Objective: To apply the Law of Cosines for finding the length of a missing side of a triangle.

Lesson 18 Law of Cosines

• When solving an oblique triangle, if two sides AND the angle BETWEEN then is know, the Law of Cosines allows us to find the third side. If all 3 sides are known we can find an angle. The equations are as follows:

The Law of Cosines

2 2 2

2 2 2

2 2 2

1) a b c 2bc cos( )

2) b a c 2ac cos( )

3) c a b 2ab cos( )

• The formula for the Law of Cosines makes use of three sides and the angle opposite one of those sides. We can use the Law of Cosines:

• a. if we know two sides and the included angle, or

• b. if we know all three sides of a triangle.

General Strategies for Using the Law of Cosines

87.0°15.017.0

c

From the model, we need to determine c, , and . We start by applying the law of cosines.

Two sides and one angles are known.SAS

To solve for the missing side in this model, we use the form:

2 2 2 2 cosc a b ab In this form, is the angle between a and b, and c is the side opposite .

87.0°15.017.0

c

ab

Using the relationship

c2 = a2 + b2 – 2ab cos

We get

c2 = 15.02 + 17.02 – 2(15.0)(17.0)cos 87.0°

= 225 + 289 – 510(0.0523)

= 487.327

So c = 22.08

Now, since we know the measure of one angle and the length of the side opposite it, we can use the Law of Sines to complete the problem.

sin87.0 sin22.1 15.0

andsin87.0 sin22.1 17.0

This gives42.7 and 50.2

Note that due to round-off error, the angles do not add up to exactly 180°.

Three sides are known.

SSS

31.4 23.2

38.6

In this figure, we need to find the three angles, , , and .

To solve a triangle when all three sides are known we must first find one angle using the Law of Cosines.

We must isolate and solve for the cosine of the angle we are seeking, then use the inverse cosine to find the angle.

We do this by rewriting the Law of Cosines equation to the following form:

2 2 2

cos2

b c abc

In this form, the square being subtracted is the square of the side opposite the angle we are looking for.

31.4 23.2

38.6

Angle to look for

Side to square and subtract

We start by finding cos .

2 2 231.4 38.6 23.2cos2(31.4)(38.6)

31.4 23.2

38.6

2 2 231.4 38.6 23.2cos2(31.4)(38.6)

From the equation

we getcos 0.7993

and36.9

31.4 23.2

38.6

36.9°

Our triangle now looks like this:

Again, since we have the measure for both a side and the angle opposite it, we can use the Law of Sines to complete the solution of this triangle.

31.4 23.2

38.6

36.9°

Completing the solution we get the following:sin sin 36.931.4 23.2

andsin sin 36.938.6 23.2

Solving these two equations we get the following:

sin 0.8126 and

sin 0.9990

54.4 87.3

Again, because of round-off error, the angles do not add up to exactly 180.

• Find x using the Law of Cosines

• x² = 10² + 14² −2(10)(14)cos(44°)• x² = 296 −280cos(44) • x² = 94.6 • x = 9.73

Practice

Solving SAS Triangles• Example.

Problem: If a = 5, c = 9, and β = 25 °, find b, α and γ

Solving SSS Triangles• Example.

Problem: If a = 7, b = 4, and c = 8, find α, β andγ

• If the angle that we know is not between the two sides, then there will be an ambiguous case.

• (There is no Angle-Side-Side theorem)

The Ambiguous Case

10 5

x25o

10 5

x25o

10 5

x25o

10 5

x25o

2 2 2 2 cosc a b ab )25cos()10(2105 222 xx

xx )13.18(10025 2 075)13.18(2 xx

We need the quadratic formula to solve this (can’t easily factor)

2)75(413.1813.18 2

x

39.674.11 orx

• Find two possible lengths for the side xPractice

7030

24ox

4.735.54 orx