Trigonometry of Right-Angled Triangles

1. GCSE Trigonometry of Right-
Angled Triangles
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 1st January 2019
Objectives:
• Find unknown sides in right-angled triangles using a side and angle.
• Find unknown angles in right-angled triangle using two sides.
• Note: ‘Exact trigonometric ratios’ and 3D Trigonometry are covered
in separate slides.

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3. 𝑥
𝑦
𝜃
(𝑎, 𝑏)
𝑟
I was in Year 9 and was trying to write
a computer program that would draw
an analogue clock (as you do).
I needed to work out the two
coordinates to draw the minute hand
between, and similarly for the hour
hand, given the current hour, minute
and desired length of each hand.
I couldn’t work out how to do it, until I
(coincidentally) learnt trigonometry in
maths the next day, and was able to
finish my program that evening!
Motivation

4. 3
4
𝑥 13
5
𝑦
Starter
Determine the length 𝑥. Determine the length 𝑦.
32 + 42 = 𝑥2
25 = 𝑥2
𝑥 = 5
?
52 + 𝑦2 = 132
25 + 𝑦2 = 169
𝑦2
= 144
𝑦 = 12
?

5. 𝟒
𝟏𝟑
𝒚
We used Pythagoras’
theorem if all 3 sides of a
right-angled triangle were
involved.
But…
But what if we had two sides
involved and an angle?
(excluding the right-angle)
𝒂
𝟑
𝟏𝟒
𝟑𝟎°
𝒙
𝟏𝟎
Unknown angle.
Unknown side.
Unknown side.

6. 30°
hypotenuse
adjacent
opposite
Names of sides relative to an angle
?
?
?
The ‘adjacent’ is the side adjacent
(i.e. next to) the angle of interest.
The ‘opposite’ is the
side opposite the
angle of interest.
The hypotenuse is the
longest side of a right-
angled triangle, and is
opposite the right-angle.

7. 60°
𝑥
𝑦
𝑧
Hypotenuse Opposite Adjacent
𝑥 𝑦 𝑧
𝑥 1 2
𝑐 𝑎 𝑏
𝜽
1
𝑥
2
20°
𝑎
𝑐
𝑏
? ? ?
? ? ?
? ? ?
Quickfire Side Naming

8. 𝜽
𝒐
𝒉
𝒂
sin 𝜃 =
𝑜
ℎ
cos 𝜃 =
𝑎
ℎ
tan 𝜃 =
𝑜
𝑎
You can remember this using:
“soh cah toa”
! sin, cos and tan are functions which take an angle and give us
the ratio between pairs of sides in a right angle triangle.
Sin/Cos/Tan
?
?
?
Recall that ratio can just mean how many
times bigger one quantity is than another.

9. 𝟒𝟎°
𝟒
𝒙
Find the value of 𝑥 (to 3sf)
Examples
𝑥 = 4 cos 40°
𝑥 = 3.06
Step 1: Determine which sides are
hyp/adj/opp.
ℎ
𝑎
Fro Tip: Put your h/a/o letters in
circles – this avoids confusion with any
variables you have used as side
lengths (in this example, 𝑥)
Step 2: Work out which
trigonometric function we need.
We used 𝑎 and ℎ. Thinking about “soh
cah toa”, we want the “cah”, i.e. cos:
cos 40° =
𝑥
4
Fro Tip: The angle always follows the sin/cos/tan. I
see a lot of students incorrectly write cos
𝑥
4
= 40°
Step 3: Rearrange the equation to
find the unknown.
𝑥 was being divided by 4, so we multiply
both sides by 4 to cancel this out.
× 4 × 4

10. 𝟐𝟎°
𝟕
𝒙
Further Example
𝐬𝐢𝐧 𝟐𝟎° =
𝒙
𝟕
𝒙 = 𝟕 𝐬𝐢𝐧 𝟐𝟎°
𝒙 = 𝟐. 𝟑𝟗
Find the value of 𝑥 (to 3sf)
ℎ
𝑜
?
Extremely Advanced Side Note: I often get asked how that calculator actually works
out, for example, sin 20°. The process is complicated and would never ordinarily be
done by a ‘human’. You won’t likely learn the technique for calculating it unless you
do Further Maths A Level, using something called Maclaurin Series. But since you
asked…
sin 𝑥 can be calculated using the infinitely long formula:
sin 𝑥 = 𝑥 −
𝑥3
3!
+
𝑥5
5!
−
𝑥7
7!
+
𝑥9
9!
− ⋯
where 5! for example means 5 × 4 × 3 × 2 × 1. But the angle 𝑥 has to be in a unit
called ‘radians’ (we can convert from degrees to radians by multiplying the angle by
𝜋
180
). We can’t substitute into an infinitely long expression, but we can stop after the
first several terms to get an accurate value, as the terms in the sum gradually
approach 0. For example, sin 20 = 0.34202 …, and I got the same (up to 10 dp) by
substituting into the above formula up to the 𝑥9
term.

