Extended. Every 3D problem at IGCSE comes down to one or two right-angled triangles hidden inside a cuboid, prism or pyramid. Find them, draw them flat, and use the tools you already have.
Work in flat right-angled triangles. The space diagonal of a cuboid is \(\sqrt{l^2 + w^2 + h^2}\). The angle between a line and a plane is the angle between the line and its projection (its 'shadow') on the plane, found in the right-angled triangle formed by dropping a perpendicular to the plane.
Fig. 1: a cuboid 8 cm by 6 cm by 5 cm. The shaded triangle ACG is right-angled at C: AC = 10 cm lies in the base, CG = 5 cm is vertical, and θ is the angle between the diagonal AG and the base ABCD
Section 01
Lengths in a cuboid.
In the lab, rotate the cuboid and switch on the base diagonal, then the space diagonal. Notice that the space diagonal, the base diagonal and a vertical edge always form a right-angled triangle standing upright inside the box.
Base diagonal: \(AC^2 = AB^2 + BC^2\) (a right-angled triangle lying flat on the base).
Space diagonal: \(AG^2 = AC^2 + CG^2\) (a right-angled triangle standing upright, right angle at \(C\)).
Together: \(AG = \sqrt{l^2 + w^2 + h^2}\) for a cuboid of length \(l\), width \(w\) and height \(h\).
Always redraw the triangle you are using as a flat 2D sketch with the right angle marked: angles that look acute or obtuse in a 3D picture may really be \(90^\circ\).
Worked example The space diagonal
In Fig. 1, \(AB = 8\) cm, \(BC = 6\) cm and \(CG = 5\) cm. Find the length of \(AG\).
Triangle \(ABC\) (flat on the base, right angle at \(B\)): \(AC^2 = 8^2 + 6^2 = 100\), so \(AC = 10\) cm.
Triangle \(ACG\) (upright, right angle at \(C\)): \(AG^2 = 10^2 + 5^2 = 125\).
A pyramid has a square base \(ABCD\) of side 6 cm. Its vertex \(V\) is directly above the centre \(M\) of the base, and each sloping edge is 10 cm. Find the height \(VM\).
Base diagonal: \(AC = \sqrt{6^2 + 6^2} = \sqrt{72}\), so \(AM = \tfrac{1}{2}\sqrt{72} = \sqrt{18} = 4.242\ldots\) cm.
Triangle \(VMA\) is right-angled at \(M\): \(VM^2 = 10^2 - (\sqrt{18})^2 = 100 - 18 = 82\).
\(VM = \sqrt{82} = 9.055\ldots\)
Answer: \(VM = 9.06\) cm (3 s.f.)
Examiner trap
Using half the side instead of half the DIAGONAL for a pyramid's sloping edge. \(M\) is the centre of the square, so \(AM\) is half of \(AC\), not half of \(AB\). (Half the side is correct only for the slant height up the middle of a face.)
Section 02
The angle between a line and a plane.
Shine a light straight down onto the plane: the line's shadow is its projection. The angle between the line and the plane is the angle between the line and that shadow. Drop a perpendicular from the top of the line to the plane to make the right-angled triangle.
Worked example Diagonal and base
In Fig. 1, find the angle \(\theta\) between the diagonal \(AG\) and the base \(ABCD\).
\(G\) is directly above \(C\), so the projection of \(AG\) on the base is \(AC\), and the angle is \(GAC\).
In triangle \(ACG\), right-angled at \(C\): \(CG = 5\) is opposite \(\theta\) and \(AC = 10\) is adjacent.
\(\tan\theta = \dfrac{5}{10}\), so \(\theta = \tan^{-1}(0.5) = 26.565\ldots^\circ\)
Answer: \(26.6^\circ\) (1 d.p.)
Worked example Sloping edge of a pyramid
For the pyramid above (base side 6 cm, sloping edges 10 cm), find the angle between the edge \(VA\) and the base.
\(V\) is above \(M\), so the projection of \(VA\) is \(MA\) and the angle is \(VAM\).
In triangle \(VMA\): \(AM = \sqrt{18}\) is adjacent and \(VA = 10\) is the hypotenuse.
\(\cos(VAM) = \dfrac{\sqrt{18}}{10} = 0.4242\ldots\), so angle \(VAM = 64.895\ldots^\circ\).
Answer: \(64.9^\circ\) (1 d.p.)
Your turn
A cube has edges of 4 cm. Find (a) the exact length of a space diagonal (b) the angle between a space diagonal and the base.
(a) \(\sqrt{4^2 + 4^2 + 4^2} = \sqrt{48} = 4\sqrt{3}\) cm. (b) The base diagonal is \(4\sqrt{2}\), so \(\tan\theta = \dfrac{4}{4\sqrt{2}}\) and \(\theta = 35.3^\circ\).
Examiner trap
Using the wrong line in the base. The angle between \(AG\) and the base is NOT angle \(GAB\): \(AB\) is not the shadow of \(AG\). Using \(\tan^{-1}(5/8) = 32.0^\circ\) is the classic wrong answer.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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Find the length of the space diagonal of a cuboid measuring 3 cm by 4 cm by 12 cm.
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The angle between a line and a plane is the angle between
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A cube has edges of 5 cm. Find the length of a diagonal of one face.
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In the cuboid of Fig. 1, which triangle contains the angle between \(AG\) and the base \(ABCD\)?
Section 03
Problems in context.
Rooms, fields with flagpoles and tents are all cuboids, rectangles and pyramids in disguise. Label the points, mark every right angle (vertical lines meet the ground at \(90^\circ\)), then work through the flat triangles one at a time, keeping full calculator values.
Worked example Flagpole in a field
A rectangular field \(ABCD\) has \(AB = 40\) m and \(BC = 30\) m. A vertical flagpole \(AP\) of height 15 m stands at corner \(A\). Find the angle of elevation of \(P\) from \(C\).
Measuring the angle of elevation from a ground line that does not lead to the foot of the pole. The angle of elevation of \(P\) from \(C\) lives in the vertical triangle \(PAC\), so you need \(AC\), the line from \(C\) to the foot \(A\).
Stage 2 · Exam
Exam-style questions
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A cuboid measures 10 cm by 4 cm by 3 cm. Find the length of its space diagonal.
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For the same cuboid (base 10 cm by 4 cm, height 3 cm), find the angle between a space diagonal and the base.
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A pyramid has a square base of side 8 cm and its vertex is directly above the centre of the base. Each sloping edge is 12 cm. Find the height of the pyramid.
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For the same pyramid (base side 8 cm, sloping edges 12 cm), find the angle between a sloping edge and the base.
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