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 plane ABCD. Give the answer to 3 significant figures.

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

Draw the right-angled triangle AFC and label the sides. The angle between AF and the plane 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.

The plane ABCD is the base of the pyramid. The line VO and the plane ABCD 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 can be used to calculate the length OC.

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

Right angle triangle (ACD) with sides 4cm x 4cm and one unknown

\[a^2 + b^2 = c^2\]

\[\text{CD}^2 + \text{AD}^2 = \text{AC}^2\]

\[4^2 + 4^2 = AC^2\]

\[32 = AC^2\]

\[AC = \sqrt{32}\]

\(\sqrt{32} \) is a surd. Do not round this answer yet.

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.

Triangle (VCO) with unknown angle y and length sq root 32/2

Use \(\tan{x} = \frac{o}{a}\)

\[\tan{x} = \frac{3}{\frac{\sqrt{32}}{2}}\]

\(\tan{x} = 1.06066 \dotsc\). Do not round this answer yet.

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

\[x = 46.7^\circ\]