How do you find the space diagonal of a rectangular box?

Use Pythagoras twice. For a box with edges l, w and h the base diagonal is the square root of l squared plus w squared, and the space diagonal is the square root of l squared plus w squared plus h squared.

When do you use the sine rule or the cosine rule in a 3D problem?

Isolate a two dimensional triangle inside the solid. If it is right angled use SOH CAH TOA and Pythagoras. If it is not right angled use the cosine rule when you know two sides and the included angle or all three sides, and the sine rule when you know a side and its opposite angle.

Year 12 Methods (Unit 3 & 4) Trigonometry using the sine and cosine rules

Angles between planes and more complex 3D problems

Q: How do you find the space diagonal of a box?

Use Pythagoras twice: base diagonal is root of l squared plus w squared, space diagonal is root of l squared plus w squared plus h squared.

Q: When do you use the sine or cosine rule in 3D?

Isolate a 2D triangle in the solid; use SOH CAH TOA for right triangles, the cosine rule for two sides and the included angle or three sides, and the sine rule for a side and its opposite angle.

Q: Why is the slant edge angle different from the dihedral angle?

The slant edge to base angle uses the half diagonal to a corner; the face to base dihedral angle uses the half edge to an edge midpoint, so the triangles and the angles differ.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), more complex three-dimensional problems apply the sine and cosine rules, right-angled-triangle trigonometry and Pythagoras inside a solid. The three central skills are the space diagonal of a box (3D Pythagoras), the angle between a line and a plane, and the angle between two planes — the dihedral angle — each found by isolating a suitable 2D triangle inside the figure.

The space diagonal of a rectangular box joins two opposite corners through the interior. It is found by Pythagoras used twice: first across the base, then up to the far corner.

The angle between a line and a plane is the angle between the line and its projection onto the plane (the shadow you get by dropping a perpendicular from the line to the plane). The line, its projection and the perpendicular form a right-angled triangle.

The angle between two planes, or dihedral angle, is measured at the planes' line of intersection. Choose a point on that line and draw one line in each plane perpendicular to the line of intersection; the dihedral angle is the angle between those two lines.

Key idea. Every 3D problem reduces to a 2D triangle: name the solid, pick the triangle that contains what you know and what you want, then use SOH–CAH–TOA, Pythagoras, or the sine or cosine rule.

Dihedral angle \(\theta=\angle VMO\): both \(OM\) and \(VM\) are perpendicular to the line of intersection \(AB\)

Space diagonal \(AG=\sqrt{l^{2}+w^{2}+h^{2}}\): base diagonal \(AC\) first, then up the vertical edge to \(G\)

Base diagonal and space diagonal of a box (Pythagoras twice):

\[\text{base diagonal}=\sqrt{l^{2}+w^{2}},\qquad \text{space diagonal}=\sqrt{l^{2}+w^{2}+h^{2}}\]

\sqrt{l^{2} + w^{2} + h^{2}}

Angle between a line and a plane (\(\theta\) between the line and its projection, of length \(p\), with perpendicular height \(o\)):

\[\tan\theta=\dfrac{o}{p},\qquad \text{where } p \text{ is the projection of the line onto the plane}\]

Sine rule, cosine rule and the triangle area (for any triangle in the solid):

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},\qquad c^{2}=a^{2}+b^{2}-2ab\cos C,\qquad \text{area}=\dfrac{1}{2}\,bc\sin A\]

c^{2} = a^{2} + b^{2} - 2 a b \cos C

Dihedral angle. At the line of intersection, draw one line in each plane perpendicular to it; the angle between those two lines (found from the right triangle they form with the perpendicular height) is the angle between the planes.

