Download Geometry_Grade 912 Non Right Angled Triangles

Non Right Angled Triangles
Non Right
Angled Triangles
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Non
Right
Angled
Triangles
NON
RIGHT
ANGLED
TRIANGLES
sin i , cos i and tan i are also useful in non-right angled triangles. This unit will show you how to use
these trigonometric ratios to find the sizes of angles and length of sides.
Try to answer these questions now, before working through the chapter.
I used to think:
How are sides labeled in relation to the labeling angles of a non right angled triangle? Draw and label a triangle.
The "cosine rule" is used when two sides of the triangle and the angle between them is given
What is this rule?
Answer these questions after you have worked through the chapter.
But now I think:
How are sides labeled in relation to the labeling angles of a non right angled triangle? Draw and label a triangle.
The "cosine rule" is used when two sides of the triangle and the angle between them is given
What is this rule?
What do I know now that I didn’t know before?
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Non Right Angled Triangles
Basics
Labeling Triangles
Everyone labels triangles the same way.
Each angle is labeled with a capital letter, and their opposite sides are labeled with the same letter, but non-capital.
For TABC
A
Opposite to C
B
Opposite to B
b
c
C
a
Opposite to A
For TLMN
Opposite to m
Opposite to n
l
M
N
n
m
L
Opposite to l
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Non Right Angled Triangles
Questions
Basics
1. Label the sides in the following triangles.
a
b
c
C
X
A
J
Y
B
H
K
W
2. Label the angles in the following the triangles.
a
b
c
m
a
c
q
p
f
g
b
r
3. Complete the labeling of the following triangles.
a
T
b
c
M
k
u
L
s
B
a
C
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Knowing More
So far we've used trigonometry in right-angled triangles, but we can actually use it to find sides and angles in
any triangle!
The Area Rule
TABC is not right-angled. The perpendicular height, ℎ, is drawn in.
Area = 1 # base # height
2
1
Area = # a # h
2
A
c
We can see: h = sin C
b
` h = b sin C
b
h
B
C
a
` Area = 1 # a # b sin C
2
`
Area = 1 ab sin C
2
So the area is half the product of two sides and the sine of their interior angle. Any two sides and their interior
angle will work:
Area = 1 ab sin C
2
Area = 1 bc sin A
2
or
or
Area = 1 ac sin B
2
Find the area of TDEF to 1 decimal place
d and f are given, so we will use Area = 1 df sin E
2
Area TDEF = 1 df sin E
2
1
= ^4h^10h sin 100c
2
F
4 cm
100c
D
10 cm
= 19.696f . 19.7cm2 ^1 decimal placeh
E
Find the area of TABC to 1 decimal place
a and b are given, so we will use Area = 1 ab sin C
2
A
83.6c
+C = 180c - 83.6c - 45.6c = 38c
18
B
4
45.6c
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C
(write this in TABC )
Area TABC = 1 ab sin C
2
1
= ^21h^18h sin 38c
2
= 116.36.. . 116.4 cm2 (1 decimal place)
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Non Right Angled Triangles
Questions
Knowing More
1. Find the area of the following triangles to 1 decimal place.
a
b
A
44c
14.3 cm
17 cm
A
35c
B
10 cm
16.4 cm
78c
B
C
C
2. TABC has +B = 103.3c, +C = 33.7c, b = 21.4 cm and c = 12.2 cm.
a
Draw the triangle.
b
Find +A to nearest degree.
c
Find the area of the triangle to 1 decimal place.
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Questions
3. Can you find the areas of the following shapes to the nearest square cm?
a
E
10 cm
A
31c
6 cm
C
D
13 cm
B
b
Hint: Draw in Diagonal AC
D
43.2c
23 cm
C
19.5 cm
12 cm
A
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93c
10 cm
B
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Knowing More
Non Right Angled Triangles
Questions
Knowing More
4. Luke wants to make a kite in the shape below. How many square cm of material will he need?
B
90 cm
C
84c
68c
48c
A
160 cm
44c
D
5. The area of the triangle TPQR is 38.6 cm2. Find +Q .
(Hint: Find +P first)
Q
10 cm
P
60.4c
8 cm
R
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Using Our Knowledge
The Sine Rule
C
In any triangle TABC we have the formulas:
a
b
A
sin A = sin B = sin C
a
b
c
a = b = c
sin A
sin B
sin C
or
B
c
The proof is different depending on whether the triangle is acute or obtuse.
