GCSE: Non-right angled triangles
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 16th November 2014
RECAP: Right-Angled Triangles
We’ve previously been able to deal with right-angled triangles, to find the area, or
missing sides and angles.
5
6
4
3
Area = 15
?
3?
Using Pythagoras:
𝑥 = 52 − 42
=3
30.96°
?
5
5
𝟏
Using trigonometry:
3
tan 𝜃 =
5
3
𝜃 = tan−1
5
= 30.96°
Using 𝑨 = 𝟐 𝒃𝒉:
1
𝐴𝑟𝑒𝑎 = × 6 × 5
2
= 15
Labelling Sides of Non-Right Angle Triangles
Right-Angled Triangles:
ℎ
𝑜
Non-Right-Angled Triangles:
𝑎
𝐶?
𝑏
𝐵?
𝑎
𝑐
?𝐴
We label the sides 𝑎, 𝑏, 𝑐 and
their corresponding OPPOSITE
angles 𝐴, 𝐵, 𝐶
OVERVIEW: Finding missing sides and angles
You have
You want
#1: Two angle-side
opposite pairs
Missing angle or Sine rule
side in one pair
#2 Two sides known
and a missing side
opposite a known
angle
Remaining side
Cosine rule
#3 All three sides
An angle
Cosine rule
#4 Two sides known
Remaining side
and a missing side not
opposite known angle
Use
Sine rule
twice
The Sine Rule
b
c
65°
A
5.02
For this triangle, try
calculating each side divided
by the sin of its opposite
angle. What do you notice in
all three cases?
10
85°
C
! Sine Rule:
𝑎
𝑏?
𝑐
=
=
sin 𝐴 sin 𝐵 sin 𝐶
B
30°
9.10
a
You have
You want
Use
#1: Two angle-side
opposite pairs
Missing angle or Sine rule
side in one pair
Examples
8
Q2
Q1
85°
8
50°
100°
45°
30°
?
15.76
?
11.27
𝒙
𝟖
=
𝐬𝐢𝐧 𝟖𝟓 𝐬𝐢𝐧 𝟒𝟓
𝒙
𝟖
=
𝐬𝐢𝐧 𝟏𝟎𝟎 𝐬𝐢𝐧 𝟑𝟎
𝟖 𝐬𝐢𝐧 𝟖𝟓
𝒙=
= 𝟏𝟏. 𝟐𝟕
𝐬𝐢𝐧 𝟒𝟓
𝒙=
𝟖 𝐬𝐢𝐧 𝟏𝟎𝟎
= 𝟏𝟓. 𝟕𝟔
𝐬𝐢𝐧 𝟑𝟎
You have
You want
Use
#1: Two angle-side
opposite pairs
Missing angle or Sine rule
side in one pair
Examples
When you have a missing angle, it’s better to ‘flip’ your formula to get
𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩
=
𝒂
𝒃
i.e. in general put the missing value in the numerator.
5
Q3
Q4
126°
40.33°
?
85°
56.11°
?
6
sin 𝜃 sin 85
=
5
6
5 sin 85
sin 𝜃 =
6
5 sin 85
−1
𝜃 = sin
6
= 56.11°
8
10
sin 𝜃 sin 126°
=
8
10
8 sin 126
sin 𝜃 =
10
8 sin 126
−1
𝜃 = sin
10
= 40.33°
Test Your Understanding
𝑄
𝑃
85°
20°
5𝑐𝑚
𝑅
Determine the length 𝑃𝑅.
𝑷𝑹
𝟓
=
𝐬𝐢𝐧 𝟐𝟎 ? 𝐬𝐢𝐧 𝟖𝟓
𝟓 𝐬𝐢𝐧 𝟐𝟎
𝑷𝑹 =
= 𝟏. 𝟕𝟐𝒄𝒎
𝐬𝐢𝐧 𝟖𝟓
𝜃
82°
12𝑚
Determine the angle 𝜃.
sin 𝜽 sin 𝟖𝟐
=
𝟏𝟎
𝟏𝟐
𝟏𝟎 sin 𝟖𝟐
?
−𝟏
𝜽 = 𝒔𝒊𝒏
𝟏𝟐
= 𝟓𝟓. 𝟔°
10𝑚
Exercise 1
Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.
Q1
Q2
15
16
10
85°
𝑥
Q3
𝑥
40°
12
30°
𝑦
30°
20
𝑦 = 56.4°
?
𝑥 = 53.1°
?
𝑥 = 23.2
?
𝑥
Q4
Q6
Q5
40°
10
35°
70°
10
5
𝛼
20
𝛼 = 16.7°
?
𝑥
𝑥 = 6.84
?
𝑥 = 5.32
?
Cosine Rule
The sine rule could be used whenever we had two pairs of sides and opposite angles involved.
