Slides: GCSE Non-Right Angled Triangles
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Transcript Slides: GCSE Non-Right Angled Triangles
GCSE: Non-right angled triangles
Dr J Frost ([email protected])
Last modified: 16th November 2014
RECAP: Right-Angled Triangles
Weβve 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:
π₯ = 52 β 42
=3
30.96°
?
5
5
π
Using trigonometry:
3
tan π =
5
3
π = tanβ1
5
= 30.96°
Using π¨ = π ππ:
1
π΄πππ = × 6 × 5
2
= 15
Labelling Sides of Non-Right Angle Triangles
Right-Angled Triangles:
β
π
Non-Right-Angled Triangles:
π
πΆ?
π
π΅?
π
π
?π΄
We label the sides π, π, π and
their corresponding OPPOSITE
angles π΄, π΅, πΆ
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:
π
π?
π
=
=
sin π΄ sin π΅ sin πΆ
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
π
π
=
π¬π’π§ ππ π¬π’π§ ππ
π
π
=
π¬π’π§ πππ π¬π’π§ ππ
π π¬π’π§ ππ
π=
= ππ. ππ
π¬π’π§ ππ
π=
π π¬π’π§ πππ
= ππ. ππ
π¬π’π§ ππ
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, itβs better to βflipβ your formula to get
π¬π’π§ π¨ π¬π’π§ π©
=
π
π
i.e. in general put the missing value in the numerator.
5
Q3
Q4
126°
40.33°
?
85°
56.11°
?
6
sin π sin 85
=
5
6
5 sin 85
sin π =
6
5 sin 85
β1
π = sin
6
= 56.11°
8
10
sin π sin 126°
=
8
10
8 sin 126
sin π =
10
8 sin 126
β1
π = sin
10
= 40.33°
Test Your Understanding
π
π
85°
20°
5ππ
π
Determine the length ππ
.
π·πΉ
π
=
π¬π’π§ ππ ? π¬π’π§ ππ
π π¬π’π§ ππ
π·πΉ =
= π. ππππ
π¬π’π§ ππ
π
82°
12π
Determine the angle π.
sin π½ sin ππ
=
ππ
ππ
ππ sin ππ
?
βπ
π½ = πππ
ππ
= ππ. π°
10π
Exercise 1
Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.
Q1
Q2
15
16
10
85°
π₯
Q3
π₯
40°
12
30°
π¦
30°
20
π¦ = 56.4°
?
π₯ = 53.1°
?
π₯ = 23.2
?
π₯
Q4
Q6
Q5
40°
10
35°
70°
10
5
πΌ
20
πΌ = 16.7°
?
π₯
π₯ = 6.84
?
π₯ = 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
π
Cosine Rule:
π2 = π 2 + π 2 β 2ππ cos π΄
π΄
115°
π
12
π
π₯
The only angle in formula is π΄, so label angle
in diagram
label sides
opposite
side π, and?so on
Howπ΄, are
labelled
(π and π can go either way).
π₯ 2 = 152 + 122 β 2 × 15 × 12 × cos 115
Calculation?
π₯ 2 = 521.14257
β¦
π₯ = 22.83
Sin or Cosine Rule?
If you were given these exam questions, which would you use?
10
π₯
π₯
10
70°
70°
15
15
Sine
οΌ
ο»
Cosine
Sineο»
10
10
7
πΌ
πΌ
70°
12
15
Sineο»
Cosine
οΌ
Cosine
οΌ
Sine
οΌ
ο»
Cosine
Test Your Understanding
e.g. 1
e.g. 2
π₯
7
47°
8
π₯
4
106.4°
7
π₯ = 8.99?
π₯ = 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
π₯
58
8
100°
70
60°
7
π¦ = 10.14
?
Q6
4
π₯
10
65°
6
π₯ = 4.398
?
π₯ = 50.22
?
π₯
Q5
6
43°
π₯
π¦
π₯ = 6.24
?
Q4
135°
Q3
3
π₯ = 9.513
?
75°
5
8
π₯
π₯ = 6.2966
?
3
Dealing with Missing Angles
You have
You want
Use
All three sides
An angle
Cosine rule
ππ = ππ + ππ β πππ ππ¨π¬ π¨
7
πΌ
4
9
πΌ = 25.2°
?
ππ = ππ + ππ β ?π × π × π × ππ¨π¬ πΆ
ππ = πππ β πππ
? ππ¨π¬ πΆ
πππ ππ¨π¬ πΆ = πππ
? β ππ
πππ
ππ¨π¬ πΆ =
πππ
πππ
? = ππ. π°
πΆ = ππ¨π¬ βπ
πππ
Label sides then
substitute into
formula.
Simplify each bit of formula.
