3D trigonometry — applying trig in three dimensions — Mathematics

Before this

This leads to

What is a vector — magnitude and direction
Volume of 3D shapes — spheres, cones, pyramids

A right angle you cannot see

Stand in one corner of a room. The floor is 5 m by 4 m, and the ceiling is 3 m high. Stretch a string from the corner of the floor where you stand to the opposite corner of the ceiling.

Two questions. How long is the string? And what angle does it make with the floor?

You cannot measure that angle with a protractor. The string rises through the middle of the room, and there is no edge under it to lay the protractor against. Every triangle you have solved so far lay flat on the page. This one stands up in the air.

Here is where the difficulty actually lives. It is not in the calculation. You already own every tool this lecture needs: Pythagoras, the three ratios, the sine rule, the cosine rule and ½ab sin C. The difficulty is in seeing. A drawing squashes three dimensions onto flat paper, and the squashing hides the right angles. This lecture teaches you to find them again.

A drawing of a box does not show true angles

Books draw a cuboid like the one below. The base is ABCD. The top is EFGH, with E directly above A, F above B, G above C and H above D. Most books letter a cuboid this way.

Top: the cuboid as books draw it. The shaded triangle ACG stands upright inside the box, but on paper its right angle at C is drawn at about 80°. Bottom: the same triangle laid flat, as it really is. Now it is an ordinary right-angled triangle.

Look at the shaded triangle ACG. AC is the diagonal of the base. CG is the vertical edge at C. AG joins a bottom corner to the opposite top corner. It is the longest line in the box, and it is called a space diagonal.

Is the angle at C a right angle? In the drawing it is about 80°. In the box it is exactly 90°. Here is the reason. CG is vertical. A vertical line makes 90° with every horizontal line through its foot, not only with the edges CB and CD. AC is a horizontal line on the base, and it passes through C. So the angle ACG is 90°.

A picture to hold in your mind

A flagpole stands at 90° to every straight path that starts at its foot. Whenever one point sits directly above another, look for this hidden right angle at the bottom.

The lower part of the figure shows triangle ACG laid flat, as it really is. Flat, it is an ordinary right-angled triangle, and you know how to solve it.

The method: pull out a flat triangle

1. Mark what you want: a length or an angle.
2. Find a triangle that contains it. Prefer a right-angled one. Look for a vertical line meeting a horizontal one.
3. Find the sides of that triangle. If one is missing, find it first from another flat triangle.
4. Draw the triangle again, flat, with its true right angle. Then solve it with the tools you already have.

Step 4 matters more than it looks. Redrawing the triangle removes the false angles of the picture. Almost every mistake in this topic starts with trusting the 3D drawing.

The longest line in a box

Take any cuboid with edges $a$ , $b$ and $c$ : AB = $a$ , BC = $b$ and the height CG = $c$ . The space diagonal AG is the hypotenuse of triangle ACG, but you do not know AC yet. So find it first, from the base.

On the base, triangle ABC has its right angle at B, because ABCD is a rectangle. So Pythagoras gives $A C^{2} = a^{2} + b^{2}$ .

Now stand the second triangle up. Triangle ACG has its right angle at C. Pythagoras again:

A G^{2} = A C^{2} + C G^{2} = a^{2} + b^{2} + c^{2}

Pythagoras in three dimensions

A cuboid with edges $a$ , $b$ and $c$ has a space diagonal $d$ with

$d^{2} = a^{2} + b^{2} + c^{2}$

It is Pythagoras used twice: once on the floor, once standing up.

For the room, $a = 5$ , $b = 4$ and $c = 3$ . So $d^{2} = 25 + 16 + 9 = 50$ , and the string is $d = 50 = 52 \approx 7.07$ m long.

Check. The string is longer than the floor diagonal, $41 \approx 6.40$ m, as the hypotenuse of triangle ACG must be. ✓

The angle between a line and a plane

Now the second question. What does "the angle the string makes with the floor" mean? A flat surface such as the floor is called a plane. The string meets it at one point, the corner where you stand. But the floor holds lines in every direction through that point. Which of them do we measure against?

