Download Section 8.1

CHAPTER 8:
Applications of Trigonometry
8.1
8.2
8.3
8.4
8.5
8.6
The Law of Sines
The Law of Cosines
Complex Numbers: Trigonometric Form
Polar Coordinates and Graphs
Vectors and Applications
Vector Operations
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8.1
The 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.
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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.
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Slide 8.1 - 4
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.)
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Slide 8.1 - 5
Solving Oblique Triangles
4. SAS: Two sides of a
triangle and the included
are known.
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Slide 8.1 - 6
Solving Oblique Triangles
5. SSS: All three sides of the
triangle are known.
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Law of Sines
The Law of Sines applies to the first three situations.
The Law of Sines
In any triangle ABC,
B
a
b
c


sin A sin B sinC
c
A
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a
b
C
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Example
In EFG, e = 4.56, E = 43º, and G = 57º. Solve
the triangle.
Solution:
Draw the triangle.
E  43º e  4.56
F ?
f ?
G  57º g  ?
We have AAS.
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Example
Solution continued
Find F: F = 180º – (43º + 57º) = 80º
Use law of sines to find the other two sides.
f
e
g
e


sin F sin E
sinG sin E
f
4.56

sin 80º sin 43º
f
4.56

sin 57º sin 43º
4.56sin 80º
f 
sin 43º
4.56sin 57º
f 
sin 43º
f  6.58
g  5.61
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Slide 8.1 - 10
Example
Solution continued
We have solved the triangle.
E  43º e  4.56
F  80º f  6.58
G  57º g  5.61
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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 acute
Case 1: No solution
b < c; side b is too
short to reach the base.
No triangle formed.
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Solving Triangles SSA
Angle B is acute
Case 2: One solution
b < c; side b just
reaches the base and is
perpendicular to it.
Angle B is acute
Case 3: Two solutions
b < c; an arc of radius b
meets the base at two
points. This is called
the ambiguous case.
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Solving Triangles SSA
Angle B is acute
Case 4: One solution
b = c; an arc of radius b
meets the base at just
one point other than B.
Angle B is acute
Case 5: One solution
b > c; an arc of radius b
meets the base at just
one point.
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Solving Triangles SSA
Angle B is obtuse
Case 6: No solution
b < c; side b is too short
to reach the base. No
triangle formed.
Angle B is obtuse
Case 7: No solution
b = c; an arc of radius b
meets the base only
point B. No triangle is
formed.
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Solving Triangles SSA
Angle B is obtuse
Case 8: One solution
b > 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.
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Example
In ABC , b = 15, c = 20, and B = 29º. Solve the
triangle.
Solution
Draw a triangle.
A?
a?
B  29º b  15
C ?
c  20
Find C.
b
c

sin B sinC
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Example
Solution continued
15
20

sin 29º sinC
20sin 29º
sinC 
 0.6464
15
There are two angles less
than 180º with a sine of
0.6464: 40º and 140º. So
there are two possible
solutions.
Possible Solution I
If C = 40º, then
A = 180º – (29º + 40º) = 111º
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Example
Solution continued
a
b
Then we find a:

sin A sin B
a
15

sin111º sin 29º
15sin111º
a
 29
sin 29º
These measures make a
triangle as shown. Thus
we have a solution.
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Example
Solution continued
Possible Solution II
If C = 140º, then
A = 180º – (29º + 140º) = 11º
Then we find a:
These measures make a
a
b
triangle as shown. Thus

sin A sin B
we have a solution.
a
15

sin111º sin 29º
15sin111º
a
 29
sin 29º
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The Area of a Triangle
The area of any ABCis one half the product of he
lengths of two sides and the sine of the included angle:
1
1
1
K  bcsin A  absinC  acsin B.
2
2
2
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Example
A 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 ft
and 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?
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Example
Solution:
Use the area formula.
1
K  absinC
2
1
K  186 ft 172 ft  sin 53º
2
K  12,775 ft 2
The area of the garden is approximately 12,775 ft2.
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