Rearrange (I use βswapsieβ
trick to swap thing youβre
subtracting and result)
Test Your Understanding
4ππ
8
7ππ
5
π
7
82
72
+ 52
=
β 2 × 7 × 5 × cos π
64 = 74 β 70 cos π
10 = 70 cos π
1
?
cos π =
7
1
π = cos β1
= ππ. ππ°
7
9ππ
π
42 = 72 + 92 β 2 × 7 × 9 × cos π
16 = 130 β 126 cos π
114 = 126 cos π
114
?
cos π =
126
114
β1
π = cos
= ππ. ππ°
126
Exercise 3
1
2
7
12
6
3
5.2
π
π½
11
5
π
13.2
6
π = 71.4°
?
π½ = 92.5°
?
8
π = 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
π₯
Sine rule
twice
Given there is just one angle involved,
you might attempt to use the cosine
rule:
ππ = ππ + ππ β π?× π × π × ππ¨π¬ ππ
π = ππ + ππ β ππ ππ¨π¬ ππ
This is a quadratic equation!
Itβs possible to solve this using the
quadratic formula (using π = π, π =
β π ππ¨π¬ ππ , π = π). However, this is a bit
fiddly and not the primary method
expected in the examβ¦
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 π΄ sin 32
=
4 ? 3
π΄ = 44.9556°
Sine rule
twice
?
πππ β ππ β ππ. ππππ = πππ. ππππ
32°
3
π₯
3: Using the sine rule a second
time allows us to find π₯
π₯
3
=
sin 103.0444? sin 32
π₯ = 5.52 π‘π 3π π
Test Your Understanding
9
?
π¦ = 6.97
π¦
61°
10
3
4
?
π¦ = 5.01
53°
π¦
Area of Non Right-Angled Triangles
3cm
Area = 0.5 x 3 x 7 x sin(59)
= 9.00cm2?
59°
7cm
!
Area =
1
π
2
π sin πΆ
Where C is the angle wedged between two sides a and b.
Test Your Understanding
1
π΄ = × 6.97 × 10 × π ππ61
?
2
= 30.48
9
6.97
61°
10
5
1
π΄ = × 5 × 5 × sin 60
2
?
25 3
=
4
5
5
Harder Examples
Q1 (Edexcel June 2014)
Q2
6
7
8
Finding angle β π΄πΆπ·:
sin π· sin 100
=
9
11
π· = 53.6829°
? β 53.6829 = 26.3171°
β π΄πΆπ· = 180 β 100
1
π΄πππ ππ Ξ = × 9 × 11 × sin 26.3171 = 21.945
2
Area of π΄π΅πΆπ· = 2 × 21.945 = 43.9 π‘π 3π π
Using cosine rule to find angle
opposite 8:
21
cos π =
84
?
π = 75.52249°
π΄πππ
1
= × 6 × 7 × sin 75.5 β¦ = ππ. π
2
Exercise 4
Q1
3
5
100°
Q3
Q2
1
3.6
1 3.8
1
π΄πππ =
Q5
5.2
3
=?
0.433
4
70°
Area = 9.04?
Q7
Q6
64°
2cm
Area = 8.03?
110°
Area = 3.11ππ
? 2
49°
π΄πππ = 29.25ππ
? 2
5
75°
8
Area = 7.39?
Q4
Q8
π is the midpoint of π΄π΅ and π the
midpoint of π΄πΆ. π΄ππ is a sector
of a circle. Find the shaded area.
1
1
× 62 × sin 60 β π 32 = 10.9ππ2
2
6
?
4.2m
3m
5.3m
Area = 6.29π
?2
Segment Area
π΄
π
ππ΄π΅ is a sector of a circle, centred at π.
Determine the area of the shaded segment.
70
π΄πππ ππ π πππ‘ππ =
× π × 102 ?= 61.0865ππ2
360
1
π΄πππ ππ π‘πππππππ = × 102 × sin 70?= 46.9846ππ2
2
70°
π΄πππ ππ π ππππππ‘ = 61.0865 β 46.9846
= 14.1 π‘π 3π π ?
π΅
Test Your Understanding
π΄ = 119π2?
π΄ = 3π β 9?
Exercise 5 - Mixed Exercises
Q1
Q2
27
80°
π₯
40°
a) π₯ = 41.37?
b) π΄πππ = 483.63
?
Q5
Q3
8
π¦
30°
130°
11
60π
18
πΌ = 17.79°
?
π§ = 26.67?
π΄πππ = 73.33
?
10
π¦ = 10.45?
π΄πππ = 37.59
?
4.6
Q6
15
5
7
52°
6ππ
12
π = 122.8°
?
πππππππ‘ππ
= 286.5π
?
Q8
Q7
π
ππ
= 12.6ππ
?
π§
πΌ
70°
90π
Q4
61°
π₯
π΄πππ = 2.15ππ2
?
π₯ = 7.89
π΄πππ = 17.25