Working a complex 3D problem

Draw and label. Sketch the solid, mark the given lengths and angles, and name the points you need.
Space diagonal. For a box, find the base diagonal \(\sqrt{l^{2}+w^{2}}\) first, then the space diagonal \(\sqrt{l^{2}+w^{2}+h^{2}}\).
Line and plane. Project the line onto the plane; the angle is between the line and its projection — solve that right triangle.
Dihedral angle. Find the line of intersection, then two lines (one per plane) both perpendicular to it; the angle between them is the dihedral angle.
Chain the triangles. Use SOH–CAH–TOA and Pythagoras for right triangles, the cosine rule for two sides and the included angle (or three sides), and the sine rule for a side and its opposite angle; a side found in one triangle often feeds the next.

Half-diagonal vs half-edge. From the centre of a square base, the distance to a corner is half the diagonal, but the distance to an edge midpoint is half the edge — the slant-edge angle and the face dihedral use different ones.

Example 1 — Space diagonal

A box has edges \(4\), \(3\) and \(12\) cm. Find its space diagonal.

Solution

Find the base diagonal first, then apply Pythagoras again up the height.

\(AC\)	\(=\)	\(\sqrt{4^{2}+3^{2}}=5\)
\(AG\)	\(=\)	\(\sqrt{5^{2}+12^{2}}=13\text{ cm}\)

A G = 13

Example 2 — Line and plane

A square pyramid has base edge \(6\) cm and height \(4\) cm. Find the angle the slant edge \(VA\) makes with the base.

Solution

The projection of \(VA\) is \(OA=\dfrac{1}{2}(6\sqrt{2})=3\sqrt{2}\).

\(\tan\theta\)	\(=\)	\(\dfrac{4}{3\sqrt{2}}\)
\(\theta\)	\(=\)	\(43.3^\circ\approx 43^\circ\)

θ \approx 43 °

Example 3 — Dihedral angle

For the same pyramid, find the dihedral angle between a sloping face and the base.

Solution

With \(M\) the edge midpoint, \(OM=\dfrac{1}{2}(6)=3\) and \(VM\perp AB\).

\(\tan\theta\)	\(=\)	\(\dfrac{VO}{OM}=\dfrac{4}{3}\)
\(\theta\)	\(=\)	\(53.1^\circ\approx 53^\circ\)

θ \approx 53 °

Example 4 — Cosine rule in 3D

In a cube of edge \(6\) cm, the face diagonals form triangle \(CFH\) with each side \(6\sqrt{2}\). Find \(\angle FCH\).

Solution

Apply the cosine rule to the equilateral triangle of face diagonals.

\(\cos C\)	\(=\)	\(\dfrac{72+72-72}{2(72)}=\dfrac{1}{2}\)
\(\angle FCH\)	\(=\)	\(60^\circ\)

∠ F C H = 60 °

Common pitfalls

Slant edge is not the face. The angle between a slant edge and the base (using the half-diagonal) is not the dihedral angle between a slant face and the base (using the half-edge).

Both lines must be perpendicular to the line of intersection. For a dihedral angle, drawing a line to the wrong point — not perpendicular to the intersection — gives the wrong angle.

Take the 3D length, not the 2D one. The space diagonal uses all three edges; do not stop at the base diagonal.

Frequently asked questions

What is the dihedral angle between two planes?

The angle between two planes at their line of intersection, measured between two lines (one in each plane) both perpendicular to that line of intersection.

How do you find the angle between a line and a plane?

Project the line onto the plane and take the angle between the line and its projection, solving the right triangle they form.

How do you find the space diagonal of a box?

Use Pythagoras twice: the base diagonal is \(\sqrt{l^{2}+w^{2}}\) and the space diagonal is \(\sqrt{l^{2}+w^{2}+h^{2}}\).

When do you use the sine or cosine rule in 3D?

Isolate a 2D triangle in the solid; use SOH–CAH–TOA for right triangles, the cosine rule for two sides and the included angle (or three sides), and the sine rule for a side and its opposite angle.

Why is the slant edge angle different from the dihedral angle?

The slant-edge-to-base angle uses the half-diagonal to a corner; the face-to-base dihedral uses the half-edge to an edge midpoint, so the triangles and the angles differ.

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