Copyright © 2009 Pearson Education, Inc.
CHAPTER 8: Applications of Trigonometry
8.1 The Law of Sines
8.2 The Law of Cosines
8.3 Complex Numbers: Trigonometric Form
8.4 Polar Coordinates and Graphs
8.5 Vectors and Applications
8.6 Vector Operations
Copyright © 2009 Pearson Education, Inc.
8.1The Law of Sines
Use the law of sines to solve triangles. Find the area of any triangle given the lengths of two
sides and the measure of the included angle.
Slide 8.1 - 4Copyright © 2009 Pearson Education, Inc.
Solving Oblique Triangles
1. AAS: Two angles of a triangle and a side opposite one of them are known.
2. ASA: Two angles of a triangle and the included side are known.
Slide 8.1 - 5Copyright © 2009 Pearson Education, Inc.
Solving Oblique Triangles
3. SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution,one solution,or two solutions. The latter is known as the ambiguous case.)
Slide 8.1 - 6Copyright © 2009 Pearson Education, Inc.
Solving Oblique Triangles
4. SAS: Two sides of a triangle and the included are known.
Slide 8.1 - 7Copyright © 2009 Pearson Education, Inc.
Solving Oblique Triangles
5. SSS: All three sides of the triangle are known.
Slide 8.1 - 8Copyright © 2009 Pearson Education, Inc.
Law of Sines
The Law of Sines applies to the first three situations.
a
sin A=
bsinB
=c
sinC
The Law of Sines
In any triangle ABC,
A
B
C
a
b
c
Slide 8.1 - 9Copyright © 2009 Pearson Education, Inc.
ExampleIn , e = 4.56, E = 43º, and G = 57º. Solve the triangle.
Solution:
Draw the triangle.
E =43º e=4.56F =? f =?G=57º g=?
We have AAS.
EFGΔ
Slide 8.1 - 10Copyright © 2009 Pearson Education, Inc.
ExampleSolution continued
Find F: F = 180º – (43º + 57º) = 80º
f
sin F=
esinE
Use law of sines to find the other two sides.
f
sin80º=
4.56sin43º
f =4.56sin80ºsin43º
f ≈6.58
g
sinG=
esinE
f
sin57º=
4.56sin43º
f =4.56sin57ºsin43º
g ≈5.61
Slide 8.1 - 11Copyright © 2009 Pearson Education, Inc.
ExampleSolution continued
We have solved the triangle.
80º 6.58
5.61
43º 4.56
57º
E e
F f
G g
= == == =
Slide 8.1 - 12Copyright © 2009 Pearson Education, Inc.
Solving Triangles SSA
When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. There are various possibilities as show in the following 8 cases.
Angle B is acuteCase 1: No solutionb < c; side b is too short to reach the base. No triangle formed.
Slide 8.1 - 13Copyright © 2009 Pearson Education, Inc.
Solving Triangles SSA
Angle B is acuteCase 2: One solutionb < c; side b just reaches the base and is perpendicular to it.
Angle B is acuteCase 3: Two solutionsb < c; an arc of radius b meets the base at two points. This is called the ambiguous case.
Slide 8.1 - 14Copyright © 2009 Pearson Education, Inc.
Solving Triangles SSA
Angle B is acuteCase 4: One solutionb = c; an arc of radius b meets the base at just one point other than B.
Angle B is acuteCase 5: One solutionb > c; an arc of radius b meets the base at just one point.
Slide 8.1 - 15Copyright © 2009 Pearson Education, Inc.
Solving Triangles SSA
Angle B is obtuseCase 6: No solutionb < c; side b is too short to reach the base. No triangle formed.
Angle B is obtuseCase 7: No solutionb = c; an arc of radius b meets the base only point B. No triangle is formed.
Slide 8.1 - 16Copyright © 2009 Pearson Education, Inc.
Solving Triangles SSA
Angle B is obtuseCase 8: One solutionb > c; an arc of radius b meets the base at just one point.
The eight cases lead us the three possibilities in the SSA situation: no solution, one solution, or two solutions.
Slide 8.1 - 17Copyright © 2009 Pearson Education, Inc.
Example
In , b = 15, c = 20, and B = 29º. Solve the triangle.
Solution
Draw a triangle.
A =? a=?B=29º b=15C =? c=20
Find C.b
sin B=
csinC
ABCΔ
Slide 8.1 - 18Copyright © 2009 Pearson Education, Inc.
ExampleSolution continued
There are two angles less than 180º with a sine of 0.6464: 40º and 140º. So there are two possible solutions.
15
sin29º=
20sinC
sinC =20sin29º
15≈0.6464
Possible Solution IIf C = 40º, then
A = 180º – (29º + 40º) = 111º
Slide 8.1 - 19Copyright © 2009 Pearson Education, Inc.
These measures make a triangle as shown. Thus we have a solution.
ExampleSolution continued
Then we find a:
a
sin111º=
15sin29º
a =15sin111ºsin29º
≈29
a
sin A=
bsinB
Slide 8.1 - 20Copyright © 2009 Pearson Education, Inc.
ExampleSolution continued
Then we find a:
a
sin111º=
15sin29º
a =15sin111ºsin29º
≈29
Possible Solution IIIf C = 140º, then
A = 180º – (29º + 140º) = 11º
a
sin A=
bsinB
These measures make a triangle as shown. Thus we have a solution.
Slide 8.1 - 21Copyright © 2009 Pearson Education, Inc.
The Area of a Triangle
The area of any is one half the product of he lengths of two sides and the sine of the included angle:
K =12bcsinA=
12absinC =
12acsinB.
ABCΔ
Slide 8.1 - 22Copyright © 2009 Pearson Education, Inc.
ExampleA university landscaping architecture department is designing a garden for a triangular area in a dormitory complex. Two sides of the garden, formed by the sidewalks in front of buildings A and B, measure 172 ftand 186 ft, respectively, and together form a 53º angle. The third side of the garden,formed by the sidewalk along Crossroads Avenue, measures 160 ft. What is the area of the garden to the nearest square foot?
Slide 8.1 - 23Copyright © 2009 Pearson Education, Inc.
Example
Solution:
Use the area formula.
The area of the garden is approximately 12,775 ft2.
K =12absinC
186 ft 172 ft 53º1
sin2
K = ⋅ ⋅ ⋅
212,775 ftK ≈