If TABC is an acute triangle
If TABC is an obtuse triangle
C
b
C
a
h
A
h
180c - A
B
D
a
b
D
c
In TACD:
h = sin A
b
` h = b sin A
In TBCD:
In TBCD:
h = sin B
a
` h = a sin B
In TACD:
A
h = sin ^180c - Ah
b
` h = b sin ^180c - Ah
` h = b sin A
Remember
sin ^180c -θ h = sin θ
sin A = sin B
a
b
` b sin A = a sin B (Both equal h)
`
If we drew the altitude from B to AC then we
could show
sin A = sin B
a
b
If we drew the altitude from B to AC then we
could show
sin A = sin C
a
c
sin A = sin C
a
c
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B
h = sin B
a
` h = a sin B
` b sin A = a sin B (Both equal h)
`
c
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Non Right Angled Triangles
Using Our Knowledge
Find the value of a in TABC to 2 decimal places
Using the Sine Rule:
A
7 cm
28c
81c
B
`
10 cm
a = b
sin A
sin B
a = 10
sin 28c
sin 81c
` a = 10 sin 28c
sin 81c
10^0.469fh
a=
= 4.753f . 4.75 cm (2 decimal places)
0.987f
a
C
Find the value of +E in TDEF to the nearest degree
E
Using the Sine Rule:
11 cm
36c
36c
sin E = sin D
e
d
` sin E = sin 36c
13
11
F
` sin E = 13 sin 36c = 0.694f
11
13 cm
` E = sin-1 ^0.694fh = 43.999fc . 44c (nearest degree)
D
Use the diagram below to answer these questions
a
L
sin +LPN = sin +LNP
LN
LP
sin
LPN
sin
55
+
c
`
=
10
9
` sin +LPN = 10 sin 55c = 0.910...
9
10 cm
32c
N
55c
Find +LPN to the nearest degree.
9 cm
P
7 cm
M
b
` +LPN = sin-1 (0.910...) = 65.52...c . 66c (nearest degree)
Find the length of LM to the nearest cm.
First, find +LPM
+LPM = 180c - +LPN
= 180c - 66c = 114c
Now, use the sine rule
LM
PM
=
sin +LPM
sin +PLM
` LM = 7
sin 114c
sin 32c
` LM = 7 sin 114c = 12.067... . 12 cm (nearest cm)
sin 32c
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Questions
Using Our Knowledge
The Sine Rule
1. Find the length y in the triangles below to 1 decimal place.
a
b
y
37c
19 cm
61c
12 cm
y
102c
42c
2. Find the size of angle a in the following triangles:
a
b
29 cm
a
45c
35 cm
120c
a
46 cm
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62 cm
Non Right Angled Triangles
Questions
Using Our Knowledge
3. A ship has sails as drawn in the diagram below:
A
D
60c 45c
25c
B
a
Find the length of AC.
b
Find the lengths of DC and AD.
85c
20 m
C
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Using Our Knowledge
The Cosine Rule
In any triangle TABC we have the formulas
C
a2 = b2 + c2 - 2bc cos A
or
a
b
b2 = a2 + c2 - 2ac cos B
or
A
B
c
c2 = a2 + b2 - 2ab cos C
The proof is different depending on if the triangle is acute or obtuse.
If TABC is an acute triangle
If TABC is an obtuse triangle
C
C
a
b
h
h
A
B
D
x
b
a
D
B
12
h2 + x2 = b2
(Pythagoras)
cos A = x
b
` x = b cos A
In TACD :
and
Subsitute
Subsitute
In TCBD : a2 = h2 + ^ x - ch2
(Pythagoras)
2
2
2
= h + x + c - 2cx
Subsitute
Subsitute
(Pythagoras)
In TCBD : a2 = h2 + ^c - xh2
2
2
2
= h + x + c - 2cx
and
A
x
c
In TACD :
c
(x - c)
(c-x)
h2 + x2 = b2
(Pythagoras)
cos A = x
b
` x = b cos A
a2 = b2 + c2 - 2cb cos A
a2 = b2 + c2 - 2cb cos A
If the ℎ is drawn from other corners we can show
If the ℎ is drawn from other corners we can show
b2 = a2 + c2 - 2ac cos B
b2 = a2 + c2 - 2ac cos B
or
or
c2 = a2 + b2 - 2ab cos C
c2 = a2 + b2 - 2ab cos C
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Non Right Angled Triangles
Using Our Knowledge
Find the value of b in TABC to 2 decimal places
Using the Cosine Rule:
8 cm
A
30c
b
B
b2 = a2 + c2 - 2ac cos B
` b2 = 52 + 82 - 2^5 h^8 h cos 30c
5 cm
` b2 = 25 + 64 - 80^0.866fh
` b2 = 19.717f
C
` b = 19.717...