11. Test Your Understanding So Far
𝟐𝟓°
𝒚
𝟓
𝑜
𝑎
𝐭𝐚𝐧 𝟐𝟓° =
𝒚
𝟓
𝒚 = 𝟓 𝐭𝐚𝐧 𝟐𝟓°
𝒚 = 𝟐. 𝟑𝟑
?
Find the value of 𝑦 (to 3sf)

12. 60 °
𝒙
12
30°
4
𝒙
Harder Examples
𝒔𝒊𝒏 𝟔𝟎 =
𝟏𝟐
𝒙
𝒙 =
𝟏𝟐
𝒔𝒊𝒏 𝟔𝟎
= 𝟏𝟑. 𝟖𝟔
𝒕𝒂𝒏 𝟑𝟎 =
𝟒
𝒙
𝒙 =
𝟒
𝒕𝒂𝒏 𝟑𝟎
= 𝟔. 𝟗𝟑
?
If the variable is in the
denominator, you can apply
something called the
‘swapsie trick’.
Notice that we can rearrange:
8
4
= 2 →
8
2
= 4
i.e. we can ‘swap’ the thing
we’re dividing by and the
result of the division. In this
trig question, we could swap
the 𝑥 and the sin 60
?

13. Test Your Understanding
8.63
?
1 2
20.2
?
Working:
sin32° =
𝐵𝐶
47
𝐵𝐶 = 24.906
tan 51° =
24.906
𝐵𝐷
𝐵𝐷 =
24.906
tan 51°
= 20.2

14. Exercise 1
Find 𝑥, giving your answers to 3 significant figures.
𝟕𝟎°
15
𝒙
1
𝟒𝟎°
22
𝒙
a b
𝟖𝟎°
20
𝒙
𝟕𝟎° 4
𝒙
𝟕𝟎°
𝒙
𝟒
e f
c
d
𝑥 = 16.9
𝑥 = 20.3
𝑥 = 7.00
𝑥 = 11.0
𝑥 = 11.7
?
?
?
?
?
(questions on provided sheet)
𝑥 = 14.1
?
𝟔𝟏°
3
𝒙
g
𝑥 = 6.19
?
Q2-7 on next slide…

15. Exercise 1 (questions on provided sheet)
I put a ladder 1.5m away from a
tree. The ladder is inclined at 70°
above the horizontal. What is the
height of the tree? 𝟒. 𝟏𝟐𝒎
Ship B is 100m east of Ship A, and
the bearing of Ship B from Ship A
is 30°. How far due North is the
ship? 𝟏𝟎𝟎 ÷ 𝐭𝐚𝐧 𝟑𝟎 = 𝟏𝟕𝟑. 𝟐𝒎
2
3
?
?
[IMC] The semicircle and
isosceles triangle have
equal areas. Find tan 𝑥.
𝝅
𝟐
N
?
[Edexcel GCSE June2012-2H Q18]
The diagram shows a
quadrilateral ABCD. AB = 16 cm.
AD = 12 cm. Angle BCD = 40°.
Angle ADB = angle CBD = 90°.
Calculate the length of CD.
16.5 cm
4
In the shape drawn below, 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷. Work out
the area of 𝐴𝐵𝐶𝐷, giving your answer to 1 dp.
89.2 cm2
5
[OCR GCSE(9-1) SAM 6H Q16aii]
Simon cuts the corners off a
square piece of card to leave the
regular octagon shown below.
O is the centre of the octagon. A
and B are vertices of the
octagon. OA = OB = 5 cm. Angle
AOB = 45°.
Work out the area of the original
square piece of card. 85.4 cm2
6
?
?
?

16. x
y
𝜃
(𝑎, 𝑏)
𝑟
So what is
𝑎, 𝑏 then?
𝒂
𝒃
sin 𝜃 =
𝑎
𝑟
→ 𝑎 = 𝑟 sin 𝜃
cos 𝜃 =
𝑏
𝑟
→ 𝑏 = 𝑟 cos 𝜃
Therefore I drew the hand between
0,0 and 𝑟 sin 𝜃 , 𝑟 cos 𝜃

17. 30 °
4
𝒙
STARTER RECAP
𝑥 =
4
tan 30
= 4 3 𝑜𝑟 6.93
?
When would we use trigonometry?
• When we have a right-angled triangle.
• When we’re involving two side lengths and an angle.
?

