Ext Maths · 6 Trigonometry

6.6 Pythagoras’ theorem and trigonometry in 3D

Before you start

Pythagoras' theorem (6.1).
SOH CAH TOA, and finding an angle with an inverse key to 1 d.p. (6.2).
Angles of elevation (6.2).
Naming an angle with three letters: angle $G A C$ is the angle at $A$ between $A G$ and $A C$ .

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 6

The only new skill: finding a flat triangle inside the solid

There are no new formulas in 3D. Every calculation is Pythagoras or SOH CAH TOA from 6.1 and 6.2, done in a right-angled triangle.
What is new is seeing that triangle inside a solid, where right angles don't look like right angles because the drawing is in perspective.
The key fact: a vertical line is perpendicular to every horizontal line that passes through its foot, not just to the edges it touches. So a vertical edge makes a right angle with a diagonal across the base.
The method every time: name the triangle with three letters, redraw it flat on its own, mark the right angle, label the lengths you know, then solve it as a 2D problem.
Most questions take two steps: find a length in one triangle, then use it in a second. Keep that first length unrounded. Even better, keep its square (such as $A C^{2} = 169$ ), which is exact.

Part 2 of 6

Naming the corners of a cuboid

Questions name a cuboid with eight letters. In this lesson, the base is $A B C D$ , and $E$ , $F$ , $G$ and $H$ are directly above $A$ , $B$ , $C$ and $D$ . On a paper, read the question's own diagram or description.
Triangle ACG stands upright inside the box, with its right angle at C.
Edges such as $A E$ , $B F$ , $C G$ and $D H$ are vertical. Lines drawn anywhere in $A B C D$ are horizontal.
Before calculating, say out loud which three letters make your triangle and where its right angle is. If you can't name the right angle, you have the wrong triangle.

Part 3 of 6

Lengths inside a cuboid: Pythagoras twice

**The base diagonal $A C$ .** Triangle $A B C$ lies flat in the base, with its right angle at $B$ .
$A C^{2} = A B^{2} + B C^{2}$
**The space diagonal $A G$ **, from a bottom corner to the opposite top corner. Triangle $A C G$ stands upright. The right angle is at $C$ , because $C G$ is vertical and $A C$ is horizontal through its foot.
$A G^{2} = A C^{2} + C G^{2}$
Put the two together and you get $A G^{2} = A B^{2} + B C^{2} + C G^{2}$ . That is a useful check, but show both steps, because the same two-step idea is what works in pyramids and other solids.
Substitute $A C^{2}$ directly into the second step rather than rounding $A C$ first. This keeps the answer exact until the last square root.

Part 4 of 6

The angle between a line and a plane

The objective names this, so learn the definition exactly. The angle between a line and a plane is the angle between the line and its projection onto the plane.
The projection is the shadow the line would cast on the plane if a light shone perpendicular to the plane.
Find it in three steps. One: the line meets the plane at a point; that is where the angle will be.
Two: from the other end of the line, drop a perpendicular onto the plane. Note where it lands, the foot.
Three: join the foot to the meeting point. That join is the projection. The line, the projection and the perpendicular form a right-angled triangle, with the right angle at the foot.
**Example: $A G$ and the base $A B C D$ .** $A G$ meets the base at $A$ . $G$ is directly above $C$ , so the perpendicular from $G$ is $G C$ , and its foot is $C$ . The projection is $A C$ .
So the angle is **angle $G A C$ **, in triangle $A C G$ : $C G$ is opposite, $A C$ is adjacent and $A G$ is the hypotenuse.
$\tan G \hat{A} C = \frac{C G}{A C}$
The common mistake is using angle $G A B$ , measured to an edge of the base. $A B$ is not the shadow of $A G$ , because $G$ is not above $A B$ .