= 4.440.. . 4.44cm (2 decimal places)
Find the value of +Q to the nearest degree
Using the Cosine Rule:
P
15 cm
q2 = p2 + r2 - 2pr cos Q
` 62 = 102 + 152 - 2^10h^15h cos Q
6 cm
R
Q
10 cm
Make cos Q the subject
2^10h^15h cos Q = 102 + 152 - 62
300 cos Q = 100 + 225 - 36 = 289
` cos Q = 289 = 0.963f
300
Solve for +Q
+Q = cos-1 ^0.963..h
= 15.56fc . 16c (nearest degree)
Find +A in the following to 1 decimal place
BC2 = DB2 + DC2 - 2 # DB # DC # cos D
D
BC2 = 252 + 212 - 2^25h^21h cos 49.2c
49.2c
25 cm
This is the
cosine rule
BC2 = 379.908f
21 cm
BC = 19.49fcm
Write this in
the diagram
C
BC2 = AB2 + AC2 - 2 # AB # AC # cos A
B
18 cm
16 cm
A
^19.49fh2 = 162 + 182 - 2^16h^18h cos A
` cos A =
162 + 182 - ^19.49fh2
= 0.347f
2^16h^18h
` +A = cos-1 ^0.347fh = 69.67fc . 69.7c
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(1 decimal place)
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Questions
Using Our Knowledge
The Cosine Rule
4. Find the length of the missing side x in each of the following to 1 decimal place.
a
b
D
P
73c
x
5 cm
29 cm
x
56c
E
Q
34c
R
15 cm
F
35 cm
5. Find the value of the angle b in each of the following to the nearest degree.
A
a
b
W
50 cm
b
C
47 cm
b
55 m
39 m
61 cm
Y
B
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23 m
X
Non Right Angled Triangles
Questions
Using Our Knowledge
6. A canyon has the following dimensions:
N
46 m
L
40c
62 m
M
30c
50 m
P
To walk around this canyon, would it be shorter to walk along LNM or LPM?
Use working to support your answer.
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Questions
Using Our Knowledge
Both Rules
7. Use the sine and cosine rules to find the missing sides and angles in these triangles to 2 decimal places.
a
A
c
b
56c
B
67c
10 cm
C
Q
b
38 cm
55 cm
P
114c
R
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Non Right Angled Triangles
Thinking More
An Ambiguous Case For The Sine Rule
Remember that sini = sin(180c-i ). So when the sine rule is used to find an angle (using sin-1 ) two answers
are possible.
For example if sini = 0.5 then i = 30c or 150c (180c - 30c).
It is our job to determine whether one or both answers are valid. This depends on whether a third angle exists or not.
Find +B (to 1 decimal place) in TABC if +C = 50c, c = 14 cm and b = 17.2 cm
According to the sine rule
sin B = sin C
b
c
` sin B = sin 50c
17.2
14
17
` sin B = .2 # sin 50c = 0.9411
14
` +B = sin-1 ^0.9411h
or
` +B = 70.2c
+B = 180c - 70.2c
= 109.8c
Add each answer to the given angle. If this sum is less than 180c then the answer is valid because the third
angle exists.
The given angle in the question is +C = 50c
` +B + +C = 70.2c + 50c
= 120.2c 1 180c
+B + +C = 109.8c + 50c
= 159.8c 1 180c
or
` valid
` valid
Both sums are less than 180c, and so both answers for +B are valid. So TABC may look like either of
these triangles:
A
A
14 cm
14 cm
17.2 cm
or
17.2 cm
B
70.2c
109.8c
B
50c
50c
C
C
A case where both answers are valid (like above) is said to be "ambiguous."
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Thinking More
Determining if Answers are Valid
Of the two possible answers, one will always be obtuse (90c to 180c) and the other will always be acute ^0c to 90ch .
• So if you know the triangle is acute, then the obtuse angle is invalid.
• If you know the missing angle is obtuse, then the acute angle is invalid.
• The sum of angles in a triangle is 180c. So if the sum of a possible answer and the given angle is 180c or more it
means that the third angle in the triangle is 0c or negative. This is impossible, so the answer would be invalid.
