Teach Me Maths

Pythagoras and Trigonometry in 3D

Find lengths and angles in cuboids and pyramids by drawing the right 2D section, including line–plane and plane–plane angles.

Turn the 3D question into a 2D triangle

Identify the required line or angle, then draw a separate triangle containing it. Label the actual lengths and establish any right angle. A line that looks diagonal on the page is not necessarily the hypotenuse of a right-angled triangle.

For a line meeting a plane at an acute angle, use the angle between the line and its perpendicular projection onto the plane. A line parallel to the plane makes 0° with it; a perpendicular line makes 90°. For the angle between two planes, take a section perpendicular to their common edge; measure the angle between the lines where that section cuts the two planes.

Explore the cuboid in three dimensions

Worked example 1

A cuboid has perpendicular edge lengths 6, 8 and 12 cm. Find its space diagonal exactly and to 3 significant figures.

  1. The base diagonal is AC=\sqrt{6^2+8^2}=10 cm. Triangle ACG is right-angled at C because CG is perpendicular to the base.

  2. Your turn. Find AG^2=10^2+12^2

    AG^2=244

  3. Answer: AG=2\sqrt{61} cm, or 15.6 cm (3 s.f.). Equivalently use AG^2=6^2+8^2+12^2

Worked example 2

In the cuboid ABCD EFGH, AB=6, BC=8 and CG=12 cm. Find the angle between AG and the base plane ABCD, to 1 decimal place.

  1. The perpendicular projection of G onto the base is C, so AG projects to AC. The required angle is \angle GAC, not an angle with a base edge.
  2. In right-angled triangle ACG, AC=10 and CG=12. Hence \tan\theta=12/10

  3. Your turn. Find the angle in degrees to 1 decimal place.

    \theta=\tan^{-1}(12/10)=50.1944\ldots^\circ

  4. Answer: 50.2^\circ (1 d.p.), measured between AG and its projection AC

Worked example 3

For the same cuboid with AB=6, BC=8, CG=12 cm, find the angle between AG and the vertical face ABFE, to 1 decimal place.

  2. GF=8 cm and GF\perp ABFE. The projection of AG is AF, so use \angle GAF. Here AF=\sqrt{6^2+12^2}=\sqrt{180} cm.

  3. Your turn. Calculate \tan^{-1}(8/\sqrt{180}) to 1 decimal place.

    \angle GAF=30.8069\ldots^\circ

  4. Answer: 30.8^\circ (1 d.p.). Always project onto the plane named in the question.

Two different sections through a pyramid

OA is half a base diagonal, so OA=5\sqrt2 cm. By contrast, OM is half a base side, so OM=5 cm. They are not interchangeable.

Worked example 4

A right square pyramid has base side 10 cm and perpendicular height VO=12 cm, with O the base centre. Find a sloping edge VA exactly.

  1. OA=\frac12\sqrt{10^2+10^2}=5\sqrt2 cm. Triangle VOA is right-angled at O

  2. Your turn. Find VA^2=12^2+(5\sqrt2)^2

    VA^2=194

  3. Answer: VA=\sqrt{194} cm. This edge is longer than the perpendicular height.

Worked example 5

For the right square pyramid with base side 10 cm and height 12 cm, find the angle between edge VA and the base, to 1 decimal place.

  1. V projects perpendicularly to O. Thus the angle is \angle VAO in triangle VOA, where OA=5\sqrt2 cm.

  2. Your turn. Calculate \tan^{-1}(12/(5\sqrt2)) to 1 decimal place.

    \angle VAO=59.4910\ldots^\circ

  3. Answer: 59.5^\circ (1 d.p.). The projection runs from the base vertex to the centre, not along a base side.

Worked example 6

For the same right square pyramid, find the acute angle between triangular face VAB and the square base, to 1 decimal place. The base side is 10 cm and the height is 12 cm.

  1. Let M be the midpoint of the common edge AB. In the base, OM\perp AB. In the isosceles face VAB, VM\perp AB. Therefore \angle VMO is the required plane–plane angle.
  2. Use right-angled triangle VOM, with OM=5 cm and VO=12 cm. Thus \tan\phi=12/5

  3. Your turn. Find the angle in degrees to 1 decimal place.

    \phi=67.3801\ldots^\circ

  4. Answer: 67.4^\circ (1 d.p.). The section must be perpendicular to the planes’ common edge.

Worked example 7

The space diagonals AG and BH of the cuboid with edges AB=6, BC=8, CG=12 cm meet at its centre O, halfway along each diagonal. Find the acute angle between them, \angle AOB, to 1 decimal place.

  2. Each full space diagonal is 2\sqrt{61} cm, so OA=OB=\sqrt{61} cm. Also AB=6 cm.
  3. \cos\theta=\frac{61+61-6^2}{2\sqrt{61}\sqrt{61}}=\frac{86}{122}=\frac{43}{61}

  4. Your turn. Evaluate \cos^{-1}(43/61) to 1 decimal place.

    \theta=45.1770\ldots^\circ

  5. Answer: 45.2^\circ (1 d.p.). The other angle at the crossing is supplementary; the question asks for the acute one.

Common mistakes

  • Using a base edge instead of a diagonal. Draw the required triangle and label its actual sides.
  • Using a sloping edge as perpendicular height. Identify the line perpendicular to the base plane.
  • Choosing the wrong line–plane angle. Use the angle between the line and its perpendicular projection onto the plane.
  • Using an arbitrary plane–plane section. Take a section perpendicular to the planes’ common edge.
  • Assuming every section is right-angled. Establish the right angle before using Pythagoras or right-angled triangle ratios.
  • Rounding a diagonal too soon. Keep exact lengths or full calculator precision for later calculations.

Now try it: Pythagoras and Trigonometry in 3D practice questions

More on this topic: Pythagoras and Trigonometry in 3D worksheet with full solutions