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Mr Barton’s Maths Notes
Trigonometry
3. 3D
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3. 3D Trigonometry
The Secret to Solving 3D Trigonometry Problems
• 3D Trigonometry is just the same as bog-standard, flat, normal trigonometry
• All we need are the skills we learnt in the last two sections:
1. Pythagoras
2. Sin, Cos and Tan
• The only difference is that is a little bit harder to spot the right-angled triangles
• But once you spot them:
- Draw them out flat
- Label your sides
- Fill in the information that you do know
- Work out what you don’t in the usual way!
• And if you can do that, then you will be able to tick another pretty tricky topic off your list!
Please Remember: You need a right-angled triangle to be able to use either Pythagoras or Sin,
Cos and Tan… and I promise that will be the last time I say it!
Example 1
The diagram below shows a record breaking wedge of Cheddar Cheese in which rectangle PQRS
is perpendicular (at 900 to) to rectangle RSTU. The distances are shown on the diagram.
Calculate: (a) The distance QT
(b) The angle QTR
P
S
Q
2.5 m
R
T
4.9 m
7.8 m
U
Working out the answer (a):
The first thing we need to figure out is what we are actually trying to work out!
We need the line QT:
P
S
Q
2.5 m
R
T
4.9 m
7.8 m
U
Now, as I said, the key to this is spotting the right-angled triangles…
Well, I can see a nice one: TQR.
That contains the length we want, and we already know how long QR is…
So now all we need to do is work out length TR…
Working Out TR:
P
Okay, if you look carefully, you
should be able to see a right-angled
triangle on the base of this wedge of
cheese
S
Q
2.5 m
R
It’s the triangle TRU:
T
T
a
c
?
4.9 m
U
4.9 m
U
7.8 m
c  4.9  7.8
2
2
b
c  8 4 .85
2
c
84.85
c  9.211 ...m
R
Well, we have two sides and we want to
work out the Hypotenuse… This looks
like a job for Pythagoras!
2
c  24.01  60.84
2
7.8 m
1. Label the sides
2. Use the formula:
c  a b
3. Put in the numbers:
2
2
2
P
Working Out TQ:
Okay, so now we have all we need to
be able to calculate TQ.
Just make sure you draw the correct
right-angled triangle!
c
?
2.5 m
b
2
c  84.8 5  6.25
2
c  91.1
2
c
R
a
c  9.21...  2.5
2
T
Q
9.21… m
T
S
91.1
c  9.54 m (2 d p )
2
Q
2.5 m
R
9.21… m
4.9 m
7.8 m
U
Once again, we have two sides and we
want to work out the Hypotenuse… This
looks like a job for Pythagoras!
1. Label the sides
2. Use the formula:
c  a b
3. Put in the numbers:
2
2
2
P
Working out the answer (b):
Again, we must be sure we know
what angle the question wants us
to find!
S
I have marked angle QTR on the
diagram

T
So now we draw our right-angled
triangle:
Q
9.54 m
O

9.21 m
A
7.8 m
U
2.5 m
T
2.5 m
R
9.21 m
4.9 m
H
9.54 m
Q
R
To calculate the size on an angle, we must
use either sin, cos or tan, which means first
we must label our sides!
Now, because we actually know all three
lengths, we can choose! I’m going for tan!
Tan θ = o ÷ a
Tan θ = 0.27144…
15.20 (1dp)
Example 2
The diagram below shows a plan of a tent that I am trying to erect before the rain comes. OP
is a vertical pole, and O is at the very centre of the rectangle QRST. The lengths and angles
are as shown on the diagram. Calculate the height of the vertical pole OP.
P
Q
T
O
48o
5m
R
12 m
S
1. Working Out OT:
You should be able to see that if we can
work out OT, we will then have a rightangled triangle which will give us OP!
So, to get started we need to use the
base of the rectangle:
Well, OT is half way along the line TR
line, so it must be… 6.5m
P
T
? c
a
5m
S
12 m b
c  5  12
2
2
2
c  2 5  144
2
H
O
?
c  169
2
c
c  13 m
480
T
169
6.5 m A
O
o = Tan θ x a
2. Working Out OP:
And now we have a right-angled triangle
where we know one length (TR), and we
know one angle (OTP)… so we can work
out any side using a bit of sin, cos or
tan!
7.22 m (2dp)
R
Good luck with
your revision!