4.8 - Three-dimensional Pythagoras and trigonometry

Find lengths and angles inside solids by bringing the right triangle into view. You will use a perpendicular projection to find a line's angle with a plane, then choose between different triangles in a pyramid.

Right triangles and space diagonals (Higher)

Can a straight rod of length $26\,\mathrm{cm}$ fit exactly between opposite corners inside a box measuring $6\,\mathrm{cm}$ by $8\,\mathrm{cm}$ by $24\,\mathrm{cm}$ ? No single face is long enough to show the whole rod. We need a triangle inside the box.

You already know

In a right-angled triangle, $c^2=a^2+b^2$ , where $c$ is the hypotenuse. You can use sine, cosine and tangent once you have identified the opposite, adjacent and hypotenuse sides.

A cuboid has three mutually perpendicular edge directions. Its perspective drawing can make a right angle look slanted. Trust the stated geometry rather than measuring the picture.

Space diagonal

A space diagonal joins opposite vertices through a solid's interior. A face diagonal joins opposite corners of one face; it stays on that face.

In cuboid $ABCDEFGH$ , the top points $E,F,G,H$ lie directly above $A,B,C,D$ . First find base diagonal $AC$ . Then use right triangle $ACG$ : the vertical $CG$ is perpendicular to the whole base, including $AC$ .

[DIAGRAM: asset_slug: m073_cuboid_projection; caption: A space diagonal and its perpendicular projection in a cuboid; description: Cuboid ABCDEFGH has E F G H directly above A B C D. AG is blue, its projection AC is sage and perpendicular height CG is clay. A separate true-shape right triangle ACG shows the right angle at C and the line-plane angle at A. The cuboid is schematic.; file: m073_cuboid_projection.png; recommended_method: matplotlib]
A space diagonal and its perpendicular projection in a cuboid

The two triangles explain why three squared edge lengths appear.

Cuboid space diagonal

For mutually perpendicular edge lengths $l,w,h$ ,

d_{\text{base}}^2=l^2+w^2,\qquad d^2=l^2+w^2+h^2

Use the positive square root for a length. Pythagoras and this derived result are absent from both tier sheets. Not in the formulae sheet: learn it.

Here is the rod from the opening question, found in two stages.

Worked example 1: a diagonal through the box

Question. A cuboid has $AB=6\,\mathrm{cm}$ , $BC=8\,\mathrm{cm}$ and $CG=24\,\mathrm{cm}$ . Find $AG$ exactly.

Plan. Find $AC$ across the base, then combine it with the vertical height.

\begin{aligned} AC^2&=6^2+8^2=100 &&\text{base right triangle}\\ AC&=10 &&\text{positive square root}\\ AG^2&=10^2+24^2=676 &&\text{right triangle }ACG\\ AG&=26 &&\text{positive square root} \end{aligned}

Why use $AC$ ? It reaches the foot of the vertical from $G$ , making a genuine right triangle with $AG$ .

Answer. $AG=26\,\mathrm{cm}$ , so the ideal thin rod fits exactly along this diagonal.

Check. $26^2=6^2+8^2+24^2$ , and the diagonal exceeds each edge length.

Now change only which length is unknown. The same relationship works backwards.

Worked example 2: a missing perpendicular edge

Question. A cuboid has base edges $6\,\mathrm{cm}$ and $8\,\mathrm{cm}$ , and space diagonal $26\,\mathrm{cm}$ . Find its height $h$ .

Plan. Subtract the two known edge squares from the diagonal square.

\begin{aligned} 26^2&=6^2+8^2+h^2 &&\text{cuboid relationship}\\ h^2&=676-36-64=576 &&\text{isolate the unknown square}\\ h&=24 &&\text{height is positive} \end{aligned}

Why subtract? The hypotenuse square already includes all three edge squares.

Answer. The height is $24\,\mathrm{cm}$ .

Check. $6^2+8^2+24^2=676$ , giving the stated diagonal $26\,\mathrm{cm}$ .