Part 5 of 6

When the plane is a side face, not the floor

The same three steps work for any plane. Try the angle between $A G$ and the right-hand face $B C G F$ .
AB sticks straight out of the face BCGF, so the shadow of AG on that face is BG.
One: $A G$ meets the face $B C G F$ at $G$ .
Two: from $A$ , the perpendicular to that face is $A B$ . It is at right angles to $B C$ and to $B F$ , two different lines in the face, so it is at right angles to the whole face. The foot is $B$ .
Three: the projection is $B G$ , and the angle is **angle $A G B$ **, with the right angle at $B$ .
The lesson from this: don't look for 'the bottom'. Look for the line that sticks straight out of the plane you were given.

Part 6 of 6

Pyramids and cones

In a right pyramid the apex $V$ is vertically above the centre $M$ of the base. So $V M$ is the height, and it is perpendicular to every line in the base through $M$ .
The shadow of a sloping edge reaches only from the corner to the centre: half the base diagonal.
The projection of a sloping edge such as $V A$ onto the base is $A M$ , from the corner to the centre. That is half of the base diagonal, not the whole of it.
So the usual route is: base diagonal by Pythagoras, halve it, then work in triangle $V A M$ with the right angle at $M$ .
Halving works neatly with squares: if $A C^{2} = 128$ , then $A M^{2} = {(\frac{A C}{2})}^{2} = \frac{128}{4} = 32$ .
A cone works the same way. The apex is above the centre of the circular base, so the radius, the vertical height and the slant height make a right-angled triangle, with the right angle at the centre.
Any ratio works once you know the sides of the flat triangle. Where you can, use lengths that were given rather than ones you calculated, so a slip earlier can't spread.

Worked examples

Example 1. $A B C D E F G H$ is a cuboid with base $A B C D$ , and $E$ , $F$ , $G$ , $H$ are vertically above $A$ , $B$ , $C$ , $D$ . $A B = 12$ cm, $B C = 5$ cm and $C G = 7$ cm. Calculate the length of $A G$ .

Show the working

Step 1, triangle $A B C$ in the base, right angle at $B$ .
$A C^{2} = 12^{2} + 5^{2} = 144 + 25 = 169$
So $A C = 13$ cm. Keep $A C^{2} = 169$ for the next step.
Step 2, triangle $A C G$ , right angle at $C$ because $C G$ is vertical and $A C$ is horizontal.
$A G^{2} = A C^{2} + C G^{2} = 169 + 49 = 218$
Square-root and round.
$A G = \sqrt{218} = 14.764 \dots cm$
Check: the space diagonal must be longer than the base diagonal ( $13$ cm) and every edge. $14.8$ cm is.

Answer: $A G = 14.8$ cm (3 s.f.).

Example 2. For the same cuboid, calculate the angle between $A G$ and the base $A B C D$ .

Show the working

$A G$ meets the base at $A$ . $G$ is directly above $C$ , so the projection of $A G$ onto the base is $A C$ . The angle is angle $G A C$ .
Redraw triangle $A C G$ flat: $A C = 13$ cm along the bottom, $C G = 7$ cm straight up from $C$ , right angle at $C$ , hypotenuse $A G$ .
From angle $G A C$ , $C G$ is opposite and $A C$ is adjacent. Both were exact, so use tan.
$\tan G \hat{A} C = \frac{7}{13}$
Inverse tan, then 1 d.p.
$G \hat{A} C = \tan^{- 1} (\frac{7}{13}) = 28.300 \dots °$

Answer: $28.3 °$ .

Example 3. For the same cuboid, calculate the angle between $A G$ and the face $B C G F$ .

Show the working

$A G$ meets the face $B C G F$ at $G$ . The perpendicular from $A$ to that face is $A B$ , since $A B$ is at right angles to both $B C$ and $B F$ . Its foot is $B$ .
So the projection of $A G$ onto the face is $B G$ , and the angle is angle $A G B$ , in triangle $A B G$ with the right angle at $B$ .
Find $B G$ in the face $B C G F$ , a rectangle $5$ cm by $7$ cm, right angle at $C$ .
$B G^{2} = 5^{2} + 7^{2} = 74$
From angle $A G B$ , $A B = 12$ is opposite and $B G = \sqrt{74}$ is adjacent.
$\tan A \hat{G} B = \frac{12}{\sqrt{74}}$
Inverse tan, then 1 d.p.
$A \hat{G} B = 54.364 \dots °$
Check another way: $\sin A \hat{G} B = \frac{A B}{A G} = \frac{12}{\sqrt{218}}$ gives the same $54.4 °$ .

