Law of Cosines
MATH 160, Precalculus
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan
Law of Cosines
Objectives
In this lesson we will learn to:
use the Law of Cosines to solve oblique triangles (SAS or
SSS),
use the Law of Cosines to model and solve real-world
problems,
use Heron’s Area Formula to find the area of a triangle.
J. Robert Buchanan
Law of Cosines
Law of Cosines
If we are given two sides of a triangle and their included angle
(SAS) or three sides of a triangle (SSS) we can use the Law of
Cosines to solve the triangle.
Law of Cosines
Alternative Form
Standard Form
cos A =
a2 = b2 + c 2 − 2bc cos A
b2 = a2 + c 2 − 2ac cos B
cos B =
c 2 = a2 + b2 − 2ab cos C
cos C =
J. Robert Buchanan
Law of Cosines
b 2 + c 2 − a2
2bc
a2 + c 2 − b 2
2ac
2
a + b2 − c 2
2ab
Example
Suppose the three sides of a triangle have lengths a = 7,
b = 3, and c = 8, solve this triangle.
J. Robert Buchanan
Law of Cosines
Example
Suppose the three sides of a triangle have lengths a = 7,
b = 3, and c = 8, solve this triangle.
Always find the largest angle first.
cos C =
C =
cos A =
A =
cos B =
B =
1
72 + 32 − 82
=−
2(7)(3)
7
◦
98.12
32 + 82 − 72
1
=
2(3)(8)
2
◦
60.00
72 + 82 − 32
13
=
2(7)(8)
14
◦
21.88
J. Robert Buchanan
Law of Cosines
Example
Suppose two sides of a triangle have lengths a = 9, b = 4.5,
and their included angle is C = 105◦ , solve this triangle.
J. Robert Buchanan
Law of Cosines
Example
Suppose two sides of a triangle have lengths a = 9, b = 4.5,
and their included angle is C = 105◦ , solve this triangle.
c 2 = 92 + (4.5)2 − 2(9)(4.5) cos 105◦ = 122.214
c = 11.06
J. Robert Buchanan
Law of Cosines
Example
Suppose two sides of a triangle have lengths a = 9, b = 4.5,
and their included angle is C = 105◦ , solve this triangle.
c 2 = 92 + (4.5)2 − 2(9)(4.5) cos 105◦ = 122.214
c = 11.06
(4.5)2 + (11.06)2 − 92
cos A =
= 0.618581
2(4.5)(11.06)
A = 51.79◦
92 + (11.06)2 − (4.5)2
cos B =
= 0.919598
2(9)(11.06)
B = 23.21◦
J. Robert Buchanan
Law of Cosines
Application
165
Minneapolis
216
368
Albany
Phoenix
Minneapolis is due west of Albany.
1
Find the bearing of Minneapolis from Phoenix.
2
Find the bearing of Albany from Phoenix.
J. Robert Buchanan
Law of Cosines
Solution
p 2 + a2 − m 2
1652 + 2162 − 3682
=
2ap
2(216)(165)
= −0.863398
cos M =
M = 149.70◦
The bearing from Minneapolis to Phoenix is N 59.70◦ E.
J. Robert Buchanan
Law of Cosines
Solution
p 2 + a2 − m 2
1652 + 2162 − 3682
=
2ap
2(216)(165)
= −0.863398
cos M =
M = 149.70◦
The bearing from Minneapolis to Phoenix is N 59.70◦ E.
p 2 + m 2 − a2
1652 + 3682 − 2162
=
2pm
2(165)(368)
= 0.955147
cos A =
A = 17.23◦
The bearing from Phoenix to Albany is N 72.77◦ E.
J. Robert Buchanan
Law of Cosines
Heron’s Area Formula
Heron’s Area Formula
Given any triangle with sides of length a, b, and c, the area of
the triangle is
p
Area = s(s − a)(s − b)(s − c)
where s =
a+b+c
.
2
Remark: the quantity s is sometimes called the
semi-perimeter.
J. Robert Buchanan
Law of Cosines
Example
Find the area of a triangle with sides of lengths a = 31, b = 52,
and c = 28.
J. Robert Buchanan
Law of Cosines
Example
Find the area of a triangle with sides of lengths a = 31, b = 52,
and c = 28.
31 + 52 + 28
= 55.5
2
p
Area =
55.5(55.5 − 31)(55.5 − 52)(55.5 − 28) ≈ 361.768
s =
J. Robert Buchanan
Law of Cosines
Homework
Read Section 5.7.
Exercises: 1, 5, 9, 13, . . . , 41, 45
J. Robert Buchanan
Law of Cosines