Imagine a light shining straight down from the ceiling. The string casts a shadow on the floor. The shadow starts at your corner and ends at the point directly below the top of the string. Mathematicians call the shadow the projection of the line onto the plane.

The angle between a line and a plane

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

To find the projection, drop a vertical from the top of the line to the plane. Then join the foot of that vertical to the point where the line meets the plane.

In the cuboid, G is directly above C. So the projection of AG on the base is AC, and the angle that AG makes with the base is the angle θ at A in triangle ACG. That is the angle marked in the figure.

Why the shadow, and not some other line on the floor? Compare three lines on the base of the 8 cm by 5 cm by 4 cm cuboid, all starting at A. Each makes a right-angled triangle with AG, so each angle is quick to find.

Line on the base	Right angle at	Angle with AG
AB, an edge (8 cm)	B	$cos^{- 1} (8/ 105) \approx 38. 7^{\circ}$
AD, an edge (5 cm)	D	$cos^{- 1} (5/ 105) \approx 60. 8^{\circ}$
AC, the projection ( $89$ cm)	C	$tan^{- 1} (4/ 89) \approx 23. 0^{\circ}$

The projection gives the smallest angle. Every other line on the floor points further away from AG. So the projection answers the honest question: how steeply does the line rise out of the plane?

Back to the room. The shadow of the string is the floor diagonal, $41$ m long, and the height is 3 m. The right angle is at the foot of the vertical, so

tan θ = \frac{3}{41} \approx 0.4685

and $θ \approx 25. 1^{\circ}$ . The string rises at about 25° from the floor.

Now see how far a drawing can mislead you. The box below turns. The triangle inside it is always the same triangle.

Explore — turn the box and watch the drawn angles

Turn the 8 cm by 5 cm by 4 cm cuboid. Watch the angle at C and the angle θ as they are drawn on the screen. Compare them with the true triangle at the bottom, which never changes.

Turn: 310°

This table shows how the same triangle ACG is drawn when the cuboid is seen from different directions. The true angles never change: 90° at C and about 23.0° at A.

Turn	Angle at C as drawn	θ as drawn
310° (book view)	about 81°	about 22°
25°	about 128°	about 18°
45°	about 155°	about 10°
200°	about 57°	about 30°
250°	about 23°	about 37°

The drawn right angle at C ranges from about 23° to about 155°. Only the flat triangle, drawn again with its true right angle, shows the real angles.

You should have seen two things. First, the drawn angle at C swings from very sharp to very wide as the box turns, yet it is always 90° in the box. Second, θ on the screen changes too. A drawing of a 3D shape is a guide to which points join which. It is never a guide to the size of an angle.

Stop and think

A student says: "The angle between AG and the base is angle GAB, because AB is an edge of the base and the base is where the angle is measured." Is the student right?

Show the answer

No. AB is a line on the base, but it is not the shadow of AG. G is directly above C, not above B, so the projection of AG is AC. Angle GAB is about 38.7°, while the angle between AG and the base is about 23.0°. The edge is chosen only because it is drawn, and that is the trap. Always find the point directly below the top of the line first.

Worked examples

Worked Example 1 — The diagonal of a cuboid

The cuboid ABCDEFGH has AB = 8 cm, BC = 5 cm and CG = 4 cm. Find (a) the length of the space diagonal AG, (b) the angle AG makes with the base ABCD, and (c) the angle the face diagonal AF makes with the edge AB.

Think first. AG sits in the upright triangle ACG, with its right angle at C. AC is missing, so the base triangle ABC comes first.

(a) On the base, $A C^{2} = 8^{2} + 5^{2} = 89$ , so $A C = 89$ . Leave it exact. Then $A G^{2} = 89 + 4^{2} = 105$ , so $A G = 105 \approx 10.2$ cm.

