Download chapter 3 additional topics in trigonometry

CHAPTER 3.2
CHAPTER 3 ADDITIONAL TOPICS IN TRIGONOMETRY
PART 2 –Law of Cosines
TRIGONOMETRY MATHEMATICS CONTENT STANDARDS:
 13.0 - Students know the law of sines and the law of cosines and apply those
laws to solve problems.
 14.0 - Students determine the area of a triangle, given one angle and the two
adjacent sides.
 19.0 - Students are adept at using trigonometry in a variety of applications
and word problems.
OBJECTIVE(S):
 Students will learn the law of cosines.
 Students will learn how to use the law of cosines to solve for a
triangle.
 Students will learn how to calculate the area of triangle using Heron’s
area formula.
 Students will learn how to apply the law of cosines to the real world.
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.
B
a
c
C
b
A
Law of Cosines
Standard Form
a2 
b2 
c2 
Alternate Form
b2  c2  a 2
cos A 
2bc
2
a  c2  b2
cos B 
2ac
2
a  b2  c2
cos C 
2ab
CHAPTER 3.2
EXAMPLE 1: Three Sides of a Triangle - SSS
Find the three angles of the triangle below:
B
a = 8 ft.
c = 14 ft.
C
A
b = 19 ft.
It is a good idea first to find the angle opposite the longest side – side b in this case.
Using the Law of Cosines, you find that
b2
=
a 2  c 2  2ac cos B
=
=
=
=

Because cos B is negative, you know that B is an obtuse angle given by B  _________.
At this point knowing that B  _________, it is simpler to use the Law of Sines to
determine A.
a
b

SinA SinB
CHAPTER 3.2
Because B is obtuse, A must be ___________, because a triangle can have at most
______ obtuse angle. So, A  _________ and

C
-
-
=
Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse.
That is
Acute
Obtuse
So, in the example above, once you found that angle B was obtuse, you knew that angles
A and C were both acute. If the largest angle is acute, the remaining two angles will be
acute also.
EXAMPLE 2: Two Sides and the Included Angle - SAS
Find the remaining two angles and the sides of the triangle below:
C
a
b = 15cm
A
c = 10 cm
Use the Law of Cosines to find the unknown side a in the figure.
a 2  b 2  c 2  2bc cos A
B
CHAPTER 3.2
Because a  _____________, you now know the ratio
a
and you can use the Law of
SinA
Sines:
a
b

SinA SinB
to solve for B.
a
b

SinA SinB
So, B = ______________________________  _______________, and C 
____________ - _______________ - _______________ = ________________.
CHAPTER 3.2
1. Solve each  .
a. a = 5.7, c = 7.6, B  240 .
B
a = 5.7
c = 7.6
C
b
A
CHAPTER 3.2
b. a = 11, b = 20, c = 15.
B
a = 11
c = 15
C
b = 20
A
DAY 1
CHAPTER 3.2
Applications
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 below (the pitcher’s mound is not
halfway between home plate and second base). How far is the pitcher’s mound from first
base?
60 ft.
60 ft.
P
h
F
f = 43 ft.
60 ft.
p = 60 ft.
45
0
H
In triangle HPF, H = 45 0 (line HP bisects the right angle at H), f = 43, and p = 60. Using
the Law of Cosines for this SAS case, you have
Therefore, the approximate distance from the pitcher’s mound to first base is
h

_________________________

_________________________
CHAPTER 3.2
EXAMPLE 4: An Application of the Law of Cosines
A ship travels 60 miles due east, then adjusts its course northward, as shown below.
After traveling 80 miles in that direction, the ship is 139 miles from its point of departure.
Describe the bearing from point B to point C?
N
W
E
S
C
b = 139 mi
a = 80 mi
A
c = 60 mi
B
You have a = 80, b = 139, and c = 60; consequently, by the Law of Cosines, you have
So, B  _______________________  ________________, and thus the bearing
measured from due north from point B to point C is ______________ - _____________
= _____________, or ________________.
CHAPTER 3.2
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 (c. 100 B.C.)
Given any triangle with sides of lengths a, b, and c, the area of the triangle is
ss  a s  bs  c
Area =
where s 
a  b  c  .
2
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.
Because s 
Area
a  b  c  = _____________ = _________, Heron’s Area Formula yields
2
=
ss  a s  bs  c
=

We have studied three different formulas for the area of triangle.
Standard Formula
Area =
Oblique Triangle
Area =
Heron’s Area Formula
Area =
CHAPTER 3.2
2.) Find the lengths of the diagonals of a parallelogram with side lengths 8 m and 5 m
and one angle being 530 .
A
8m
B
5m
D
DAY 2
5m
8m
C