gcse: non-right angled triangles
DESCRIPTION
GCSE: Non-right angled triangles. Dr J Frost ([email protected]) . Last modified: 2 nd November 2013. Recap. We’ve previously been able to deal with right-angled triangles, to find the area, or missing sides and angles. 5. 6. 3. 4. Area = 15. ? . 30.96 °. ? . 5. 5. 3. - PowerPoint PPT PresentationTRANSCRIPT
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GCSE: Non-Right Angled Triangles
Dr J Frost ([email protected])www.drfrostmaths.com
Last modified: 31st August 2015
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RECAP: Right-Angled TrianglesWe’ve previously been able to deal with right-angled triangles, to find the area, or missing sides and angles.
54
3
6
5
Area = 15
5
3
30.96°
?
? ?
Using Pythagoras: Using : Using trigonometry:
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Labelling Sides of Non-Right Angle Triangles
Right-Angled Triangles:
h𝑜
𝑎
Non-Right-Angled Triangles:
𝑎𝑏
𝑐
𝐶
𝐴𝐵?
?
?
We label the sides and their corresponding OPPOSITE angles
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OVERVIEW: Finding missing sides and angles
You have You want Use#1: Two angle-side opposite pairs
Missing angle or side in one pair
Sine rule
#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 and a missing side not opposite known angle
Remaining side Sine rule twice
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The Sine Rule
65°
85°
30°
105.02
9.10
For this triangle, try calculating each side divided by the sin of its opposite angle. What do you notice in all three cases?
! Sine Rule:
c
C
b
B
a
A
?
You have You want Use#1: Two angle-side opposite pairs
Missing angle or side in one pair
Sine rule
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Examples
45°
8
11.27
85°
?
Q1
You have You want Use#1: Two angle-side opposite pairs
Missing angle or side in one pair
Sine rule
100°
8
15.7630° ?
Q2
50°
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sin𝜃5 =
sin 856
Examples
85°
6
5
56.11°?
Q3 8
When you have a missing angle, it’s better to ‘flip’ your formula to get
i.e. in general put the missing value in the numerator.
40.33°
10
126°?
Q4
sin𝜃8 =
sin 126 °10
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Test Your Understanding
𝑃𝑄
𝑅
85 °20 °
5𝑐𝑚
Determine the length .
82 °𝜃 10𝑚
12𝑚
Determine the angle .
? ?
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Exercise 1Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.
Q1
85 °
𝑥
15
40 °
𝑥=23.2?
Q2
𝑥 30 °
𝑥=53.1 °?
1610Q3
30 °
𝑦=56.4 °?
𝑦12
40 °
𝑥
10
𝑥=6.84?
Q4
20
Q5
𝛼=16.7 °?
𝛼20
1035 °Q6
𝑥=5.32?
70 °𝑥
5
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Cosine RuleThe 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
12
115°𝑥
Cosine Rule:
𝑎𝐴
𝑏
𝑐
The only angle in formula is , so label angle in diagram , label opposite side , and so on ( and can go either way).
How are sides labelled ?
Calculation?
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Sin or Cosine Rule?If you were given these exam questions, which would you use?
Sine Cosine Sine Cosine
Sine Cosine
10
15
𝑥70 °
10
15
𝑥
70 °
10
15
7𝛼
Sine Cosine
10
70 °12
𝛼
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Test Your Understanding
𝑥
7 8
e.g. 1
𝑥=6.05
47 °
e.g. 2
7
4106.4 °
𝑥
𝑥=8.99? ?
You have You want UseTwo sides known and a missing side opposite a known angle
Remaining side Cosine rule
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Exercise 2
Use the cosine rule to determine the missing angle/side. Quickly copy out the diagram first.
5 𝑥
7
60 °
𝑥=6.24?
100 °5 8
𝑦𝑦=10.14?
135 °58
70𝑥
𝑥=50.22?
6𝑥
643 °
𝑥=4.398?
Q1 Q2 Q3
Q4 Q5
10
3
8
𝑥
65 °
𝑥=9.513? 𝑥 3
54 75 °
𝑥=6.2966?
