Maths · 7 Trigonometry

7.6 Pythagoras’ theorem and trigonometry in 3D

Before you start

Pythagoras' theorem and identifying the hypotenuse (7.1).
The three trigonometric ratios and finding an angle with an inverse function (7.2).
That the perpendicular distance from a point to a line is the shortest distance (7.2).

Syllabus objectives covered (1)

Carry out calculations and solve problems in three dimensions using Pythagoras’ theorem and trigonometry, including calculating the angle between a line and a plane.

Part 1 of 3

The one skill this whole subtopic needs

There is no new theorem here. Everything is Pythagoras and SOH CAH TOA, exactly as in 7.1 and 7.2.
The only new skill is finding the right-angled triangle hidden inside the solid, and then redrawing it flat on its own.
That redrawing step is what makes these questions manageable. A 3D diagram is hard to read; a 2D triangle with three labels is not.
So the method is always: identify the triangle, draw it separately, label everything you know, then solve it as an ordinary 2D problem.
Most 3D problems take two steps: use Pythagoras once to find a length inside the solid, then use that length in a second triangle.
Carry the unrounded value from the first step into the second. Rounding halfway is the most common way to lose the accuracy mark in a question you have otherwise done correctly.
Better still, keep the first result in surd form or as a squared value. If step one gives $d^{2} = 74$ , you can often use $74$ directly in step two without ever taking a root.

Part 2 of 3

Diagonals of a cuboid

The classic case. A cuboid has two kinds of diagonal, and telling them apart is half the battle.
The face diagonal runs corner to corner across one flat face. The space diagonal runs from one corner of the solid to the opposite corner, through the middle.
Find the base diagonal first, then use it as a side of the vertical triangle.
Step 1. Find the face diagonal of the base using the length and the width.
$d^{2} = l^{2} + w^{2}$
Step 2. That base diagonal, the vertical height, and the space diagonal form a second right-angled triangle — and the angle between the base diagonal and the vertical edge is a right angle because the height is perpendicular to the base.
Apply Pythagoras again in that triangle.
$D^{2} = d^{2} + h^{2} = l^{2} + w^{2} + h^{2}$
So the space diagonal of a cuboid is $\sqrt{l^{2} + w^{2} + h^{2}}$ — but derive it in two steps rather than quoting it, because the two-step method is what transfers to pyramids and cones where no such shortcut exists.
The same two-step idea handles pyramids. Find a length in the base, then work in a vertical triangle containing the apex.

Part 3 of 3

The angle between a line and a plane

This is named in the objective, so expect it. The definition is precise and worth learning as a procedure.
The angle between a line and a plane is the angle between the line and its projection onto that plane.
The projection is the 'shadow' the line would cast straight down onto the plane. Finding it is the whole technique.
Practically, three steps. One: drop a perpendicular from the far end of the line down to the plane.
Two: join the foot of that perpendicular back to where the line meets the plane. That joining line is the projection.
Three: the line, its projection and the perpendicular form a right-angled triangle. The angle you want is between the line and the projection — at the point where the line meets the plane.
In a cuboid, the angle between the space diagonal and the base is found this way: the projection of the space diagonal onto the base is the base diagonal.
So the triangle contains the base diagonal (adjacent), the height (opposite) and the space diagonal (hypotenuse) — and $\tan θ = h / d$ gives the angle directly.
The commonest error is measuring to the wrong line — using a vertical edge instead of the projection. Ask yourself: which line in the plane does the sloping line lie directly above?

Worked examples

Example 1. A cuboid measures $8$ cm by $6$ cm by $5$ cm. Calculate the length of its space diagonal.

Show the working

Step 1: find the diagonal of the base rectangle, using the $8$ cm and $6$ cm edges.
$d^{2} = 8^{2} + 6^{2} = 64 + 36 = 100$
That gives $d = 10$ cm, though it is often neater to keep $d^{2} = 100$ for the next step.
Step 2: the base diagonal, the $5$ cm height and the space diagonal form a right-angled triangle, since the height is perpendicular to the base.
$D^{2} = d^{2} + h^{2} = 100 + 25 = 125$
Take the square root.
$D = \sqrt{125} = 11.2 cm$
Check for plausibility: the space diagonal must be longer than any edge and longer than the base diagonal of 10 cm. At 11.2 cm it is.

Answer: $11.2$ cm (3 s.f.).

Example 2. For the same cuboid ( $8$ cm by $6$ cm by $5$ cm), calculate the angle between the space diagonal and the base.

Show the working

Identify the projection of the space diagonal onto the base. It is the base diagonal, which we found to be $10$ cm.
The angle wanted is between the space diagonal and that base diagonal, at the corner where they meet.
In the right-angled triangle, the base diagonal is adjacent to the angle and the $5$ cm height is opposite it.
Opposite and adjacent means tangent.
$\tan θ = \frac{5}{10} = 0.5$
Take the inverse tangent and give the answer to one decimal place, as the syllabus requires for angles.
$θ = \tan^{- 1} (0.5) = 26.6 °$

Answer: $26.6 °$ .

Example 3. A square-based pyramid has a base of side $10$ cm and a vertical height of $12$ cm from the centre of the base to the apex. Calculate the angle between a sloping edge and the base.

