confidential1 geometry law of sines and law of cosines
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CONFIDENTIAL 2
Warm up
Use the given trigonometric ratio to determine which angle of the triangle is /A.
2
1 17 cm
15 cm
8 cma) cos A = 15/17
b) sin A = 15/17
c) tan A = 1.875
1) 22) 13) 1
CONFIDENTIAL 3
Law of Sines and Law of Cosines
In this lesson, you will learn to solve any triangle. To do so, you will need to
calculate trigonometric ratios for angle measures up to 180˚.
CONFIDENTIAL 4
Finding Trigonometric Ratios for Obtuse Angles
Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.
A) sin 135˚ B) tan 98 ˚ C) cos 108 ˚
sin 135˚ = 0.71sin 135˚ = 0.71 tan 98tan 98˚= -0.72 cos cos 108˚=-0.31
CONFIDENTIAL 5
Now you try!
1) Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.
a) tan 175˚ b) cos 92˚ c) sin 160˚
1a) -0.091b) -0.031c) 0.34
CONFIDENTIAL 6
h
c
b a
C
BA
You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.
In ∆ABC, let h represent the length of the altitude from C to AB.
From the diagram, sin A = (h), and sin B = (h).
By solving for h, you find that h = bsin A
And h = a.sinB .So b sin A = a.sinB,
And (sin A/a) = (sin B/b).
You can use another altitude to show that these ratios
Equal (sin C/c).
‾b a
CONFIDENTIAL 7
The Law of Sines
Theorem 1.1
For any ABC with side lengths a, b, and c.
sin A
a =
sin B
b =
sin C
c.
ab
cA B
C
CONFIDENTIAL 8
You can use the Law of Sines to solve a triangle if you are given
• two angle measures and any side length (ASA or AAS) or
• two side length and a non-include angle measure (SSA).
CONFIDENTIAL 9
Use the Law of Sines
Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
32
105
F
D
E
Law of Sines
A) DF
sin D
EF =
sin E
DF
sin 105
18 =
sin 32
DF DFsin 105 18sin32
DF = 18sin 32
sin 105 9.9
Substitute the given values.
Cross Products Property
Divide both sides by sin 105˚.
Next page
CONFIDENTIAL 10
B) mS
sin T
RS =
sin S
RT
sin 75
7 =
sin S
5
sin S = 5 sin 75
7
mS sin-15 sin 75
7 44
75
5
7
S
T
R
Law of Sines
Substitute the given values.
Multiply both sides by 5.
Use the inverse sine function to find m/s.
CONFIDENTIAL 11
Now you try!
2) Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
L
18
7.6
4.3
10
622
67
125
4450
39
88
J
KPN
M
Y
ZX
C
B
AmX
AC
MLNP
d)c)
b)a)
2a) 22b) 1
CONFIDENTIAL 12
The Law of Sine cannot be used to solve every triangle. If you know two side lengths and the include angle measure or if you know all three side lengths. You cannot use the Law of Sines.
Instead, you can apply the Law of Cosines.
CONFIDENTIAL 13
Theorem 1.2
The Law of Cosines
C
BAc
b a
For any ABC with side lengths a, b, and c: a2 = b2 + c2 - 2bc cosA b2 = a2 + c2 - 2ac cosB
c2 = a2 + b2 - 2ab cosC
CONFIDENTIAL 14
You can use the Law of Cosines to solve a triangle if you are given
• two side lengths and the included angle measure (SAS) or
• three side lengths (SSS).
CONFIDENTIAL 15
Using the Law of CosinesFind each measure. Round lengths to the nearest tenth
and angle measures to the nearest degree.
9
14
62
C
B
A
Law of Cosines
Substitute the given values.
Simplify.
A ) BC
BC2 = AB2 + AC2 - 2(AB)(AC)cos A = 142 + 92 - 2(14)(9)cos 62
BC2 158.6932
BC 12.6 Find the square root of both sides.
Next page
CONFIDENTIAL 16
B ) mR
ST2 = RS2 + RT2 - 2(RS)(RT)cos R
92 = 42 + 72 - 2(4)(7)cos R 81 = 16 + 49 -56 cos R 16 = -56 cos R
cos R = -16
56
m R = cos-1 -16
56 107
9
74
T
S
R
Law of Cosines
Substitute the given values.
Simplify
Subtract 65 from both sides.
Solve for cos R
Use the inverse Cosine function to find m/R.
CONFIDENTIAL 17
Now you try!3) Find each measure. Round lengths to the nearest tenth
and angle measures to the nearest degree.
10.5
9.6
5.934
4
10
8
15
10
21
16
18
R
QP
Z
Y
X
L
K
JF
D
E
d) m Rc) YZ
b) m Ka ) DE
3a) 23b) 1
CONFIDENTIAL 18
Engineering Application
The Leaning Tower of Pisa is 56 m tall. In 1999, the tower made a 100˚ angle with the ground. To stabilize the tower, an engineer considered attaching a cable from the top of the tower to a point
that is 40 m from the base. How long would the cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.
40 m
100
56 m
BA
C
Law of Cosines
Substitute the given values.
Find the square root of both sides.
Step 1 Find the length of the cable. AC2 = AB2 + BC2 - 2(AB)(BC)cos B
= 402 + 562 - 2(40)(56)cos 100
AC2 5513.9438
AC 74.3 m
Simplify.
Next page
CONFIDENTIAL 19
Step 2 Find the measure of the angle the cable the cable would make.
sin A
BC =
sin B
AC
sin A
56
sin 100
74.2559
sin A 56 sin 100
74.2559
mA sin-156sin 100
74.2559 48
Law of Sines
Substitute the Calculated values.