(b) The projection of AG is AC, so θ is the angle at A in triangle ACG. The height CG is opposite θ, and AC is adjacent:

tan θ = \frac{4}{89} \approx 0.4240

so $θ \approx 23. 0^{\circ}$ .

(c) AF lies in the front face ABFE, a flat rectangle. Triangle ABF has its right angle at B, with AB = 8 and BF = 4. So $tan ∠ F A B = \frac{4}{8} = 0.5$ , and $∠ F A B \approx 26. 6^{\circ}$ . This part is a flat problem hiding in a box. Once you see which face it lives in, it is ordinary trigonometry.

Check. For (b), use the sine instead: $sin θ = 4/ 105 \approx 0.3904$ gives 23.0° again. ✓

In part (b), the dangerous moment is naming the sides of triangle ACG. In a 3D drawing, the hypotenuse is easy to miss. Find the mistake in this student's working.

Spot the error — the angle between a space diagonal and the base

A cuboid ABCDEFGH has AB = 6 cm, BC = 3 cm and CG = 2 cm. A student finds the angle θ between AG and the base. Tap the line where the mistake is.

This box shows a student's working for the angle between the space diagonal AG and the base of a 6 cm by 3 cm by 2 cm cuboid. One line contains a mistake.

1. AC² = 6² + 3² = 45, so AC = √45 ≈ 6.708 cm.
2. AG² = 45 + 2² = 49, so AG = 7 cm.
3. G is directly above C, so the projection of AG on the base is AC.
4. In triangle ACG, CG is opposite θ and AG is adjacent, so tan θ = 2 ÷ 7. (This is the mistake.)
5. θ = tan⁻¹(2 ÷ 7) ≈ 15.9°.
6. Answer: AG makes an angle of about 15.9° with the base.

The right angle of triangle ACG is at C, so AG, which faces it, is the hypotenuse. The adjacent side is AC. Correct: tan θ = 2 ÷ √45, or sin θ = 2 ÷ 7, and both give θ ≈ 16.6°.

The two answers differ by less than 1°, so a rough check would not catch this. Step 4 catches it: redraw the triangle flat and name the hypotenuse first.

Worked Example 2 — A square-based pyramid

A pyramid VABCD has a square base of side 10 cm with centre O. The top point V, called the apex, is 12 cm directly above O. M is the midpoint of AB. Find (a) the length VM, (b) the angle the edge VA makes with the base, and (c) the angle the face VAB makes with the base.

Two upright triangles inside the pyramid, both with the right angle at O. Triangle VOM (idea colour) goes up the middle of a face: OM = 5 cm. Triangle VOA (grey-blue) goes up an edge: OA is half the diagonal of the base, 5√2 ≈ 7.07 cm. The height VO = 12 cm is in both.

Think first. V is directly above O, so every line from O along the base makes 90° with VO. That gives two upright triangles, VOM and VOA.

(a) OM goes from the centre to the middle of a side, so OM = 5 cm. In triangle VOM, $V M^{2} = 1 2^{2} + 5^{2} = 169$ , so VM = 13 cm exactly.

(b) The projection of VA is OA, half the diagonal of the base. The diagonal is $102$ , so $O A = 52 \approx 7.07$ cm. In triangle VOA, the right angle is at O:

tan ∠ V A O = \frac{12}{5 2} \approx 1.697

so the edge makes about 59.5° with the base.

(c) A sloping face is not a single line, so which line on it do we use? Use VM, which runs straight up the middle of the face, at 90° to the base edge AB. It is the steepest line on the face: a ball placed on the face would roll down along it. Its projection is OM, so $tan ∠ V MO = \frac{12}{5} = 2.4$ , and the face makes about 67.4° with the base.

Check. The face (67.4°) is steeper than the edge (59.5°). That is right. VM and VA rise the same 12 cm, but VM covers only 5 cm across the base, while VA covers 7.07 cm. ✓

Worked Example 3 — A triangle with no right angle

Use the cuboid of Worked Example 1 (8 cm by 5 cm by 4 cm). Join A to F and A to H. These are two face diagonals. Then join F to H across the top. Find the angle FAH and the area of triangle AFH.

