras twice: first across the base to get the base diagonal, then in the vertical triangle containing that diagonal and the height.
- Do not round the intermediate length. Carry the surd or the full calculator value into the second stage, and round only the final answer.
- For an angle between a line and a plane, identify the projection of the line onto the plane by dropping a perpendicular from its upper end. The required angle is between the line and that projection.
- Label the sides of your two dimensional triangle relative to the angle you want (opposite, adjacent, hypotenuse) and pick the ratio accordingly. If no right angle exists in the triangle you need, use the cosine rule instead.
Worked example
A cuboid ABCDEFGH has AB = 8 cm, BC = 6 cm and the vertical edge CG = 5 cm. Find the length of the space diagonal AG, and the angle AG makes with the base ABCD. Give answers to 1 decimal place.
- Identify the two-stage route. AG runs from a base corner to the opposite top corner, so first find the base diagonal AC, then work in the vertical triangle ACG.
- Apply Pythagoras in the base triangle ABC, which is right-angled at B: AC^2 = 8^2 + 6^2 = 64 + 36 = 100, so AC = 10 cm exactly.
- Sketch triangle ACG separately. It is right-angled at C, because CG is vertical and AC lies in the horizontal base. The two shorter sides are AC = 10 and CG = 5.
- Apply Pythagoras again: AG^2 = 10^2 + 5^2 = 100 + 25 = 125, so AG = root 125 = 11.180... = 11.2 cm to 1 decimal place.
- For the angle, note that AC is the projection of AG onto the base, so the required angle is GAC, at vertex A in triangle ACG.
- Relative to angle GAC, CG = 5 is opposite and AC = 10 is adjacent, so tan(GAC) = 5/10 = 0.5.
- Therefore GAC = tan inverse of 0.5 = 26.565... = 26.6 degrees to 1 decimal place.
Practice questions
Try each question, then tap to reveal the answer.
Q1A cuboid has edges 3 cm, 4 cm and 12 cm. Calculate the length of its space diagonal.Show answer
Answer: Space diagonal = root(3^2 + 4^2 + 12^2) = root(9 + 16 + 144) = root 169 = 13 cm.
Q2State what is meant by the angle between a line and a plane.Show answer
Answer: It is the angle between the line and its projection onto the plane, where the projection is found by dropping a perpendicular from the end of the line onto the plane.
Q3A square-based right pyramid has a base of side 10 cm and a vertical height of 12 cm. Calculate the distance from the centre of the base to a base vertex.Show answer
Answer: The base diagonal is root(10^2 + 10^2) = root 200 = 14.14 cm, and the centre is halfway along it, so the distance is 7.07 cm (exactly 5 root 2).
Q4For the pyramid in the previous question, calculate the length of a slant edge from the apex to a base vertex.Show answer
Answer: Slant edge = root(12^2 + (5 root 2)^2) = root(144 + 50) = root 194 = 13.9 cm to 1 decimal place.
Q5Explain why you should not round the base diagonal before calculating the space diagonal.Show answer
Answer: The rounded value carries an error that is then squared and combined with the height, so the final answer can be wrong in the digit you are asked to give. Keeping the exact value or the full calculator value avoids this.
Q6A vertical mast is 30 m tall. A guy rope runs from the top to a point 16 m from the base on level ground. Calculate the angle the rope makes with the ground.Show answer
Answer: tan(angle) = 30/16 = 1.875, so the angle is 61.9 degrees to 1 decimal place.
Q7A cone has a base radius of 5 cm and a slant height of 13 cm. Calculate its vertical height and the angle between the slant height and the base.Show answer
Answer: Height = root(13^2 - 5^2) = root 144 = 12 cm. cos(angle) = 5/13, so the angle is 67.4 degrees to 1 decimal place.
Exam-style questions
Written in the style of a GCSE Further Maths exam paper, with a full mark scheme.
The diagram shows a cuboid PQRSTUVW in which PQ = 12 cm, QR = 9 cm and the vertical edge RV = 8 cm. (a) Calculate the length of the diagonal PR of the base. (b) Calculate the length of PV, giving your answer to 3 significant figures. (c) Calculate the angle that PV makes with the base PQRS, giving your answer to 1 decimal place.
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A right pyramid has a rectangular base ABCD with AB = 16 cm and BC = 12 cm. The apex E is vertically above the centre of the base, and each slant edge EA = EB = EC = ED = 26 cm. (a) Calculate the vertical height of the pyramid. (b) Calculate the angle between the edge EA and the base. (c) Calculate the angle AEC.
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See real GCSE Further Maths past-paper questions, with official mark schemes →
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