1. 6.2
LAW OF COSINES

2. 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 the area of a triangle.

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

4. Let's consider types of triangles with the three pieces of information shown below.
We can't use the Law of Sines on these because we don't have an angle and a side opposite it. We need another method for SAS and SSS triangles.
SAS
AAA
You may have a side, an angle, and then another side
You may have all three angles.
This case doesn't determine a triangle because similar triangles have the same angles and shape but "blown up" or "shrunk down"
SSS
You may have all three sides

5. Triangle Side Length Restriction
In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
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6. Proof of the Law of Cosines
Prove that c2= a2 + b2 – 2ab cos C
In triangle CBD,
cos C = x / a
Then, x= a cos C (Eq #1)
Using Pythagorean Theorem
h2 = a2 - x2 (Eq #2)
In triangle BDA,
c2 = h2 + (b – x)2
c2 = h2 + b2 – 2bx + x2 (Eq #3)
Substitute with h2 Eq #2
c2 = a2 - x2 + b2 – 2bx + x2
Combine like Terms
c2 = a2 + b2 – 2bx
Finally, substitute x with Eq #1
c2 = a2 + b2 – 2ba cos C
Therefore: c2 = a2 + b2 – 2ab cos C C A B c
a h b x D
b - x
Prove:
c2 = a2 + b2 – 2ab cos C

7. Introduction

8. Using the Law of Cosines to Solve a Triangle (SAS)
Example Solve triangle ABC if
A = 42.3°, b = 12.9 meters, and
c = 15.4 meters.

9. Using the Law of Cosines to Solve a Triangle (SAS)
B must be the smaller of the two remaining angles
since it is opposite the shorter of the two sides b and c.
Therefore, it cannot be obtuse. We use the Law of
Sines to find B.
10.47 m
𝟖𝟏.𝟕°
𝟓𝟔.𝟎°
Caution If we had chosen to find C rather than B, we would not have known whether C equals 81.7° or its supplement, 98.3°.

10. Law of Cosines
Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is,
cos  > for 0 <  < 90
cos  < for 90 <  < 180.
If the largest angle is acute, the remaining two angles are acute also.
Acute
Obtuse

11. Using the Law of Cosines to Solve a Triangle (SAS)
Step 1: Use the Law of Cosines to find the side opposite the given angle.
Step 2: Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute.
Step 3: Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180°.

12. Using the Law of Cosines to Solve a Triangle (SSS)
Example Solve triangle ABC if a = 9.47 ft, b =15.9 ft, and c = 21.1 ft.
Solution We solve for C, the largest angle, first. If
cos C < 0, then C will be obtuse. C
𝑏=15.9
109.9°
𝑎=9.47 B A
𝑐=21.1

13. Using the Law of Cosines to Solve a Triangle (SSS)
We will use the Law of Sines to find B.
sin 𝐵 = sin
sin 𝐵= 15.9∙ sin
sin 𝐵=
sin − =45.1° C
𝑏=15.9
109.9°
𝑎=9.47
𝐴=180°−109.9°−45.1°≈25.0°
45.1°
25.0° B A
𝑐=21.1

14. Using the Law of Cosines to Solve a Triangle (SSS)
Use the Law of Cosines to find the angle opposite the longest side.
Use the Law of Sines to find either of the two remaining acute angles.
Find the third angle by subtracting the measures of the angles found in steps 1 and 2 from 180°.

15. Example 1
USING THE LAW OF COSINES IN AN APPLICATION (SAS)
A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that AC = 259 m, BC = 423 m, and angle ACB measures 132°40′. Find the distance AB.
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16. The distance between the two points is about 628 m.
Example 1
USING THE LAW OF COSINES IN AN APPLICATION (SAS)
Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.
𝐴𝐵 2 = 𝐴𝐶 𝐵𝐶 2 −2 𝐴𝐶 𝐵𝐶 cos 𝐶
𝐴𝐵 2 = 𝟐𝟓𝟗 𝟐 + 𝟒𝟐𝟑 𝟐 −2 𝟐𝟓𝟗 𝟒𝟐𝟑 cos 𝟏𝟑𝟐°𝟒𝟎′
The distance between the two points is about 628 m.
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17. Example 3 – An Application of the Law of Cosines
The pitcher’s mound on a women’s softball field is 43 feet
from home plate and the distance between the bases is 60
feet, as shown in Figure (The pitcher’s mound is not
halfway between home plate and second base.) How far is
the pitcher’s mound from first base?
Figure 6.13

18. Example 3 – Solution
In triangle HPF, H = 45 (line HP bisects the right angle
at H), f = 43, and p = 60.
Using the Law of Cosines for this SAS case, you have
h2 = f 2 + p2 – 2fp cos H
= – 2(43)(60) cos 45 
So, the approximate distance from the
pitcher’s mound to first base is
 feet.

19. Area Formula
The law of cosines can be used to derive a formula for the area of a triangle given the lengths of three sides known as Heron’s Formula.
Heron’s Formula
If a triangle has sides of lengths a, b, and c
and if the semiperimeter is
Then the area of the triangle is

20. Example 5 – Using Heron’s Area Formula
Find the area of a triangle having sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters.
Solution:
Because s = (a + b + c)/2
= 168/2
= 84,
Heron’s Area Formula yields
 square meters.

21. Example 5
USING HERON’S FORMULA TO FIND AN AREA (SSS)
The distance “as the crow flies” from Los Angeles to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of Earth.)
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22. Using Heron’s formula, the area  is
Example 5
USING HERON’S FORMULA TO FIND AN AREA (SSS) (continued)
The semiperimeter s is
Using Heron’s formula, the area  is
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23. When to use the Law of Sines and the Law of Cosines
Four possible cases can occur when solving an oblique triangle.
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