Think first. Triangle AFH cuts diagonally through the box. None of its angles is a right angle, so the right-angle ratios cannot solve it. But each side is a face diagonal, and each face is a flat rectangle. Find the three sides, then use the cosine rule.

From the faces: $A F^{2} = 8^{2} + 4^{2} = 80$ (front face), $A H^{2} = 5^{2} + 4^{2} = 41$ (left face) and $F H^{2} = 8^{2} + 5^{2} = 89$ (top face).

FH faces the angle at A, so it takes the minus sign. The top of the fraction is $80 + 41 - 89 = 32$ . The bottom is $28041 = 23280 \approx 114.5$ .

cos ∠ F A H = \frac{32}{2 3280} \approx 0.2794

so $∠ F A H \approx 73. 8^{\circ}$ . For the area, use the two sides that meet at A. Here $\frac{1}{2} 8041 \approx 28.64$ .

Area \approx 28.64 sin 73. 8^{\circ} \approx 27.5

The area is about 27.5 cm².

Check. The same rule gives the angle at F: $cos ∠ A F H = \frac{80 + 89 - 41}{2 80 89} \approx 0.7585$ , so about 40.7°. The third angle is then about 65.6°. The largest angle, 73.8°, faces the longest side, $89$ . ✓

Three traps

Trap 1 — Measuring against an edge instead of the projection

The angle between a line and a plane is measured against the line's shadow on the plane, not against whichever edge is drawn nearby. Find the point directly below the top of the line. Join it to the point where the line meets the plane. That is the projection.

Trap 2 — Trusting the drawing

On paper, a right angle in an upright triangle can be drawn at any size. Decide where the right angle is by reasoning ("CG is vertical, AC is horizontal"), never by looking. Then name the hypotenuse, the side facing the right angle.

Trap 3 — Half the side or half the diagonal?

In a square-based pyramid, the distance from the centre O to the middle of a side is half the side. The distance from O to a corner is half the diagonal. The first belongs to a face (VM); the second belongs to an edge (VA). Mixing them up gives the wrong angle with no warning.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A box measures 12 cm by 4 cm by 3 cm. (a) Will a pencil 12.5 cm long fit inside it, lying straight? (b) Find the angle a space diagonal makes with the base.

Hint 1
The longest straight line in a box is a space diagonal. Use Pythagoras in three dimensions.

Hint 2
For (b), the projection of the space diagonal is the base diagonal, $1 2^{2} + 4^{2}$ .

Full answer
(a) $d^{2} = 144 + 16 + 9 = 169$ , so $d = 13$ cm exactly. The pencil fits, along a space diagonal, with 0.5 cm to spare. (b) The base diagonal is $160 \approx 12.65$ cm, and the height is 3 cm. So $tan θ = 3/ 160 \approx 0.2372$ , and $θ \approx 13. 3^{\circ}$ . Check: $sin θ = 3/13 \approx 0.2308$ gives 13.3° too. ✓
A real problem, nearly 4 000 years old. Problem 56 of the Rhind Mathematical Papyrus, an Egyptian text, describes a pyramid with a square base of side 360 cubits and a height of 250 cubits. (a) Find the angle each face makes with the base. (b) Find the angle each edge makes with the base.

Hint 1
Use the two upright triangles of Worked Example 2: one up the middle of a face, one up an edge.

Hint 2
For the face, the run across the base is half the side, 180 cubits. For the edge, it is half the diagonal, $1802$ .

Full answer
(a) $tan θ = 250/180 \approx 1.389$ , so the face makes about 54.2° with the base. (b) Half the diagonal is $1802 \approx 254.6$ cubits, so $tan ϕ = 250/254.6 \approx 0.9821$ , and the edge makes about 44.5°. Check: the edge is less steep than the face (Trap 3), because it covers more ground for the same height. ✓ The scribe described the slope as a run of $\frac{180}{250}$ of a cubit for each cubit of height. With 7 palms in a cubit, that is $7 \times \frac{180}{250} = 5.04 = 5 \frac{1}{25}$ palms, exactly the answer the papyrus gives.
A cube has edges of 6 cm. From one corner A, two face diagonals run across two different faces, to corners P and Q. Find the angle PAQ and the area of triangle APQ.

