Slides: GCSE Non-Right Angled Triangles

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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:
๐‘ฅ = 52 − 42
=3
30.96°
?
5
5
๐Ÿ
Using trigonometry:
3
tan ๐œƒ =
5
3
๐œƒ = tan−1
5
= 30.96°
Using ๐‘จ = ๐Ÿ ๐’ƒ๐’‰:
1
๐ด๐‘Ÿ๐‘’๐‘Ž = × 6 × 5
2
= 15
Labelling Sides of Non-Right Angle Triangles
Right-Angled Triangles:
โ„Ž
๐‘œ
Non-Right-Angled Triangles:
๐‘Ž
๐ถ?
๐‘
๐ต?
๐‘Ž
๐‘
?๐ด
We label the sides ๐‘Ž, ๐‘, ๐‘ and
their corresponding OPPOSITE
angles ๐ด, ๐ต, ๐ถ
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:
๐‘Ž
๐‘?
๐‘
=
=
sin ๐ด sin ๐ต sin ๐ถ
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
๐’™
๐Ÿ–
=
๐ฌ๐ข๐ง ๐Ÿ–๐Ÿ“ ๐ฌ๐ข๐ง ๐Ÿ’๐Ÿ“
๐’™
๐Ÿ–
=
๐ฌ๐ข๐ง ๐Ÿ๐ŸŽ๐ŸŽ ๐ฌ๐ข๐ง ๐Ÿ‘๐ŸŽ
๐Ÿ– ๐ฌ๐ข๐ง ๐Ÿ–๐Ÿ“
๐’™=
= ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ•
๐ฌ๐ข๐ง ๐Ÿ’๐Ÿ“
๐’™=
๐Ÿ– ๐ฌ๐ข๐ง ๐Ÿ๐ŸŽ๐ŸŽ
= ๐Ÿ๐Ÿ“. ๐Ÿ•๐Ÿ”
๐ฌ๐ข๐ง ๐Ÿ‘๐ŸŽ
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
๐ฌ๐ข๐ง ๐‘จ ๐ฌ๐ข๐ง ๐‘ฉ
=
๐’‚
๐’ƒ
i.e. in general put the missing value in the numerator.
5
Q3
Q4
126°
40.33°
?
85°
56.11°
?
6
sin ๐œƒ sin 85
=
5
6
5 sin 85
sin ๐œƒ =
6
5 sin 85
−1
๐œƒ = sin
6
= 56.11°
8
10
sin ๐œƒ sin 126°
=
8
10
8 sin 126
sin ๐œƒ =
10
8 sin 126
−1
๐œƒ = sin
10
= 40.33°
Test Your Understanding
๐‘„
๐‘ƒ
85°
20°
5๐‘๐‘š
๐‘…
Determine the length ๐‘ƒ๐‘….
๐‘ท๐‘น
๐Ÿ“
=
๐ฌ๐ข๐ง ๐Ÿ๐ŸŽ ? ๐ฌ๐ข๐ง ๐Ÿ–๐Ÿ“
๐Ÿ“ ๐ฌ๐ข๐ง ๐Ÿ๐ŸŽ
๐‘ท๐‘น =
= ๐Ÿ. ๐Ÿ•๐Ÿ๐’„๐’Ž
๐ฌ๐ข๐ง ๐Ÿ–๐Ÿ“
๐œƒ
82°
12๐‘š
Determine the angle ๐œƒ.
sin ๐œฝ sin ๐Ÿ–๐Ÿ
=
๐Ÿ๐ŸŽ
๐Ÿ๐Ÿ
๐Ÿ๐ŸŽ sin ๐Ÿ–๐Ÿ
?
−๐Ÿ
๐œฝ = ๐’”๐’Š๐’
๐Ÿ๐Ÿ
= ๐Ÿ“๐Ÿ“. ๐Ÿ”°
10๐‘š
Exercise 1
Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.
Q1
Q2
15
16
10
85°
๐‘ฅ
Q3
๐‘ฅ
40°
12
30°
๐‘ฆ
30°
20
๐‘ฆ = 56.4°
?
๐‘ฅ = 53.1°
?
๐‘ฅ = 23.2
?
๐‘ฅ
Q4
Q6
Q5
40°
10
35°
70°
10
5
๐›ผ
20
๐›ผ = 16.7°
?
๐‘ฅ
๐‘ฅ = 6.84
?
๐‘ฅ = 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
๐‘
Cosine Rule:
๐‘Ž2 = ๐‘ 2 + ๐‘ 2 − 2๐‘๐‘ cos ๐ด
๐ด
115°
๐‘
12
๐‘Ž
๐‘ฅ
The only angle in formula is ๐ด, so label angle
in diagram
label sides
opposite
side ๐‘Ž, and?so on
How๐ด, are
labelled
(๐‘ and ๐‘ can go either way).
๐‘ฅ 2 = 152 + 122 − 2 × 15 × 12 × cos 115
Calculation?
๐‘ฅ 2 = 521.14257
…
๐‘ฅ = 22.83
Sin or Cosine Rule?
