MA 6.2 Law of Cosines

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6.2
Law of Cosines

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What You Should Learn

• Use the Law of Cosines to solve oblique

triangles (SSS or SAS).

• Use the Law of Cosines to model and solve

real-life problems.

• Use Heron’s Area Formula to find areas of

triangles.

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Introduction

Two cases remain in the list of conditions needed to solve
an oblique triangle—SSS and SAS.

When you are given three sides (SSS) or two sides and
their included angle (SAS), none of the ratios in the Law of
Sines would be complete. In such cases you can use the
Law of Cosines.

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Introduction

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Example 1 – Three Sides of a Triangle—SSS

Find the three angles of the triangle shown in the figure.

Solution:
It is a good idea first to find the angle opposite the longest
side—side b in this case. Using the alternative form of the
Law of Cosines, you find that

Alternative form

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Example 1 – Solution

Because cos B is negative, you know that B is an obtuse angle
given by B ≈ 116.80°.

cont’d

Substitute for a, b, and c.

Simplify.

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Example 1 – Solution

At this point it is simpler to use the Law of Sines to determine A.

Because B is obtuse and a triangle can have at most one
obtuse angle, you know that A must be acute. So,
A≈ 22.08° and

cont’d

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Applications

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Example 3 – An Application of the Law of Cosines

The pitcher’s mound on a softball field is 43 feet from home
plate. The distance between the bases is 60 feet, as shown in
Figure 6.9. (The pitcher’s mound is not halfway between home
plate and second base.) How far is the pitcher’s mound from first
base?

Figure 6.9

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Example 3 – Solution

In triangle HPF, H = 45° (line segment HP bisects the right angle
at H), f = 43, and p = 60.

Using the Law of Cosines for this SAS case, you have

h2 = f2 + p2 – 2fp cos H

= 432 + 602 – 2(43)(60) cos 45°

≈1800.33.

cont’d

Law of Cosines

Substitute for H, f,
and p.

Simplify.

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Example 3 – Solution

So, the approximate distance from the pitcher’s mound to first
base is

h

≈ 42.43 feet.

cont’d

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Heron’s Area Formula

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Heron’s Area Formula

The Law of Cosines can be used to establish the following
formula for the area of a triangle. This formula is called
Heron’s Area Formula after the Greek mathematician
Heron (ca. 100 B.C.).

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Example 5 – Using Heron’s Area Formula

Find the area of the triangle shown below.

Solution:
Because

= 84

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Example 5 – Solution

Heron’s Area Formula yields

Area

≈ 1131.89 square meters.

cont’d

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Heron’s Area Formula