However, sometimes there may only be one angle involved. We then use something called the
cosine rule.
15
𝑏
Cosine Rule:
𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴
𝐴
115°
𝑐
12
𝑎
𝑥
The only angle in formula is 𝐴, so label angle
in diagram
label sides
opposite
side 𝑎, and?so on
How𝐴, are
labelled
(𝑏 and 𝑐 can go either way).
𝑥 2 = 152 + 122 − 2 × 15 × 12 × cos 115
Calculation?
𝑥 2 = 521.14257
…
𝑥 = 22.83
Sin or Cosine Rule?
If you were given these exam questions, which would you use?
10
𝑥
𝑥
10
70°
70°
15
15
Sine


Cosine
Sine
10
10
7
𝛼
𝛼
70°
12
15
Sine
Cosine

Cosine

Sine


Cosine
Test Your Understanding
e.g. 1
e.g. 2
𝑥
7
47°
8
𝑥
4
106.4°
7
𝑥 = 8.99?
𝑥 = 6.05?
You have
You want
Use
Two sides known and a missing
side opposite a known angle
Remaining side
Cosine rule
Exercise 2
Use the cosine rule to determine the missing angle/side. Quickly copy out the diagram first.
Q1
Q2
5
5
𝑥
58
8
100°
70
60°
7
𝑦 = 10.14
?
Q6
4
𝑥
10
65°
6
𝑥 = 4.398
?
𝑥 = 50.22
?
𝑥
Q5
6
43°
𝑥
𝑦
𝑥 = 6.24
?
Q4
135°
Q3
3
𝑥 = 9.513
?
75°
5
8
𝑥
𝑥 = 6.2966
?
3
Dealing with Missing Angles
You have
You want
Use
All three sides
An angle
Cosine rule
𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐 − 𝟐𝒃𝒄 𝐜𝐨𝐬 𝑨
7
𝛼
4
9
𝛼 = 25.2°
?
𝟒𝟐 = 𝟕𝟐 + 𝟗𝟐 − ?𝟐 × 𝟕 × 𝟗 × 𝐜𝐨𝐬 𝜶
𝟏𝟔 = 𝟏𝟑𝟎 − 𝟏𝟐𝟔
? 𝐜𝐨𝐬 𝜶
𝟏𝟐𝟔 𝐜𝐨𝐬 𝜶 = 𝟏𝟑𝟎
? − 𝟏𝟔
𝟏𝟏𝟒
𝐜𝐨𝐬 𝜶 =
𝟏𝟐𝟔
𝟏𝟏𝟒
? = 𝟐𝟓. 𝟐°
𝜶 = 𝐜𝐨𝐬 −𝟏
𝟏𝟐𝟔
Label sides then
substitute into
formula.
Simplify each bit of formula.
Rearrange (I use ‘swapsie’
trick to swap thing you’re
subtracting and result)
Test Your Understanding
4𝑐𝑚
8
7𝑐𝑚
5
𝜃
7
82
72
+ 52
=
− 2 × 7 × 5 × cos 𝜃
64 = 74 − 70 cos 𝜃
10 = 70 cos 𝜃
1
?
cos 𝜃 =
7
1
𝜃 = cos −1
= 𝟖𝟏. 𝟕𝟗°
7
9𝑐𝑚
𝜃
42 = 72 + 92 − 2 × 7 × 9 × cos 𝜃
16 = 130 − 126 cos 𝜃
114 = 126 cos 𝜃
114
?
cos 𝜃 =
126
114
−1
𝜃 = cos
= 𝟐𝟓. 𝟐𝟏°
126
Exercise 3
1
2
7
12
6
3
5.2
𝜃
𝛽
11
5
𝜃
13.2
6
𝜃 = 71.4°
?
𝛽 = 92.5°
?
8
𝜃 = 111.1°
?
Using sine rule twice
You have
You want
Use
#4 Two sides known
Remaining side
and a missing side not
opposite known angle
4
32°
3
𝑥
Sine rule
twice
Given there is just one angle involved,
you might attempt to use the cosine
rule:
𝟑𝟐 = 𝒙𝟐 + 𝟒𝟐 − 𝟐?× 𝒙 × 𝟒 × 𝐜𝐨𝐬 𝟑𝟐
𝟗 = 𝒙𝟐 + 𝟏𝟔 − 𝟖𝒙 𝐜𝐨𝐬 𝟑𝟐
This is a quadratic equation!
It’s possible to solve this using the
quadratic formula (using 𝒂 = 𝟏, 𝒃 =
− 𝟖 𝐜𝐨𝐬 𝟑𝟐 , 𝒄 = 𝟕). However, this is a bit
fiddly and not the primary method
expected in the exam…
Using sine rule twice
You have
You want
Use
#4 Two sides known
Remaining side
and a missing side not
opposite known angle
!
