3D Pythagoras and Trigonometry

Mathematics revision notes · Cambridge (CIE) · 0580

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1. Working in three dimensions

Reduce it to two dimensions

  • Nothing new is needed here: the tools are still Pythagoras' theorem and SOHCAHTOA
  • The whole difficulty is finding the right-angled triangle inside the solid, so the first move is always to locate one
  • Once found, redraw that triangle flat on the page rather than leaving it drawn at an angle inside the solid
A cone of base radius 5 cm and perpendicular height 12 cm with its slant edge labelled l cm, and beside it the right-angled triangle formed by the height, the radius and the slant edge redrawn flat, with sides 12 cm, 5 cm and l cm and the right angle between the 12 cm and 5 cm sides
  • A 3D drawing distorts angles, so a triangle that looks right-angled in the picture may not be, and one that looks skewed often is
  • Many problems need two triangles in succession, where the answer from the first becomes a side of the second
  • Keep the full unrounded value between the two stages and round only at the very end

The space diagonal

  • The longest straight line inside a cuboid runs between opposite corners, and it is what a "longest rod that fits in the box" question asks for
  • Reach it in two steps: Pythagoras across the base to get the base diagonal, then Pythagoras again using that diagonal and the height
A cuboid 2 cm by 2 cm by 6 cm with A and B at opposite corners: the diagonal AB is drawn in red, and the flat right-angled triangle across the base is drawn in purple, because that base diagonal is the missing side of the triangle containing AB
Source: 3D Pythagoras & trigonometry by Save My Exams
The two triangles from that cuboid drawn flat, one above the other: the first has sides 2 and 6 giving a squared of 40, and the second uses root 40 with the 2 cm height to give x squared as 44, so x is 6.63 cm, with the note that the root need not be worked out in between
Source: 3D Pythagoras & trigonometry by Save My Exams
  • There is a one-step version, d² = x² + y² + z², but it is not on the list of formulas, so learn it or use the two-step route
  • Either route earns the method marks: the schemes credit the three squares added in one line, and the first face diagonal on its own earns a method mark

Worked example

The longest rod in a box

A box is a cuboid measuring 5 cm by 12 cm by 9 cm. Find the length of the longest straight rod that fits inside it, correct to 3 significant figures.

A cuboid drawn in three dimensions with its hidden back edges dashed. The front bottom edge is labelled 12 cm, the depth running back is labelled 5 cm and the vertical edge is labelled 9 cm. A straight line is drawn inside the box from the front bottom left corner to the opposite top back corner, and it carries no length label. The figure is marked NOT TO SCALE.

Solution:

  • The longest line joins two opposite corners, so first find the diagonal across the 5 cm by 12 cm base
  • Base diagonal² = 5² + 12² = 25 + 144 = 169, so the base diagonal is √169 = 13 cm
  • Now use that diagonal and the 9 cm height as the two shorter sides of a second right-angled triangle
  • Rod² = 13² + 9² = 169 + 81 = 250
  • Rod = √250 = 15.8113…
  • The longest rod is 15.8 cm to 3 significant figures
  • On the non-calculator paper, leave the length as √250 cm; to show that a rod of a given length does not fit, compare squares: a 16 cm rod does not fit, because 16² = 256 is more than 250

2. The angle between a line and a plane

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
  • 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
A right-angled triangle 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 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 that far corner of the base
  • The angle is opposite the height and adjacent to the base diagonal, so use tangent
  • tan x° = 9 ÷ 13
  • x = tan⁻¹(9 ÷ 13) = 34.6951…
  • The angle is 34.7° to 1 decimal place
  • Without a calculator you cannot find the angle itself, so a non-calculator question asks for a ratio instead: here tan x° = 9/13 exactly

Exam tip

Name the angle before you calculate it

Before working out any lengths, name the angle with three letters: the upper end of the line, then the point where the line meets the plane (the vertex, in the middle), then the foot of the perpendicular. Mark it on the diagram too. This angle is set in 7 of the 21 papers, and most of their schemes give a mark for identifying the correct angle on its own, so that mark survives a wrong length later.

Worked example

Angle between a pyramid's slant edge and the base

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 x° = 10 ÷ 5.6568…
  • x = tan⁻¹(1.7677…) = 60.5037…
  • The angle is 60.5° to 1 decimal place
  • Using half an edge, 4 cm, instead of half a diagonal would give tan⁻¹(10 ÷ 4) = 68.2°, which is wrong