Trigonometry in 3 dimensions - Higher

The trigonometric ratios can be used to solve 3-dimensional problems which involve calculating a length or an angle in a right-angled triangle.

It may be necessary to use Pythagoras' theorem and trigonometry to solve a problem.

Example

The shape ABCDEFGH is a cuboid.

Cuboid (ABCDEFGH) measuring 2cm x 3cm x 6cm

Length AB is 6 cm, length BG is 3 cm and length FG is 2 cm.

The length of the diagonal AF is 7 cm.

Calculate the angle between AF and the base ABCD. Give the answer to 3 significant figures.

ABCD is the base of the cuboid. The line FC and the base ABCD form a right angle.

Draw the right-angled triangle AFC and label the sides. The angle between AF and AC is x.

Triangle (ACF) with unknown angle, x and side, a

Use \sin{x} = \frac{o}{h}

\sin{x} = \frac{3}{7}

\sin{x} = 0.428571 \dotsc. Do not round this answer yet.

To calculate the angle use the inverse \sin button on the calculator ( \sin^{-1}).

x = 25.4^\circ

Question

The shape ABCDV is a square-based pyramid. O is the midpoint of the square base ABCD.

Pyramid (ABCDV) with height 3cm

Lengths AD, DC, BC and AB are all 4 cm.

The perpendicular height of the pyramid (OV) is 3 cm.

Calculate the angle between VC and the plane ABCD. Give the answer to 3 significant figures.

ABCD is the base of the pyramid. The line VO and the line OC form a right angle.

Draw the right-angled triangle OVC and label the sides. The angle between VC and the plane is y.

Triangle (VCO) with unknown angle y and height, 3cm

It is not possible to use trigonometry to calculate the angle y because the length of another side is required.

Pythagoras' theorem can be used to calculate the length OC.

Draw the right-angled triangle ACD and label the sides.

A right-angled triangle with points ACD. Both D to C and A to D measure 4 cm. C to A is blank.

c^2 = a^2 + b^2

AC^2 = 4^2 + 4^2

AC^2 = 16 + 16

AC^2 = 32

AC = \sqrt{32}

The length AC is \sqrt{32}~cm

The point O is in the centre of the length AC so OC is half of the length AC.

The length OC is \frac{\sqrt{32}}{2} cm.

A right-angled triangle with points VCO. VO labelled o measures 3 cm, OC labelled a measures 32 squared over 2. VC is labelled h.

Use \tan{y} = \frac{o}{a}

\tan{y} = \frac{3}{\frac{\sqrt{32}}{2}}

\tan{y} = 1.06066 \dotsc. Do not round this answer yet.

To calculate the angle use the inverse \tan button on the calculator (  \tan^{-1})

y = 46.7^\circ