Hint 1
Find all three sides of triangle APQ. Each one is a diagonal of a 6 cm square face.

Hint 2
All three sides are $72$ . What kind of triangle is that?

Full answer
AP, AQ and PQ are each a diagonal of a square face, so each is $6^{2} + 6^{2} = 72$ cm. By the cosine rule, $cos ∠ P A Q = \frac{72 + 72 - 72}{2 \times 72} = \frac{1}{2}$ , so the angle is exactly 60°. The triangle is equilateral. Area $= \frac{1}{2} \times 72 \times sin 6 0^{\circ} = 183 \approx 31.2$ cm². Check: a 3D drawing usually shows this angle as nearer 45° or 90°, but reasoning says 60°, and three equal sides confirm it. ✓
A stretch. A vertical tower TP stands on level ground, with P at its foot. From a point A due south of P, the angle of elevation of T is 32°. From B, 50 m due east of A, the angle of elevation of T is 24°. Find the height of the tower.

Hint 1
There are three right angles: at P in triangle TPA, at P in triangle TPB, and at A on the ground, because east is at 90° to north. Call the height $h$ .

Hint 2
Write $P A = h \div tan 3 2^{\circ}$ and $PB = h \div tan 2 4^{\circ}$ . Then use Pythagoras on the ground: $P B^{2} = P A^{2} + 5 0^{2}$ .

Full answer
From the two upright triangles, $P A = h \times \frac{1}{t a n 3 2 ^{\circ}} \approx 1.6003 h$ and $PB = h \times \frac{1}{t a n 2 4 ^{\circ}} \approx 2.2460 h$ . On the ground, triangle PAB has its right angle at A, so $P B^{2} = P A^{2} + 5 0^{2}$ . This gives $5.0447 h^{2} = 2.5611 h^{2} + 2500$ , so $2.4836 h^{2} = 2500$ , $h^{2} \approx 1006.6$ and $h \approx 31.7$ m. Check: then PA ≈ 50.8 m and PB ≈ 71.3 m, and $tan^{- 1} (31.7 \div 50.8) \approx 3 2^{\circ}$ and $tan^{- 1} (31.7 \div 71.3) \approx 2 4^{\circ}$ . ✓ Three triangles, two upright and one on the ground, solved one at a time.

The bigger idea

People have measured slopes in three dimensions for a very long time. The Rhind Mathematical Papyrus was copied by a scribe called Ahmes about 1550 BC. It measures the slope of a pyramid face by its seked: how many palms you move across for each cubit you rise. That is the run divided by the rise, taken from exactly the triangle VOM of Worked Example 2. Modern surveys put the Great Pyramid's seked at 5½ palms, which means $tan θ = 7 \div 5.5$ and a face angle of about 51.8°.

The method has not changed in 3 500 years: find the flat triangle, then measure it. Next, the vectors lectures turn a direction in space into an arrow you can add and stretch. At Pre-University, vectors give the angle between a line and a plane directly. The flat triangles you pulled out today are what those formulas describe.

Key concepts in this lesson

A 3D problem is solved by finding flat triangles inside the solid and solving them one at a time.
A vertical line makes 90° with every horizontal line through its foot, so upright triangles have hidden right angles that a drawing does not show.
The space diagonal $d$ of a cuboid with edges $a$ , $b$ and $c$ satisfies $d^{2} = a^{2} + b^{2} + c^{2}$ , which is Pythagoras used twice.
The angle between a line and a plane is the angle between the line and its projection, the shadow cast by a light straight above.
The slope of a face is measured along the line straight up the face, at 90° to its base edge, and it is steeper than the edges.
When a triangle inside a solid has no right angle, the sine rule, the cosine rule and ½ab sin C solve it as they do on paper.