18. 𝒂
3
5
But what if the angle is unknown?
sin 𝑎 =
3
5
𝑜
ℎ
𝑎 = sin−1
3
5
= 36.9°
(si𝑛−1
) (si𝑛−1
)
Step 1: Determine which
sides are hyp/adj/opp.
Step 2: Work out which
trigonometric function we
need.
Step 3: Rearrange the
equation to find the
unknown.
We can use the same process…
Remember that the angle
always goes after the
sin/cos/tan.
We want to get the angle 𝑎 on
its own. But how do we get rid
of the 𝑠𝑖𝑛 on front of it?
We know we get rid of something in an equation by doing the opposite.
The opposite of sin is ‘inverse sin’, written si𝑛−1
. Use the shift button
to get it on your calculator.

19. cos−1
4
5
cos−1
5
4
sin−1
4
5
sin−1
5
4
What is the missing angle?
𝟓
𝟒
𝒂

20. cos−1
1
2
sin−1 2 tan−1 2 tan−1
1
2
What is the missing angle?
𝟏
𝟐
𝒂

21. cos−1
3
5
sin−1
3
5
tan−1
3
5
sin−1
5
3
What is the missing angle?
𝟓
𝟑
𝒂

22. cos−1
2
3
sin−1
2
3
sin−1
3
2
tan−1
2
3
What is the missing angle?
𝟑
𝟐
𝒂

23. Test Your Understanding
[Edexcel GCSE Nov2007-4I Q25, Nov2007-6H Q14]
PQR is a right-angled triangle.
PR = 12 cm. QR = 4.5 cm. Angle 𝑃𝑅𝑄 = 90°.
Work out the value of 𝑥.
Give your answer correct to one decimal place.
tan−1 4.5
12
= 20.6 cm
?
[Edexcel IGCSE Jan2014(R)-3H Q17] The diagram
shows triangle ABC. D is the point on AB, such
that CD is perpendicular to AB.
AC = 8.3 cm. AD = 4.7 cm. BD = 7.5 cm.
Calculate the size of angle ABC.
Give your answer correct to 1 decimal place.
𝑪𝑫 = 𝟖. 𝟑𝟐 − 𝟒. 𝟕𝟐 = 𝟔. 𝟖𝟒𝟏𝟎𝟓
∠𝑨𝑩𝑪 = 𝐭𝐚𝐧−𝟏
𝟔. 𝟖𝟒𝟏𝟎𝟓
𝟕. 𝟓
= 𝟒𝟐. 𝟒°
?

24. Exercise 2
𝜃
7
4
𝜃
3
5
𝜃
2
1
𝜃
5
Find 𝜃, giving your answer to 3sf.
4
𝜃 = 55.2°
𝜃 = 31.0°
𝜃 = 30.0°
𝜃 = 51.3°
7
60° 𝜃
10
𝜃 = 37.3°
?
?
?
?
?
1
b
c
d
3
[Edexcel GCSE June2014-2H Q15b]
The diagram shows the positions of
three turbines A, B and C.
Calculate the bearing of C from A.
Give your answer correct to the
nearest degree.
2
217°
?
𝜃
11
7
𝜃 = 50.5°
?
e
(questions on provided sheet)
a

25. Exercise 2
𝜃
3
4
13
1
1
1
1
1
𝜃1
𝜃2
𝜃3
The angles 𝜃1, 𝜃2, … form a
sequence. Give the formula for
the 𝑛th term of the sequence.
𝜽𝒏 = 𝐭𝐚𝐧−𝟏
𝟏
𝒏
𝜽 = 𝐜𝐨𝐬−𝟏
𝟓
𝟏𝟑
= 𝟔𝟕. 𝟒°
?
?
N
4
(questions on provided sheet)
[Edexcel IGCSE May2015-3H Q15]
𝐴𝐵𝐶𝐷 is a trapezium. 𝐴𝐵 = 25 cm.
𝐵𝐶 = 24 cm. 𝐶𝐷 = 10 cm.
Angle 𝐴𝐵𝐶 = angle 𝐵𝐶𝐷°
Calculate the size of angle 𝐶𝐷𝐴.
Give your answer correct to 3
significant figures.
Solution: 𝟏𝟐𝟐°
5
?
[Edexcel GCSE(9-1) Nov 2017 2F Q22, Nov 2017 2H Q7]
𝐴𝐵𝐶𝐷 is a trapezium. Work out the size of angle 𝐶𝐷𝐴.
Give your answer correct to 1dp.
6
Solution: 𝟑𝟐. 𝟑°
?

26. Related Slides
In other slides on www.drfrostmaths.com we’ll cover:
1. Exact Trigonometric Ratios
e.g. What is the exact value of sin 60°? (without
using a calculator)
2. 3D Trigonometry & Pythagoras