If you were given these exam questions, which would you use?
10
๐‘ฅ
๐‘ฅ
10
70°
70°
15
15
Sine
๏ƒผ
๏ƒป
Cosine
Sine๏ƒป
10
10
7
๐›ผ
๐›ผ
70°
12
15
Sine๏ƒป
Cosine
๏ƒผ
Cosine
๏ƒผ
Sine
๏ƒผ
๏ƒป
Cosine
Test Your Understanding
e.g. 1
e.g. 2
๐‘ฅ
7
47°
8
๐‘ฅ
4
106.4°
7
๐‘ฅ = 8.99?
๐‘ฅ = 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
๐‘ฅ
58
8
100°
70
60°
7
๐‘ฆ = 10.14
?
Q6
4
๐‘ฅ
10
65°
6
๐‘ฅ = 4.398
?
๐‘ฅ = 50.22
?
๐‘ฅ
Q5
6
43°
๐‘ฅ
๐‘ฆ
๐‘ฅ = 6.24
?
Q4
135°
Q3
3
๐‘ฅ = 9.513
?
75°
5
8
๐‘ฅ
๐‘ฅ = 6.2966
?
3
Dealing with Missing Angles
You have
You want
Use
All three sides
An angle
Cosine rule
๐’‚๐Ÿ = ๐’ƒ๐Ÿ + ๐’„๐Ÿ − ๐Ÿ๐’ƒ๐’„ ๐œ๐จ๐ฌ ๐‘จ
7
๐›ผ
4
9
๐›ผ = 25.2°
?
๐Ÿ’๐Ÿ = ๐Ÿ•๐Ÿ + ๐Ÿ—๐Ÿ − ?๐Ÿ × ๐Ÿ• × ๐Ÿ— × ๐œ๐จ๐ฌ ๐œถ
๐Ÿ๐Ÿ” = ๐Ÿ๐Ÿ‘๐ŸŽ − ๐Ÿ๐Ÿ๐Ÿ”
? ๐œ๐จ๐ฌ ๐œถ
๐Ÿ๐Ÿ๐Ÿ” ๐œ๐จ๐ฌ ๐œถ = ๐Ÿ๐Ÿ‘๐ŸŽ
? − ๐Ÿ๐Ÿ”
๐Ÿ๐Ÿ๐Ÿ’
๐œ๐จ๐ฌ ๐œถ =
๐Ÿ๐Ÿ๐Ÿ”
๐Ÿ๐Ÿ๐Ÿ’
? = ๐Ÿ๐Ÿ“. ๐Ÿ°
๐œถ = ๐œ๐จ๐ฌ −๐Ÿ
๐Ÿ๐Ÿ๐Ÿ”
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
4๐‘๐‘š
8
7๐‘๐‘š
5
๐œƒ
7
82
72
+ 52
=
− 2 × 7 × 5 × cos ๐œƒ
64 = 74 − 70 cos ๐œƒ
10 = 70 cos ๐œƒ
1
?
cos ๐œƒ =
7
1
๐œƒ = cos −1
= ๐Ÿ–๐Ÿ. ๐Ÿ•๐Ÿ—°
7
9๐‘๐‘š
๐œƒ
42 = 72 + 92 − 2 × 7 × 9 × cos ๐œƒ
16 = 130 − 126 cos ๐œƒ
114 = 126 cos ๐œƒ
114
?
cos ๐œƒ =
126
114
−1
๐œƒ = cos
= ๐Ÿ๐Ÿ“. ๐Ÿ๐Ÿ°
126
Exercise 3
1
2
7
12
6
3
5.2
๐œƒ
๐›ฝ
11
5
๐œƒ
13.2
6
๐œƒ = 71.4°
?
๐›ฝ = 92.5°
?
8
๐œƒ = 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
๐‘ฅ
Sine rule
twice
Given there is just one angle involved,
you might attempt to use the cosine
rule:
๐Ÿ‘๐Ÿ = ๐’™๐Ÿ + ๐Ÿ’๐Ÿ − ๐Ÿ?× ๐’™ × ๐Ÿ’ × ๐œ๐จ๐ฌ ๐Ÿ‘๐Ÿ
๐Ÿ— = ๐’™๐Ÿ + ๐Ÿ๐Ÿ” − ๐Ÿ–๐’™ ๐œ๐จ๐ฌ ๐Ÿ‘๐Ÿ
This is a quadratic equation!
It’s possible to solve this using the
quadratic formula (using ๐’‚ = ๐Ÿ, ๐’ƒ =
− ๐Ÿ– ๐œ๐จ๐ฌ ๐Ÿ‘๐Ÿ , ๐’„ = ๐Ÿ•). 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 ๐ด sin 32
=
4 ? 3
๐ด = 44.9556°
Sine rule
twice
?
