Triangles inside a solid

Lengths and angles in 3D

In 4 of 20 recent exams
Paper 1 or 2
2 to 4 marks

How it is set in the exam

Mostly Paper 2 (calculator) (9 of 11 parts), 2–4 marks a part; on Paper 1 (no calculator) as a show-that with the lengths in terms of an angle. Typical (4 of 20 exam versions). A 2-mark distance or diagonal usually comes first, then the angle.

Spot it in the exam

Exam phrases to spot (5)

“find the angle between the edge [VA] and the base ABCD”
“find the angle between the line and the plane”
“find the size of the angle at the apex of the face”
“find the length of the diagonal [AC]”
“the solid fits exactly inside the sphere; find its height in terms of θ”

In disguise (3)

only one slant edge is given by coordinates: in a right pyramid on a square, a rectangle or a regular polygon (the only bases set) the other slant edges have the same length, by symmetry
an inscribed solid with a letter for an angle: the cut through the axis of symmetry holds a right triangle whose hypotenuse is the sphere's radius (from the centre) or its diameter (the diagonal of the rectangle through a cylinder's axis; markschemes give the method mark for the two diameters)
two lines that cross inside a solid, such as two space diagonals of a cuboid: the triangle they make with an edge

Variants (7)

the angle between a slant edge and the base: height and half the diagonal, then tan (Example 2.1)
the diagonal of the base, as a 2-mark step before the angle (Example 2.1)
the apex angle of a face, with the slant edge from coordinates: half the face, or the cosine rule (Example 2.2)
the height from a slant edge and half the diagonal (Example 2.2)
a cone or a cylinder inside a sphere or a hemisphere: lengths in terms of an angle (Example 2.3, Your turn 2)
the apex angle given and one slant edge from coordinates: the other edge by symmetry, the base edge, then a line–plane angle (Your turn 3)
the angle between two crossing diagonals (Your turn 4)

Looks like it, but isn't (4)

the triangle's sides are all known already (stated in the question, or found in an earlier part, such as a diameter already worked out), so none has to come from coordinates, symmetry or a half-diagonal in this part, and there is no 3D step, even in a 3D figure: Find a side or an angle: choose the rule
an observer outside the solid looks up at its top: Angles of elevation and depression
the length is a step to a volume or surface area asked next (a cone's slant height before its curved surface, a height from coordinates before a volume), and the solid is not inside a sphere or a hemisphere: Volume and surface area
the area of one face is asked, by $\frac{1}{2}ab\sin C$ (the area of all the faces of a solid is a surface area: Volume and surface area): Area of a triangle: ½ab sin C, forwards and backwards

Worked example

A right pyramid from coordinates: the angle between an edge and the base

Paper 2 (calculator)

Question

The right pyramid $VABCD$ has a rectangular base $ABCD$ with $AB = 9$ cm and $BC = 6$ cm. $X$ is the centre of the base, and the apex $V$ is directly above $X$ . Using a coordinate system with units in centimetres, $X$ has coordinates $(1, -2, 3)$ and $V$ has coordinates $(3, 4, 6)$ .
(a) Find $VX$ . [2]
(b) Find the length of the diagonal $[AC]$ . [2]
(c) Find, in radians, the angle between the edge $[VA]$ and the base $ABCD$ . [3]

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Which method?

right-angled (a height and its projection; the cut through a cone or sphere)
What you know
the two sides next to the right angle
Use
$\tan$ , or Pythagoras for the hypotenuse
right-angled
What you know
the hypotenuse and one other side
Use
$\sin$ or $\cos$ , or Pythagoras for the third side
a face with two equal edges
What you know
the edges and the base
Use
split it at the midpoint of the base into two right triangles: $\sin\frac{\theta}{2} = \frac{\text{half the base}}{\text{edge}}$ . The cosine rule also scores
any triangle
What you know
three sides
Use
the cosine rule (Find a side or an angle: choose the rule)
any triangle
What you know
two sides and the angle between them
Use
the cosine rule for the third side; for equal edges, the half-triangle is quicker

The flat triangle	What you know	Use
right-angled (a height and its projection; the cut through a cone or sphere)	the two sides next to the right angle	$\tan$ , or Pythagoras for the hypotenuse
right-angled	the hypotenuse and one other side	$\sin$ or $\cos$ , or Pythagoras for the third side
a face with two equal edges	the edges and the base	split it at the midpoint of the base into two right triangles: $\sin\frac{\theta}{2} = \frac{\text{half the base}}{\text{edge}}$ . The cosine rule also scores
any triangle	three sides	the cosine rule (Find a side or an angle: choose the rule)
any triangle	two sides and the angle between them	the cosine rule for the third side; for equal edges, the half-triangle is quicker

Steps

Find the flat triangle that contains the unknown:
a face; a cut through the apex or the axis; for the angle between a line and a plane, the right triangle made by the line, the perpendicular from its top to the plane, and the projection.
Draw it flat and label it.
Mark the right angle, which is at the foot of the perpendicular (the centre of the base when the perpendicular is the height of a right pyramid or cone; the midpoint of the base edge when a face with two equal edges is split).
Get its sides from the 3D data:
the distance formula between given points; half the base diagonal (Pythagoras on the base); edges that are equal by symmetry; for a solid inside a sphere, the sphere's radius as the hypotenuse (or its diameter, across the rectangle through a cylinder's axis).
Solve:
right-angled trig first (the table above); the cosine rule when no right angle is available.
Answer
in the question's unit (degrees if it shows °; otherwise radians, with the unit written), or exactly in terms of the letters for a show-that. GDC: angle from a ratio.

Common mistakes

The angle at the wrong vertex.
The angle between an edge and the base is where the edge meets the base (at $A$ in triangle $VXA$ ), not at the apex.
Half a side instead of half a diagonal.
For a slant edge, the projection runs to a corner: half the diagonal. Half a side (half the other side, on a rectangle) is for the line from the apex to the middle of a base edge.
Reading angles off the 3D picture.
A right angle in a 3D sketch rarely looks like one: draw the triangle flat, and mark the right angle from the geometry (at the foot of the perpendicular).

More mistakes (6)

Treating a slant edge as the height.
The height is at right angles to the base; a slant edge is the hypotenuse.
The cosine rule with the sides in the wrong places
the side opposite the angle is the one on its own ( $10$ in Example 2.2).
Equal edges used but not stated
before the cosine rule, write $VB = VC$ (a right pyramid) and $BC = 10$ ; the mark for recognising them needs the labelled sides or the lengths in context. For the half-face, write the half base, $BN = 5$ .
An inscribed solid's height taken as $OM$
(the distance from the centre $O$ of the sphere to the centre $M$ of a circle on it). Read the height off the cut through the axis: a cone with its apex $V$ on the sphere and $O$ between $V$ and $M$ has height $VO + OM = r + OM$ (Example 2.3), or $r - OM$ if $M$ is between $V$ and $O$ ; a cylinder with its top and bottom circles on the sphere has height $2 \times OM$ ; a cylinder standing on the flat face of a hemisphere has height $OM$ itself, the distance from $O$ to the centre of its top circle (Your turn 2).
The calculator in the wrong mode
when a given angle is in degrees ( $50°$ in Your turn 3), $22\sin 25$ in radian mode gives $-2.91$ , a negative length. Set the mode to match the question before you start (check the settings): the method marks survive, but the answer mark is lost.
Leaving out the unit
of an angle: " $0.913$ " or " $52.3$ " alone could be read in the wrong unit, so write the unit and the marker reads it the way you meant. When the question shows degrees, answer in degrees.

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