Three-dimensional trigonometry

Solve 3D problems by breaking them into triangles: Pythagoras in three dimensions, the angle between a line and a plane, the angle between two planes, and problems needing a pronumeral across several triangles.

The method

In three dimensions Pythagoras' theorem extends to a box with sides $a$, $b$ and $c$: the space diagonal has length $\sqrt{a^2 + b^2 + c^2}$.

The angle between a line and a plane

Drop a perpendicular from a point on the line to the plane. The angle between the line and the plane is the angle between the line and its projection (its "shadow") on the plane.

Example 1●●

A box is $4$ cm by $3$ cm by $12$ cm high. Find the length of a space diagonal and the angle it makes with the base.

Base diagonal: $\sqrt{4^2 + 3^2} = 5$ cm.

Pythagoras on the base.

Space diagonal: $\sqrt{5^2 + 12^2} = 13$ cm.

The space diagonal rises 12 above the base diagonal, which is its projection.

$\tan\theta = \dfrac{12}{5}$, so $\theta \approx 67^\circ 23'$.

The angle between two planes

Two planes meet in a line. Take a point on that line and draw a line in each plane perpendicular to it. The angle between those two lines is the angle between the planes.

Example 2●●

A square pyramid has base side $6$ m and vertical height $4$ m. Find the angle between a triangular face and the base.

The distance from the centre to the midpoint of an edge is $3$ m.

Faces meet the base along a base edge. From the midpoint of that edge, draw lines perpendicular to it in each plane: one to the centre of the base, one up the face to the apex.

$\tan\theta = \dfrac{4}{3}$, so $\theta \approx 53^\circ 8'$.

When no triangle can be solved

Sometimes every triangle is missing information. Then call the unknown $h$, write the other lengths in terms of $h$ in separate triangles, and link them in one last triangle to form an equation.

Example 3●●●

From $A$, due south of a tower, the angle of elevation of the top is $30^\circ$. From $B$, due east of the tower, it is $45^\circ$. If $AB = 100$ m, find the height $h$ of the tower.

$AF = h\sqrt3$ and $BF = h$.

Let F be the foot of the tower. In the vertical triangles, tan 30° = h/AF and tan 45° = h/BF.

$AB^2 = AF^2 + BF^2$: $\;10\,000 = 3h^2 + h^2 = 4h^2$.

South and east are perpendicular, so triangle AFB has a right angle at F.

$h^2 = 2500$, so $h = 50$ m.

Flashcards

Space diagonal of an $a imes b imes c$ box

$\sqrt{a^2 + b^2 + c^2}$

Angle between a line and a plane

The angle between the line and its projection on the plane.

Angle between two planes

Use lines in each plane perpendicular to their common line.

No triangle can be solved

Introduce a pronumeral and link the triangles with an equation.

Practice

Find the length of a space diagonal of a $4 \times 3 \times 12$ box.
Find the angle between that space diagonal and the base, to the nearest degree.
The angle between a line and a plane is the angle between the line and:
A square pyramid has base side $6$ m and height $4$ m. Find the angle between a face and the base, to the nearest degree.
In Example 3, find the height of the tower in metres.
In Example 3, why is the angle at the foot of the tower in triangle $AFB$ a right angle?
Find the length of the diagonal of a $3$ by $4$ rectangular base.
The angle between two planes is measured between two lines that are:
A cube has edges $\dfrac{10}{\sqrt3}$ cm. Find the length of a space diagonal.
A vertical pole meets level ground at its foot $F$. Which angle must be $90^\circ$?