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3D Pythagoras and Trigonometry

Pythagoras and Trigonometry

Finding it

  • The syllabus names this skill directly, and it is the part of the topic students find hardest to see
  • A plane is a flat surface, such as the base of a solid, and the angle is measured between the line and the plane itself, not between the line and an edge
A cuboid ABCDEFGH with a line PQ drawn from the base to the top face: dropping a perpendicular from Q to the plane creates a right-angled triangle, and three different angles x, y and z are labelled to show that the angle depends on which plane is named
Source: 3D Pythagoras & trigonometry by Save My Exams
  • Build the right-angled triangle in three steps
    • Drop a perpendicular from the top of the line straight down onto the plane
    • Join where it lands to the point where the line meets the plane
    • Those two, with the original line, form a right-angled triangle whose right angle is on the plane
The right-angled triangle taken out of the cuboid and drawn flat: the 3 cm side is the opposite, root 52 cm is the adjacent and root 61 cm the hypotenuse, so the angle c can be found by any of the three ratios
Source: 3D Pythagoras & trigonometry by Save My Exams
  • The angle you want sits where the line meets the plane
A cuboid ABCDEFGH measuring 4 cm by 3 cm by 6 cm with two right-angled triangles picked out: BEF in purple gives the length a of BF across the top face, and ABF in red then uses that with the 3 cm height to give the angle c between BF and the diagonal AF
Source: 3D Pythagoras & trigonometry by Save My Exams
  • A useful image: hold the line like a fishing rod and let the line drop vertically to the surface, then look at the triangle that forms
  • Where the foot of the perpendicular is the centre of a base, the horizontal side is half a diagonal, not half an edge
Worked example

Angle between a diagonal and the base

A cuboid measures 5 cm by 12 cm by 9 cm. Find the angle between the space diagonal and the base, correct to 1 decimal place.

The same cuboid, 12 cm along the front, 5 cm deep and 9 cm high, with two lines drawn from the front bottom left corner: a dashed line across the base to the far bottom corner, and a solid line to the opposite top corner. The angle between them at the base corner is marked x degrees. The figure is marked NOT TO SCALE.

Solution:

  • The space diagonal rises from one corner of the base to the opposite top corner
  • Dropping vertically from the top corner lands on the far corner of the base, so the horizontal side is the base diagonal, 13 cm
  • The vertical side is the height, 9 cm, and the right angle is at the base corner
  • The angle is opposite the height and adjacent to the base diagonal, so use tangent
  • tan θ = 9 ÷ 13
  • θ = tan⁻¹(9 ÷ 13) = 34.6951…
  • The angle is 34.7° to 1 decimal place
Exam tip

Redraw each right-angled triangle flat, on its own, with its sides labelled. The angle between a line and a plane is measured to the plane, so the perpendicular must drop onto the surface, not onto an edge; and in a pyramid whose apex sits above the centre, the horizontal side of that triangle is half a diagonal of the base, not half a side. Never measure from a 3D drawing — it distorts every angle in it.

Worked example

A pyramid's slant edge

A pyramid has a square base of side 8 cm, and its apex is 10 cm vertically above the centre of the base. Find the angle between a slant edge and the base, correct to 1 decimal place.

A pyramid on a square base of side 8 cm, with the four slant edges meeting at an apex above the middle. A dashed vertical line from the apex to the centre of the base is labelled 10 cm, and a second dashed line runs from that centre out to the front right corner, meeting the vertical at a marked right angle. The angle between that dashed line and the slant edge, at the corner, is marked x degrees. The figure is marked NOT TO SCALE.

Solution:

  • The apex drops vertically to the centre of the base, so the horizontal side runs from the centre to a corner, which is half a diagonal
  • The full base diagonal is √(8² + 8²) = √128, so half of it is √(4² + 4²) = √32 = 5.6568…
  • The vertical side is the height, 10 cm, and the right angle is at the centre of the base
  • The angle sits at the corner, opposite the height and adjacent to the half-diagonal, so use tangent
  • tan θ = 10 ÷ 5.6568…
  • θ = tan⁻¹(1.7677…) = 60.5037…
  • The angle is 60.5° to 1 decimal place
  • Note the common trap: using half an edge, 4 cm, instead of half a diagonal gives a different and wrong answer