9 - 4 Law of Cosines
 
 
 
To find the law of cosines, we will use the diagram above.
side a is the distance between BC.  Using the distance formula yields:
a2 = (c Cos A - b)2 + (c Sin A - 0)2
a2 = c2 Cos2 A -2bc Cos A + b2 + c2 Sin2 A
a2 = c2(Cos2 A + Sin2 A) - 2bc Cos A + b2
Since Cos2 A + Sin2 A = 1, we get:
a2 = b2 + c2 - 2bc Cos A
We have derived the law of Cosines for side a.  It has three forms which are useful:
b2 = a2 + c2 - 2ac Cos B
c2 = a2 + b2 - 2ab Cos C
The law of cosines is useful when given two sides and an included angle (SAS)
or when given all three sides
(SSS)
Demo: Law of Cosines (Manipula Math)
The following chart gives a good review of the last two chapters for non-right triangles:
 
Given                         Use
SAS law of Cosines
SSS law of Cosines
ASA or AAS law of Sines
SSA ambiquous case 
(0, 1, or 2 triangles)
 
Sample Problems
1)  Solve the triangle with sides a = 3, b = 5, c = 7.  Round to the nearest tenth of a degree.
Solution:
Find the largest angle of the triangle first.  This will be / C because the longest side is c.
Use this form:
c2 = a2 + b2 - 2ab Cos C
49 = 9 + 25 - 2(3)(5) Cos C
49 = 34 - 30 Cos C
15 = -30 Cos C
-1/2 = Cos C
Cos-1 (-1/2) = C
120 = C
/ C = 120o
Now that we have an angle, we can switch to the law of sines.(Easier to use)
Find / B
(Sin B)/b = (Sin C)/c
(Sin B)/5 = (Sin 120)/7
7 Sin B = 5 Sin 120
Sin B = (5 Sin 120)/7
Sin B =  0.6185895741317
B = 38.2
/ B = 38.2o
To find / A, subtract from 180
180 - (120 + 38.2) = 180 - 158.2 = 21.8
/ A = 21.8o
 
2)  Solve the triangle if a = 3, b = 7 and / C = 37o 
Solution
We are given two sides and the included angle.  We must find the third side.  The missing side is c.
Use the form:
c2 = a2 + b2 - 2ab Cos C
c2 = 9 + 49 - 2(3)(7) Cos 37
c2 = 58 - 42 Cos 37
c2 = 24.457308
c = 4.9
Now use the law of sines and find the smallest angle.  The smallest angle is definitely an acute angle.  The law of sines can not distinquish between acute and obtuse because both angles give a positive answer.
The smallest angle is opposite side a, the shortest side.
(Sin A)/3 = (Sin 37)/4.9
4.9 Sin A = 3 Sin 37
Sin A = (3 Sin 37)/ 4.9
Sin A = .36845817
A = 21.6o
To find / B, subtract from 180
180 - (21.6 + 37) = 121.4
/ B = 121.4o
 
3)  A farmer has a triangular field with sides 120 yards, 170 yards, and 220 yards.  Find the area of the field in square yards.  Then find the number of acres if  1 acre = 4840 square yards.
Solution:
We need to find an angle so we can use the area formula
let a = 120, b = 170, c = 220.  Find / C
c2 = a2 + b2 - 2ab Cos C
48400 = 14400 + 28900 - 40800 Cos C
5100 = -40800 Cos C
-5100/40800 = Cos C
Cos-1(-5100/40800) = C
97.2o = C
Now find the area
K = (1/2)(ab Sin C)
K = (1/2)(120)(170) Sin 97.2
K = 10120 square yards
The number of acres is found by:
10120/4840 = 2.1
2.1 acres
 
4)  Find the area of the following figure:
Solution:
Divide the quadrilateral into two triangles:
 
You can find the area of the triangle on the left:
K = (1/2)(14)(12)Sin 110
K = 78.9 square units.
To find the area of the triangle on the right, we need to find x, and z.
To find x, we can use the law of cosines
x2 = 196 + 144 - 2(14)(12) Cos 110
x2 = 454.918
x = 21.3
To find z, we first have to find y.  We can use the law of sines for the left triangle:
(Sin y)/14 = (Sin 110)/21.3
21.3 Sin y = 14 Sin 110
Sin y = (14 Sin 110)/21.3
Sin y = .6176383
y = 38.1o
To find z, subtract 130 - 38.1 = 91.9
We can now find the area of the right side triangle:
K = (1/2)(21.3)(8)(Sin 91.9)
K = 85.2
Add the two areas:
85.2 + 78.9 = 164.1
The area is 164.1 square units
We can now try some applications to navigation and surveying!!