Show the working

The sloping edge runs from a base corner to the apex. Its projection onto the base is the line from that corner to the centre of the base — which is half the base diagonal.
Find the full base diagonal of the $10$ cm square first.
$d^{2} = 10^{2} + 10^{2} = 200 \Rightarrow d = 14.14 cm$
Half of it is $7.071$ cm. Keep the unrounded value on the calculator rather than writing $7.07$ and reusing that.
Now the right-angled triangle: half-diagonal adjacent, vertical height opposite, sloping edge as the hypotenuse.
$\tan θ = \frac{12}{7.071}$
Evaluate and take the inverse tangent.
$θ = \tan^{- 1} (1.697) = 59.5 °$
Note the projection is half the diagonal, because the apex sits above the centre of the base. Using the whole diagonal is the standard error here.

Answer: $59.5 °$ .

Where people go wrong

Mistake: Trying to work directly on the 3D diagram.

Instead: Redraw the relevant triangle flat, on its own, with its lengths labelled.

A perspective drawing distorts lengths and angles. A separate 2D sketch turns a hard problem into an ordinary one.

Mistake: Measuring the angle between a line and a plane to a vertical edge.

Instead: Measure it to the projection of the line on the plane.

The definition is the angle between the line and its shadow on the plane. The vertical edge is the perpendicular, not the projection.

Mistake: Using the whole base diagonal in a pyramid.

Instead: Use half of it, because the apex is above the centre of the base.

The projection of a sloping edge reaches only from the corner to the centre, which is half the diagonal.

Mistake: Rounding the intermediate length before the second step.

Instead: Carry the unrounded value, or keep it as a squared value.

3D problems are two-step by nature, so an early rounding is carried into the final answer and loses the accuracy mark.

Mistake: Assuming any two lines meeting in a solid form a right angle.

Instead: Check that one is genuinely perpendicular to the other before using Pythagoras.

Pythagoras and SOH CAH TOA both need a real right angle. In a cuboid the height is perpendicular to the base, which is why the method works there.

Check yourself

State the definition of the angle between a line and a plane.
Answer
The angle between the line and its projection onto the plane.
A cuboid measures $3$ cm by $4$ cm by $12$ cm. Find the length of its space diagonal.
Answer
Base diagonal $= \sqrt{9 + 16} = 5$ ; space diagonal $= \sqrt{25 + 144} = \sqrt{169} = 13$ cm.
For that cuboid, what is the angle between the space diagonal and the base?
Answer
$\tan θ = 12 / 5$ , so $θ = \tan^{- 1} (2.4) = 67.4 °$ .
Why is a 3D problem usually solved in two steps?
Answer
The first step finds a length inside the solid — often a face diagonal — which then becomes a side of the second right-angled triangle.
In a square-based pyramid, what is the projection of a sloping edge onto the base?
Answer
The line from that base corner to the centre of the base, which is half the base diagonal.
Why should you avoid rounding after the first step of a 3D calculation?
Answer
The rounded value is used in the second step, so the error is carried through into the final answer and costs the accuracy mark.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A rectangular box $A B C D E F G H$ has $A B = 15$ cm, $B C = 8$ cm and a height of $6$ cm. $A$ is a bottom corner and $G$ is the top corner diagonally opposite it.

(a) Calculate the length of the diagonal of the base. 2 marks

Mark scheme

M1 15² + 8² = 289 seen
A1 17 cm

(b) Calculate the length of $A G$ . 2 marks

Mark scheme

M1 their base diagonal² + 6² used; FT their (a)
A1 18.0 awrt (cm)

(c) Calculate the angle between $A G$ and the base, correct to one decimal place. 3 marks

Mark scheme

B1 identifies the base diagonal as the projection of AG on the base
M1 tan θ = 6/17 oe; FT their (a)
A1 19.4° awrt

Question 2. A pyramid has a square base $P Q R S$ of side $14$ cm. The apex $T$ is vertically above the centre of the base, and $T P = 20$ cm.

(a) Show that the distance from $P$ to the centre of the base is $7 \sqrt{2}$ cm. 3 marks No calculator

Mark scheme

M1 base diagonal² = 14² + 14² = 392 seen
M1 halves the diagonal, or uses 7² + 7² directly
A1 reaches 7√2 with correct working; nfww

(b) Calculate the vertical height of the pyramid. 3 marks

Mark scheme

M1 recognises a right-angled triangle with hypotenuse 20 and one side 7√2
M1 √(400 - 98) seen
A1 17.4 awrt (cm)

(c) Calculate the angle between the edge $T P$ and the base. 2 marks

Mark scheme

M1 cos θ = 7√2 / 20, or tan θ = their height / 7√2 oe
A1 60.4° awrt; FT their (b)

Question 3. A vertical mast $T M$ stands on horizontal ground. $A$ and $B$ are two points on the ground, with $M A = 40$ m and $M B = 55$ m, and angle $A M B = 90 °$ . The angle of elevation of the top of the mast $T$ from $A$ is $32 °$ .

(a) Calculate the height of the mast. 2 marks

Mark scheme

M1 TM = 40 tan 32° oe
A1 25.0 awrt (m)

(b) Calculate the distance $A B$ . 2 marks

Mark scheme

M1 √(40² + 55²) oe, using the right angle at M
A1 68.0 awrt (m)

(c) Calculate the angle of elevation of $T$ from $B$ , correct to one decimal place. 2 marks

Mark scheme

M1 tan θ = their TM / 55 oe
A1 24.4° awrt; FT their (a)

Words to know

Face diagonal: A diagonal drawn across one flat face of a solid.
Space diagonal: A diagonal running through the inside of a solid, from one corner to the opposite corner.
Projection: The 'shadow' a line casts on a plane, found by dropping a perpendicular from its far end.
Angle between a line and a plane: The angle between the line and its projection onto that plane.

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