The Law of Cosine

To get the Law of Cosines the thing to do is simply solve the SAS problem by using analytic geometry. I'm sure the ancients didn't do it this way, but we can, and it certainly makes it a lot easier. So what you do is first locate your triangle at a convenient place in the coordinate system. Put one vertex at the origin and one side on the x-axis to make things simplest.

Now we are ready to use the distance formula.

c2 = a2 + b2 - 2ab cos(C)

b2 = a2 + c2 - 2ac cos(B)

a2 = b2 + c2 - 2bc cos(A)

(Law of Cosines)

So now we have the Law of Cosines. Since we could have just as well done this with the letters switched around any way we wanted, we can write it in three different ways. 

Example 1. Solve the triangle with sides a = 3, b = 5, c = 7.  Round to the nearest tenth of a degree.

Givens

a = 3

b = 5

c = 7

Find the largest angle of the triangle first.  This will be Angle C because the longest side is c.

Using Pythagorean's Theorem:

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

Angle C = 120o

Now that we have an angle, we can switch to the law of sines.

Find Angle B

(Sin B)/b = (Sin C)/c

(Sin B)/5 = (Sin 1200)/7

7 Sin B = 5 Sin 1200

Sin B = (5 Sin 1200)/7

Sin B =  0.6185895741317

B = 38.20

Angle B = 38.2o

To find Angle A, subtract from 1800

1800 - (1200 + 38.20)

= 1800 - 158.20

= 21.80

Angle A = 21.8o

Example 2.  Solve the triangle if a = 3, b = 7 and Angle C = 37o 

We are given two sides and the included angle.  We must find the third side.  The missing side is c.

Using Pythagorean's Theorem:

c2 = a2 + b2 - 2ab Cos C

c2 = 9 + 49 - 2(3)(7) Cos 370

c2 = 58 - 42 Cos 370

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 distinguish 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 370)/4.9

4.9 Sin A = 3 Sin 370

Sin A = (3 Sin 370) / 4.9

Sin A = .36845817

A = 21.6o

To find Angle B, subtract from 1800,

1800 - (21.60 + 370) = 121.40

Angle B = 121.4o

Example 3.  A farmer has a triangular field with sides 120 meters, 170 meters, and 220 meters.  Find the area of the field in square meters. 

We need to find an angle so we can use the area formula

let a = 120, b = 170, c = 220.  Find Angle 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 97.20

K = 10120 square meters