The opening numbers have a neat feature: both triangles have whole-number sides, $6,8,10$ and $10,24,26$ . Three-dimensional distances can therefore be exact whole numbers too.

Watch out

The line $AG=6+8+24=38$ measures a route along edges, not a straight diagonal. Add squares and take the square root: $AG=\sqrt{6^2+8^2+24^2}=26$ .

Try the forward calculation with different edge lengths. The two-triangle route and the three-square route should agree.

The angle between a line and a plane (Higher)

Imagine shining light straight down onto the base of the box. The diagonal rod casts a straight shadow across the base. That shadow gives the direction we need for the angle.

Perpendicular projection

To project a point onto a plane, drop a perpendicular to the plane. For a line meeting the plane, join its intersection point to the projected position of another point on the line. This segment is the line's perpendicular projection.

For $AG$ , point $A$ is already in the base and $G$ projects onto $C$ . The projection is $AC$ . The angle between the line and the plane is the smaller angle between $AG$ and $AC$ , namely $\angle GAC$ . A line lying in or parallel to a plane has angle $0^\circ$ ; a line perpendicular to it has angle $90^\circ$ .

In right triangle $ACG$ , height $CG$ is opposite this angle, projection $AC$ is adjacent, and $AG$ is the hypotenuse.

Right-triangle ratios

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Choose the ratio containing the lengths you know. Use inverse functions in degree mode for an angle. These ratios are absent from both tier sheets. Not in the formulae sheet: learn it.

Use the original box to see how steeply the rod rises above its shadow.

Worked example 3: inclination to the base

Question. In a cuboid, $AB=6\,\mathrm{cm}$ , $BC=8\,\mathrm{cm}$ and $CG=24\,\mathrm{cm}$ . Find the angle between $AG$ and base plane $ABCD$ , to $1$ decimal place.

Plan. Find projection $AC$ , then use height divided by projection in tangent.

\begin{aligned} AC&=\sqrt{6^2+8^2}=10 &&\text{base diagonal}\\ \tan\theta&=\frac{24}{10}=2.4 &&\text{opposite over adjacent}\\ \theta&=\tan^{-1}(2.4) &&\text{inverse tangent}\\ &\approx67.4^\circ &&\text{round at the end} \end{aligned}

Why choose $\angle GAC$ ? Its base arm $AC$ is the perpendicular projection of $AG$ , rather than an arbitrary base edge.

Answer. The angle is $67.4^\circ$ .

Check. $AG=26\,\mathrm{cm}$ , so $\sin^{-1}(24/26)$ gives the same unrounded angle.

Change only the box's height from $24\,\mathrm{cm}$ to $12\,\mathrm{cm}$ . The projection stays the same, so the rod should be less steep.

Worked example 4: a lower box

Question. The same cuboid now has height $12\,\mathrm{cm}$ . Find the angle between its space diagonal and its base, to $1$ decimal place.

Plan. Keep the base projection $10\,\mathrm{cm}$ and change the opposite side.

\begin{aligned} AC&=\sqrt{6^2+8^2}=10 &&\text{unchanged base}\\ \tan\phi&=\frac{12}{10}=1.2 &&\text{new height}\\ \phi&=\tan^{-1}(1.2)\approx50.2^\circ &&\text{degree mode} \end{aligned}

Why keep $AC$ ? Only the perpendicular height changed; the diagonal's projection did not.

Answer. The angle is $50.2^\circ$ .

Check. This is less than $67.4^\circ$ , as a lower box requires. Using $\cos^{-1}(10/\sqrt{244})$ confirms it.

You can also use an inclination to find a missing length. Keep the same projection triangle, but rearrange the ratio instead of applying an inverse function.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Shape, space and measure

Handling data

4.8 - Three-dimensional Pythagoras and trigonometry

Right triangles and space diagonals (Higher)

The angle between a line and a plane (Higher)

Continue this lesson.

Learn faster with Nova