Answer: $54.4 °$ .

Example 4. $V A B C D$ is a pyramid with a horizontal square base $A B C D$ of side $8$ cm. The apex $V$ is vertically above the centre $M$ of the base, and $V A = 11$ cm. Calculate the height $V M$ and the angle between $V A$ and the base.

Show the working

Base diagonal first, in triangle $A B C$ .
$A C^{2} = 8^{2} + 8^{2} = 128$
$M$ is the midpoint of $A C$ , so $A M$ is half of $A C$ . Halving a length quarters its square.
$A M^{2} = \frac{128}{4} = 32$
Triangle $V A M$ has its right angle at $M$ and hypotenuse $V A = 11$ . You want a shorter side, so subtract.
$V M^{2} = 11^{2} - 32 = 121 - 32 = 89 \Rightarrow V M = \sqrt{89} = 9.433 \dots cm$
The projection of $V A$ onto the base is $A M$ , so the angle is angle $V A M$ . Use cos, because $A M = \sqrt{32}$ is adjacent and the given $V A = 11$ is the hypotenuse.
$\cos V \hat{A} M = \frac{\sqrt{32}}{11} \Rightarrow V \hat{A} M = 59.052 \dots °$
Using the whole diagonal instead of half is the standard error here. The apex is above the centre, not above the far corner.

Answer: $V M = 9.43$ cm (3 s.f.); the angle is $59.1 °$ .

Where people go wrong

Mistake: Working straight from the 3D drawing without redrawing the triangle.

Instead: Name the triangle, redraw it flat and mark its right angle before calculating.

Perspective drawings distort angles, so right angles look acute or obtuse. A flat sketch shows the triangle as it really is.

Mistake: Measuring the angle between a line and a plane to an edge of the plane, such as angle $G A B$ instead of angle $G A C$ .

Instead: Measure it to the projection: the line from where the line meets the plane to the foot of the perpendicular.

The definition is the angle with the line's shadow on the plane. $G$ is above $C$ , so the shadow of $A G$ runs to $C$ , not to $B$ .

Mistake: Using the whole base diagonal for a pyramid's sloping edge.

Instead: Use half of it, from the corner to the centre.

The apex is above the centre of the base, so the perpendicular lands at $M$ .

Mistake: Rounding the first length, such as $B G = 8.60$ , before the second step.

Instead: Carry the square ( $B G^{2} = 74$ ) or the full calculator value.

3D problems are two-step, so an early rounding is fed into the final answer and can cost the accuracy mark.

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.
In the cuboid $A B C D E F G H$ , explain why angle $A C G$ is a right angle.
Answer
$C G$ is vertical and $A C$ is a horizontal line through its foot $C$ . A vertical line is perpendicular to every horizontal line through its foot.
Without a calculator, work out the space diagonal of a cuboid measuring $2$ cm by $3$ cm by $6$ cm.
Answer
Base diagonal squared $= 4 + 9 = 13$ ; space diagonal squared $= 13 + 36 = 49$ , so the diagonal is $7$ cm.
In cuboid $A B C D E F G H$ (with $G$ above $C$ ), name the angle between $A G$ and the base $A B C D$ .
Answer
Angle $G A C$ , because $A C$ is the projection of $A G$ onto the base.
In a right pyramid $V A B C D$ with centre of base $M$ , what is the projection of edge $V B$ onto the base?
Answer
$B M$ , the line from $B$ to the centre, which is half of the diagonal $B D$ .
A cone has base radius $5$ cm and vertical height $12$ cm. Calculate its slant height and the angle between a slant line and the base.
Answer
Slant $= \sqrt{25 + 144} = 13$ cm. The angle has $\tan θ = \frac{12}{5}$ , so $θ = 67.4 °$ .