๐Ÿ๐Ÿ–๐ŸŽ − ๐Ÿ‘๐Ÿ − ๐Ÿ’๐Ÿ’. ๐Ÿ—๐Ÿ“๐Ÿ“๐Ÿ” = ๐Ÿ๐ŸŽ๐Ÿ‘. ๐ŸŽ๐Ÿ’๐Ÿ’๐Ÿ’
32°
3
๐‘ฅ
3: Using the sine rule a second
time allows us to find ๐‘ฅ
๐‘ฅ
3
=
sin 103.0444? sin 32
๐‘ฅ = 5.52 ๐‘ก๐‘œ 3๐‘ ๐‘“
Test Your Understanding
9
?
๐‘ฆ = 6.97
๐‘ฆ
61°
10
3
4
?
๐‘ฆ = 5.01
53°
๐‘ฆ
Area of Non Right-Angled Triangles
3cm
Area = 0.5 x 3 x 7 x sin(59)
= 9.00cm2?
59°
7cm
!
Area =
1
๐‘Ž
2
๐‘ sin ๐ถ
Where C is the angle wedged between two sides a and b.
Test Your Understanding
1
๐ด = × 6.97 × 10 × ๐‘ ๐‘–๐‘›61
?
2
= 30.48
9
6.97
61°
10
5
1
๐ด = × 5 × 5 × sin 60
2
?
25 3
=
4
5
5
Harder Examples
Q1 (Edexcel June 2014)
Q2
6
7
8
Finding angle ∠๐ด๐ถ๐ท:
sin ๐ท sin 100
=
9
11
๐ท = 53.6829°
? − 53.6829 = 26.3171°
∠๐ด๐ถ๐ท = 180 − 100
1
๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ Δ = × 9 × 11 × sin 26.3171 = 21.945
2
Area of ๐ด๐ต๐ถ๐ท = 2 × 21.945 = 43.9 ๐‘ก๐‘œ 3๐‘ ๐‘“
Using cosine rule to find angle
opposite 8:
21
cos ๐œƒ =
84
?
๐œƒ = 75.52249°
๐ด๐‘Ÿ๐‘’๐‘Ž
1
= × 6 × 7 × sin 75.5 … = ๐Ÿ๐ŸŽ. ๐Ÿ‘
2
Exercise 4
Q1
3
5
100°
Q3
Q2
1
3.6
1 3.8
1
๐ด๐‘Ÿ๐‘’๐‘Ž =
Q5
5.2
3
=?
0.433
4
70°
Area = 9.04?
Q7
Q6
64°
2cm
Area = 8.03?
110°
Area = 3.11๐‘๐‘š
? 2
49°
๐ด๐‘Ÿ๐‘’๐‘Ž = 29.25๐‘๐‘š
? 2
5
75°
8
Area = 7.39?
Q4
Q8
๐‘ƒ is the midpoint of ๐ด๐ต and ๐‘„ the
midpoint of ๐ด๐ถ. ๐ด๐‘ƒ๐‘„ is a sector
of a circle. Find the shaded area.
1
1
× 62 × sin 60 − ๐œ‹ 32 = 10.9๐‘๐‘š2
2
6
?
4.2m
3m
5.3m
Area = 6.29๐‘š
?2
Segment Area
๐ด
๐‘‚
๐‘‚๐ด๐ต is a sector of a circle, centred at ๐‘‚.
Determine the area of the shaded segment.
70
๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘ ๐‘’๐‘๐‘ก๐‘œ๐‘Ÿ =
× ๐œ‹ × 102 ?= 61.0865๐‘๐‘š2
360
1
๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘ก๐‘Ÿ๐‘–๐‘Ž๐‘›๐‘”๐‘™๐‘’ = × 102 × sin 70?= 46.9846๐‘๐‘š2
2
70°
๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐‘ ๐‘’๐‘”๐‘š๐‘’๐‘›๐‘ก = 61.0865 − 46.9846
= 14.1 ๐‘ก๐‘œ 3๐‘ ๐‘“ ?
๐ต
Test Your Understanding
๐ด = 119๐‘š2?
๐ด = 3๐œ‹ − 9?
Exercise 5 - Mixed Exercises
Q1
Q2
27
80°
๐‘ฅ
40°
a) ๐‘ฅ = 41.37?
b) ๐ด๐‘Ÿ๐‘’๐‘Ž = 483.63
?
Q5
Q3
8
๐‘ฆ
30°
130°
11
60๐‘š
18
๐›ผ = 17.79°
?
๐‘ง = 26.67?
๐ด๐‘Ÿ๐‘’๐‘Ž = 73.33
?
10
๐‘ฆ = 10.45?
๐ด๐‘Ÿ๐‘’๐‘Ž = 37.59
?
4.6
Q6
15
5
7
52°
6๐‘๐‘š
12
๐œƒ = 122.8°
?
๐‘ƒ๐‘’๐‘Ÿ๐‘–๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ
= 286.5๐‘š
?
Q8
Q7
๐œƒ
๐‘„๐‘… = 12.6๐‘๐‘š
?
๐‘ง
๐›ผ
70°
90๐‘š
Q4
61°
๐‘ฅ
๐ด๐‘Ÿ๐‘’๐‘Ž = 2.15๐‘๐‘š2
?
๐‘ฅ = 7.89
๐ด๐‘Ÿ๐‘’๐‘Ž = 17.25
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