Q6
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Dealing with Missing Angles
7
49
𝛼
𝛼=25.2 °? 𝟒𝟐=𝟕𝟐+𝟗𝟐− (𝟐×𝟕×𝟗×𝐜𝐨𝐬𝜶 )
Label sides then substitute into formula.
Simplify each bit of formula.
Rearrange (I use ‘swapsie’ trick to swap thing you’re subtracting and result)
? ? ?
?
𝒂𝟐=𝒃𝟐+𝒄𝟐−𝟐𝒃𝒄𝐜𝐨𝐬 𝑨
You have You want UseAll three sides An angle Cosine rule
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Test Your Understanding
𝜃5
82=72+52− (2×7×5× cos𝜃 )?
8
7𝜃7𝑐𝑚
4𝑐𝑚
9𝑐𝑚
42=72+92− (2×7×9× cos𝜃 )?
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𝜃
76
6
𝜃=71.4 °?
12
513.2
𝛽
𝛽=92.5 °?
5.211
8
𝜃
𝜃=111.1 °?
Exercise 3
1 2 3
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Using sine rule twiceYou have You want Use#4 Two sides known and a missing side not opposite known angle
Remaining side Sine rule twice
4
𝑥
3 32 °
Given there is just one angle involved, you might attempt to use the cosine rule:
This is a quadratic equation!It’s possible to solve this using the quadratic formula (using ). However, this is a bit fiddly and not the primary method expected in the exam…
?
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Using sine rule twiceYou have You want Use#4 Two sides known and a missing side not opposite known angle
Remaining side Sine rule twice
4
𝑥
3 32 °
𝟏𝟖𝟎−𝟑𝟐−𝟒𝟒 .𝟗𝟓𝟓𝟔=𝟏𝟎𝟑 .𝟎𝟒𝟒𝟒
1: We could use the sine rule to find this angle.
2: Which means we would then know this angle.
3: Using the sine rule a second time allows us to find
𝑥sin 103.0444=
3sin 32
!
?
?
?
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Test Your Understanding
61 °
53 °
10
9
𝑦
34
𝑦
𝑦=6.97
𝑦=5.01
?
?
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Area of Non Right-Angled Triangles
Area = Where C is the angle wedged between two sides a and b.
59°
3cm
7cm
Area = 0.5 x 3 x 7 x sin(59) = 9.00cm2
!
?
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Test Your Understanding
61 ° 10
9
6.97
5 5
5
𝐴=12×6.97×10×𝑠𝑖𝑛61
𝐴=12×5×5×sin 60
?
?
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Harder Examples
Q1 (Edexcel June 2014)
Finding angle :
Area of
67
8Using cosine rule to find angle opposite 8:
? ?
Q2
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Exercise 4
64 ° 49 °
8 .7𝑐𝑚Q5
5100°
Q1
Area = 7.39? 8
3Q2
𝐴𝑟𝑒𝑎=√34
=0.433?
1 1
1 5.2
3.63.8
Q3
75°
Area = 9.04?
Q4
Area = 8.03
5
70°
? Q6
𝐴𝑟𝑒𝑎=29.25𝑐𝑚2 is the midpoint of and the midpoint of . is a sector of a circle. Find the shaded area.
( 12×62×sin 60)− 16 𝜋 (32 )=10.9𝑐𝑚2
?
?
Q7
3cm
2cm
110°
Area = ? Q8
3m
4.2m
5.3m
Area = ?
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Segment Area
𝑂
𝐴
𝐵
70 °
10𝑐𝑚 is a sector of a circle, centred at .Determine the area of the shaded segment.
? ?
?
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𝐴=3𝜋 −9?
Test Your Understanding
𝐴=119𝑚2?
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Exercise 5 - Mixed ExercisesQ1
8 0°
𝑥
27
40 °
b) ? ?
8 𝑦
10
70 °
𝑦=10.45?
Q2
?
Q3
𝛼=17.79 °?
𝛼18
1130 °𝑧
? ?
𝑄𝑅=12.6𝑐𝑚?
130 °90𝑚
60𝑚
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟?
Q4
Q5
Q6
4.615
12
𝜃
𝜃=122.8 °?
6𝑐𝑚52 °
𝐴𝑟𝑒𝑎=2.15𝑐𝑚2?
Q7
61 °57
𝑥
𝑥=7.89
Q8
? ?