Exam questions

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

Question 1. $A B C D E F G H$ is a cuboid. The base $A B C D$ is horizontal, and $E$ , $F$ , $G$ and $H$ are vertically above $A$ , $B$ , $C$ and $D$ . $A B = 9$ cm, $B C = 6$ cm and $C G = 4$ cm.

(a) Calculate the length of $A C$ . 2 marks

Mark scheme

M1 9² + 6² oe, or 117 seen
A1 10.8 awrt (cm)

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

Mark scheme

M1 their AC² + 4² oe, or 9² + 6² + 4² oe
A1 11.5 awrt (cm)

(c) Calculate the angle between $A G$ and the base $A B C D$ . 3 marks

Mark scheme

B1 identifies angle GAC, or AC as the projection of AG; may be implied by a correct ratio
M1 tan GAC = 4 / their AC, or sin GAC = 4 / their AG oe
A1 20.3 awrt

Question 2. A pyramid has a horizontal rectangular base $P Q R S$ with $P Q = 16$ cm and $Q R = 12$ cm. The diagonals of the base meet at $M$ . The apex $V$ is vertically above $M$ , and $V M = 13$ cm.

(a) Without using a calculator, show that $P R = 20$ cm. 2 marks No calculator

Mark scheme

M1 16² + 12² = 256 + 144 = 400 seen
A1 √400 = 20 stated with no errors

(b) Calculate the length of the edge $V P$ . 2 marks

Mark scheme

M1 PM = 10 used with 13² + 10² oe
A1 16.4 awrt (cm)

(c) Calculate the angle between the edge $V P$ and the base $P Q R S$ . 2 marks

Mark scheme

M1 tan VPM = 13/10 oe, or a correct sin or cos using their VP
A1 52.4 awrt

Question 3. A horizontal rectangular courtyard $A B C D$ has $A B = 30$ m and $B C = 18$ m. A vertical flagpole $D T$ , of height $12$ m, stands at the corner $D$ .

(a) Calculate the distance $D B$ . 2 marks

Mark scheme

M1 30² + 18² oe, or 1224 seen
A1 35.0 awrt (m)

(b) Calculate the angle of elevation of the top of the flagpole, $T$ , from $B$ . 2 marks

Mark scheme

M1 tan θ = 12 / their DB oe
A1 18.9 awrt; FT their (a)

(c) Without calculating, explain why the angle of elevation of $T$ from $A$ is greater than the angle of elevation of $T$ from $C$ . 1 mark No calculator

Mark scheme

B1 A is closer to D (AD = 18 m) than C is (CD = 30 m), and the height is the same, so the line of sight from A is steeper oe

How this is examined

The angle between a line and a plane, found by naming the projection: A 3-mark Calculate at the end of a cuboid or pyramid question, often 'the angle between $A G$ and the base'. Naming the correct angle (such as angle $G A C$ ) or using the correct projection, a correct trig ratio in that right-angled triangle, and the angle to 1 d.p. A ratio from the wrong triangle, such as $\tan G \hat{A} B$ , does not earn the method mark, however good the arithmetic.
Length of a space diagonal or a pyramid height in two Pythagoras steps: A 2- to 4-mark Calculate, often split into (a) the base diagonal and (b) the space diagonal or height. Each correct sum of squares with the numbers in, such as $9^{2} + 6^{2}$ and then $117 + 4^{2}$ . For a pyramid, half the diagonal (or its square quartered) as the base of the vertical triangle.
Elevation in 3D: find the horizontal distance first: A mast or flagpole at one corner of a horizontal rectangle, asking for the angle of elevation from another corner. The horizontal distance across the rectangle by Pythagoras, then tan of height over that distance, to 1 d.p.

Words to know

Space diagonal: A line through the inside of a solid joining two corners that are not on the same face, such as $A G$ in a cuboid.
Projection: The shadow of a line on a plane: the join from where the line meets the plane to the foot of the perpendicular from its other end.
Angle between a line and a plane: The angle between the line and its projection onto the plane.
Right pyramid: A pyramid whose apex is vertically above the centre of its base.

All Ext Maths lessons Next: 7.2 Vectors in two dimensions