Find +L (to 1 decimal place) in TLMN if +N = 80c, n = 48.6 cm and l = 40.5 cm
According to the sine rule
sin L = sin N
l
n
` sin L = sin 80c
40.5
48.6
40
` sin L = .5 # sin 80c = 0.8206f
48.6
` +L = sin-1 ^0.8206fh
` +L = 55.1522fc . 55.2c
or
+L = 180c - 55.1522
= 124.8477 . 124.8c
or
+L + +N = 124.8c + 80c
= 204.8c 2 180c
` invalid
Add each answer to the given angle ^+N = 80ch .
` +L + +N = 55.2c + 80c
= 135.2c 1 180c
` valid
So only the first answer is valid and +L = 55.2c. This is not an ambiguous case.
+L = 55.2c and TLMN looks like this
M
40.5cm
48.6cm
80c
55.2c
N
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Non Right Angled Triangles
Questions
Thinking More
1. In triangle TABC a = 19 cm, b = 15 cm and +B = 20c.
a
Find two possible values for +A to the nearest degree.
b
Check if these answers are valid. Is this an ambiguous case?
c
Draw a rough sketch of the triangle(s).
d
Find the third angle in the triangle(s) and add them to the diagram in c .
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Questions
2. In TPQR , +R = 40c, p = 20.3 cm and r = 13.9 .
a
Find the two possible values for +P to the nearest degree.
b
Is this an ambiguous case (are both angles valid)?
c
Which answer is valid if TPQR is acute? Which answer is invalid?
d
Which answer is valid if +P is obtuse? Which answer is invalid?
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Thinking More
Non Right Angled Triangles
Questions
Thinking More
3. In triangle TJKL, +K = 70c, k = 51.4cm and l = 42cm .
a
Find the valid size(s) of +L .
b
Find +J .
c
Find the length of j.
d
Draw TJKL .
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Non Right Angled Triangles
Basics:
Basics:
C
1. a
A
3. a
b
a
A
b
Answers
B
c
w
X
b
c
B
T
b
Y
C
a
u
S
x
y
s
t
W
U
J
c
M
c
k
k
h
L
l
H
K
j
m
2. a
K
B
a
Knowing More:
c
C
1. a Area TABC = 81.5cm2
A
b
R
b
q
p
b
Area TABC = 74.2cm2
2. a
B
103.3c
P
r
C
c
f
+A = 43c
c
Area TABC = 89.0cm2
M
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21.4cm
b
F
22
33.7c
G
m
g
12.2cm
Q
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A
Non Right Angled Triangles
Thinking More:
Knowing More:
3. a Total area = 53.563 . 54cm2
b
Answers
or
1. a +A = 26c
Total area = 213cm2
Both answers are valid; therefore this is an
ambiguous case.
b
4. Total area = 12919cm2
+A = 154c
C
c
134c
19cm
5. +Q = 44.8c
1. a
b
26c
20c
B
Using Our Knowledge:
15cm
A
y = 9.2cm
154c
15cm
y = 12.7cm
20c
3. a
d
AC = 13.2m (to 1 d.p.)
B
19cm
2. a a = 35.9c
a = 40c
c
6c
C
b
A
c
or
+C = 134c
or
2. a +C = 70c
+C = 6c
+C = 110c
CD . 10.0m and AD . 6.0m
b
Both answers are valid; therefore this is an
ambiguous case.
4. a
x = 19.6cm (to 1 d.p.)
c
b
x = 12.5cm (to 1 d.p.)
If ∆PQR is acute, the angles are in the
range 0c - 90c, + P = 70c is the valid
answer and + P = 110c is invalid.
d
If ∆PQR is obtuse, the angles are in the
range 90c - 180c, + P = 110c is the valid
answer and + P = 70c is invalid.
b
5.
a
b = 49c (to nearest degree)
b
b = 21c (to nearest degree)
3. a +L = 50c
6. It would be shorter to walk along LPM
than LNM
b
+ J = 60c
c
j = 47.3cm
7. a c = 10.99cm (to 2 d.p.)
d
K
b = 9.89cm (to 2 d.p.)
b
70c
42cm
q = 27.20cm (to 2 d.p.)
+Q = 26.86c (to 2 d.p.)
J
+R = 39.14c (to 2 d.p.)
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47.3cm
60c
50c
51.4cm
L
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Non Right Angled Triangles
24
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Notes
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