Multiply both sides by 56.
Use the inverse sine function to find m/A.
40 m
100
56 m
BA
C
CONFIDENTIAL 20
Now you try!
4) Another engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it
make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.
CONFIDENTIAL 22
Use a calculator to find each trigonometric ratio.
Round to the nearest hundredth.
1) sin 100˚ 2) cos 167˚ 3) tan 92˚
4) tan 141˚ 5) sin 150˚ 6) cos 133˚
CONFIDENTIAL 23
Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
15
70
36
S
T
Ra) RT7)
CONFIDENTIAL 24
Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
10
67
R Q
P
a) m Q
11
131
8
CB
A
b) AB8) 9)
CONFIDENTIAL 25
10) A carpenter makes a triangular frame by joining three pieces of wood that are 20 cm, 24 cm, and 30 cm
long. What are the measures of the angles of the triangle? Round to the nearest degree.
CONFIDENTIAL 26
Finding Trigonometric Ratios for Obtuse Angles
Use a calculator to find each trigonometric ratio. Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.Round to the nearest hundredth.
A sin 135˚ B tan 98 ˚ C cos 108 ˚A sin 135˚ B tan 98 ˚ C cos 108 ˚
sin 135˚ = 0.71sin 135˚ = 0.71 tan 98tan 98˚= -0.72 cos cos 108˚=-0.31
Let’s review
CONFIDENTIAL 27
You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.
In ∆ABC, let h represent the length of the altitude from C to AB.
From the diagram, sin A = (h/b), and sin B = (h/a).
By solving for h, you find that h = bsin A
And h = asin B .So b sin A = asin B,
And (sin A/a) = (sin B/b).
You can use another altitude to show that these ratios
Equal (sin C/c).
‾
h
c
b a
C
BA
CONFIDENTIAL 28
The Law of Sines
Theorem 1.1
For any ABC with side lengths a, b, and c.
sin A
a =
sin B
b =
sin C
c.
ab
cA B
C
CONFIDENTIAL 29
You can use the Law of Sines to solve a triangle if you are given
• two angle measures and any side length (ASA or AAS) or
• two side length and a non-include angle measure (SSA).
CONFIDENTIAL 30
Use the Law of Sines
Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
32
105
F
D
E
Law of Sines
A) DF
sin D
EF =
sin E
DF
sin 105
18 =
sin 32
DF DFsin 105 18sin32
DF = 18sin 32
sin 105 9.9
Substitute the given values.
Cross Products Property
Divide both sides by sin 105˚.
Next page
CONFIDENTIAL 31
B) mS
sin T
RS =
sin S
RT
sin 75
7 =
sin S
5
sin S = 5 sin 75
7
mS sin-15 sin 75
7 44
75
5
7
S
T
R
Law of Sines
Substitute the given values.
Multiply both sides by 5.
Use the inverse sine function to find m/s.
CONFIDENTIAL 32
The Law of Sine cannot be used to solve every triangle. If you know two side lengths and the include angle measure or if you know all three side lengths. You cannot use the Law of Sines. Instead, you can apply the Law of Cosines.
CONFIDENTIAL 33
Theorem 1.2
The Law of Cosines
C
BAc
b a
For any ABC with side lengths a, b, and c: a2 = b2 + c2 - 2bc cosA b2 = a2 + c2 - 2ac cosB
c2 = a2 + b2 - 2ab cosC
CONFIDENTIAL 34
You can use the Law of Cosines to solve a triangle if you are given
• two side lengths and the included angle measure (SAS) or
• three side lengths (SSS).
CONFIDENTIAL 35
Using the Law of Cosines
Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.
9
14
62
C
B
A
Law of Cosines
Substitute the given values.
Simplify.
A ) BC
BC2 = AB2 + AC2 - 2(AB)(AC)cos A = 142 + 92 - 2(14)(9)cos 62
BC2 158.6932
BC 12.6 Find the square root of both sides.
Next page
CONFIDENTIAL 36
B ) mR
ST2 = RS2 + RT2 - 2(RS)(RT)cos R
92 = 42 + 72 - 2(4)(7)cos R 81 = 16 + 49 -56 cos R 16 = -56 cos R
cos R = -16
56
m R = cos-1 -16
56 107
9
74
T
S
R
Law of Cosines
Substitute the given values.
Simplify
Subtract 65 from both sides.
Solve for cos R
Use the inverse Cosine function to find m/R.
CONFIDENTIAL 37
Engineering Application
The Leaning Tower of Pisa is 56 m tall. In 1999, the tower made a 100˚ angle with the ground. To stabilize the tower, an engineer considered attaching a cable from the top of the tower to a point that is 40 m from the base. How long would the cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.
40 m
100
56 m
BA
C
Law of Cosines
Substitute the given values.
Find the square root of both sides.
Step 1 Find the length of the cable. AC2 = AB2 + BC2 - 2(AB)(BC)cos B
= 402 + 562 - 2(40)(56)cos 100
AC2 5513.9438
AC 74.3 m
Simplify.
Next page
CONFIDENTIAL 38
Step 2 Find the measure of the angle the cable the cable would make.
sin A
BC =
sin B
AC
sin A
56
sin 100
74.2559
sin A 56 sin 100
74.2559
mA sin-156sin 100
74.2559 48
Law of Sines
Substitute the Calculated values.
Multiply both sides by 56.
Use the inverse sine function to find m/A.
40 m
100
56 m
BA
C
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