2: Which means we would then
know this angle.
4
1: We could use the sine
rule to find this angle.
sin 𝐴 sin 32
=
4 ? 3
𝐴 = 44.9556°
Sine rule
twice
?
𝟏𝟖𝟎 − 𝟑𝟐 − 𝟒𝟒. 𝟗𝟓𝟓𝟔 = 𝟏𝟎𝟑. 𝟎𝟒𝟒𝟒
32°
3
𝑥
3: Using the sine rule a second
time allows us to find 𝑥
𝑥
3
=
sin 103.0444? sin 32
𝑥 = 5.52 𝑡𝑜 3𝑠𝑓
Test Your Understanding
9
?
𝑦 = 6.97
𝑦
61°
10
3
4
?
𝑦 = 5.01
53°
𝑦
Area of Non Right-Angled Triangles
3cm
Area = 0.5 x 3 x 7 x sin(59)
= 9.00cm2?
59°
7cm
!
Area =
1
𝑎
2
𝑏 sin 𝐶
Where C is the angle wedged between two sides a and b.
Test Your Understanding
1
𝐴 = × 6.97 × 10 × 𝑠𝑖𝑛61
?
2
= 30.48
9
6.97
61°
10
5
1
𝐴 = × 5 × 5 × sin 60
2
?
25 3
=
4
5
5
Harder Examples
Q1 (Edexcel June 2014)
Q2
6
7
8
Finding angle ∠𝐴𝐶𝐷:
sin 𝐷 sin 100
=
9
11
𝐷 = 53.6829°
? − 53.6829 = 26.3171°
∠𝐴𝐶𝐷 = 180 − 100
1
𝐴𝑟𝑒𝑎 𝑜𝑓 Δ = × 9 × 11 × sin 26.3171 = 21.945
2
Area of 𝐴𝐵𝐶𝐷 = 2 × 21.945 = 43.9 𝑡𝑜 3𝑠𝑓
Using cosine rule to find angle
opposite 8:
21
cos 𝜃 =
84
?
𝜃 = 75.52249°
𝐴𝑟𝑒𝑎
1
= × 6 × 7 × sin 75.5 … = 𝟐𝟎. 𝟑
2
Exercise 4
Q1
3
5
100°
Q3
Q2
1
3.6
1 3.8
1
𝐴𝑟𝑒𝑎 =
Q5
5.2
3
=?
0.433
4
70°
Area = 9.04?
Q7
Q6
64°
2cm
Area = 8.03?
110°
Area = 3.11𝑐𝑚
? 2
49°
𝐴𝑟𝑒𝑎 = 29.25𝑐𝑚
? 2
5
75°
8
Area = 7.39?
Q4
Q8
𝑃 is the midpoint of 𝐴𝐵 and 𝑄 the
midpoint of 𝐴𝐶. 𝐴𝑃𝑄 is a sector
of a circle. Find the shaded area.
1
1
× 62 × sin 60 − 𝜋 32 = 10.9𝑐𝑚2
2
6
?
4.2m
3m
5.3m
Area = 6.29𝑚
?2
Segment Area
𝐴
𝑂
𝑂𝐴𝐵 is a sector of a circle, centred at 𝑂.
Determine the area of the shaded segment.
70
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 =
× 𝜋 × 102 ?= 61.0865𝑐𝑚2
360
1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = × 102 × sin 70?= 46.9846𝑐𝑚2
2
70°
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 61.0865 − 46.9846
= 14.1 𝑡𝑜 3𝑠𝑓 ?
𝐵
Test Your Understanding
𝐴 = 119𝑚2?
𝐴 = 3𝜋 − 9?
Exercise 5 - Mixed Exercises
Q1
Q2
27
80°
𝑥
40°
a) 𝑥 = 41.37?
b) 𝐴𝑟𝑒𝑎 = 483.63
?
Q5
Q3
8
𝑦
30°
130°
11
60𝑚
18
𝛼 = 17.79°
?
𝑧 = 26.67?
𝐴𝑟𝑒𝑎 = 73.33
?
10
𝑦 = 10.45?
𝐴𝑟𝑒𝑎 = 37.59
?
4.6
Q6
15
5
7
52°
6𝑐𝑚
12
𝜃 = 122.8°
?
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
= 286.5𝑚
?
Q8
Q7
𝜃
𝑄𝑅 = 12.6𝑐𝑚
?
𝑧
𝛼
70°
90𝑚
Q4
61°
𝑥
𝐴𝑟𝑒𝑎 = 2.15𝑐𝑚2
?
𝑥 = 7.89
𝐴𝑟𝑒𝑎 = 17.25