Transcript bat03_ppt_07 - University of South Alabama

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 10- 1
Applications of
Trigonometry
Chapter 7
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7.1
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

To solve a triangle means to find the lengths of all its
sides and the measures of all its angles.

Any triangle that is not a right triangle is called oblique.

Any triangle, right or oblique, can be solved if at least
one side and any other two measures are known.
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Slide 10- 4
5
Possible
Triangles
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Slide 10- 5
The Law of Sines

In any triangle ABC,
B
a
b
c


.
sin A sin B sin C
a
c
A
b
C
Thus in any triangle, the sides are proportional to the
sines of the opposite angles.
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Slide 10- 6
Solving a Triangle (AAS)

In triangle ABC, A = 57, B = 43, and b = 11.2. Solve
B
the triangle.
43
a
c
C = (180°  (57° + 43°))
57
C
A
11.2
C = 180°  100° = 80°
Find a and c by using the law of sines:
c
11.2
a
11.2


sin 80 sin 43
sin 57 sin 43
11.2sin 80
11.2sin 57
c
a
sin 43
sin 43
c  16.2
a  13.8
Therefore,
A  57
a  13.8
B  43
b  11.2
C  80
c  16.2
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Slide 10- 7
Solving a Triangle (SSA)

If the given angle is acute, then there may be:
 a) no solution,
 b) one solution, or
 c) two solutions

If the given angle is obtuse, then there may be:
 a) no solution, or
 b) one solution
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Slide 10- 8
Example

In triangle ABC, b = 8.6, c = 6.2, and C = 35.
Solve the triangle.
B
6.2
a

Solution:
sin B sin 35

8.6
6.2
8.6sin 35
sin B 
6.2
sin B  .7956
B  52.7 , 127.3
35
A?
C
8.6
A  180  B  C
A  180  52.7  35  92.3
or
A  180  127.3  35  17.7
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Slide 10- 9
Example continued
a
6.2

sin 92.3 sin 35
6.2sin 92.3
a
sin 35
a  10.8

a
6.2
or

sin17.7 sin 35
6.2sin17.7
a
sin 35
a  3.3
There are two solutions:
A  92.3
a  10.8
B  52.7
b  8.6
C  35
c  6.2
or
A  17.7
a  3.3
B  127.3
b  8.6
C  35
c  6.2
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Slide 10- 10
The Area of a Triangle

The area K of any ABC is one half the product of the
lengths of two sides and the sine of the included angle:
1
1
1
K  bc sin A  ab sin C  ac sin B.
2
2
2
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Slide 10- 11
Example

Find the area of the triangle, ABC with A = 72, b = 16,
and c = 10.
Solution:
B
1
K  bc sin A
2
1
K  (16)(10)sin 72
2
K  76.1
10
72
A
16
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C
Slide 10- 12
7.2
The Law of Cosines

Use the law of cosines to solve triangles.

Determine whether the law of sines or the law of
cosines should be applied to solve a triangle.
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The Law of Cosines

In any triangle ABC,
a 2  b 2  c 2  2bc cos A,
b  a  c  2ac cos B,
2
2

2
b
a
2
c  a  b  2ab cos C.
2
C
A
2
c
B
Thus, in any triangle, the square of a side is the sum of
the squares of the other two sides, minus twice the
product of those sides and the cosine of the included
angle. When the included angle is 90, the law of
cosines reduces to the Pythagorean theorem.
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Slide 10- 14
Solving Triangles (SAS)

Solve ABC if a = 4, c = 6, and B = 105.2.
C
b
4
A
105.2
B
6
b 2  42  62  2(4)(6)cos105.2
b  64.585
b  8.0
2
sin A sin105.2

4
8
4sin105.2
sin A 
8
sin A  .4825
A  28.8 or 151.2
A  151.2, so A  28.8
Therefore,
A  28.8
a4
B  105.2
b  8.0
C  46
c6
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Slide 10- 15
Solving Triangles (SSS)

Solve ABC if a = 15, b = 11, and c = 8.

Solution: Solve for A first.
152  112  82  2(11)(8)cos A
112  82  152
cos A 
2(11)(8)
B
8
A
cos A  .227
15
11
C
A  103.1
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Slide 10- 16
Solving Triangles (SSS) continued

sin B sin103.1

11
15
11sin103.1
sin B 
15
sin B  .7142
C  180  A  B
C  180  103.1  45.6
C  31.3
B  45.6
B
A  103.1
a  15
B  45.6
b  11
C  31.3
c 8
8
A
15
11
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C
Slide 10- 17
7.3
Complex Numbers:
Trigonometric Form

Graph complex numbers.

Given a complex number in standard form, find
trigonometric, or polar, notation; and given a
complex number in trigonometric form, find standard
notation.

Use trigonometric notation to multiply and divide
complex numbers.

Use DeMoivre’s theorem to raise complex numbers to
powers.

Find the nth roots of a complex number.
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Absolute Value of a Complex Number

The absolute value of a complex number a + bi is
a  bi  a 2  b 2 .
Example: Find the absolute value of:
2
2

4

5
i

4

5
 41
a) 4 + 5i
2
2
b) 1  i  1  i  (1)  (1) 
2
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Slide 10- 19
Trigonometric Notation for Complex
Numbers

a + bi = r(cos + i sin )
Example: Find trigonometric notation for 1  i.

Solution: First, find r.
r  a 2  b2
sin  
r  (1) 2  (1) 2
5

4
r 2

1  2

2
2
cos 
1  2

2
2
Thus,
5
5 

1  i  2  cos
 i sin
 or 2  cos 225  i sin 225
4
4 

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
Slide 10- 20
Complex Numbers: Multiplication

For any complex numbers r1  cos1  i sin1  and
r2  cos 2  i sin  2  ,
r1  cos1  i sin 1   r2  cos 2  i sin  2  
r1r2 cos 1   2   i sin 1   2   .
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Slide 10- 21
Example

Multiply and express in standard notation.
4(cos50  i sin 50 ) and 2(cos10  i sin10 )
4(cos50  i sin 50 )  2(cos10  i sin10 )
 8  cos(50  10)  i sin(50  10) 
 8(cos 60  i sin 60)
1
3
 8  i

2 
2
 4  4i 3
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Slide 10- 22
Complex Numbers: Division

For any complex numbers r1  cos1  i sin1  and
r2  cos 2  i sin  2  , r2  0,
r1  cos1  i sin 1 
r2  cos 2  i sin  2 
r1
 cos 1   2   i sin 1   2   .
r2
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Slide 10- 23
Example

Divide and express in standard notation.
16(cos 70  i sin 70 ) and 4(cos 40  i sin 40 )
16(cos 70  i sin 70 ) 16
=  cos(70  40)  i sin(70  40) 
4(cos 40  i sin 40 )
4
 4(cos30  i sin 30)
 3 1
 4
 
 2 2
2 32
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Slide 10- 24
DeMoivre’s Theorem

For any complex number r  cos  i sin  and
any natural number n,
 r  cos  i sin    r n  cos n  i sin n .
n
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Slide 10- 25
Example

Find (1  i)5.

Solution: First, find trigonometric notation for 1  i.
1  i  2  cos 225  i sin 225

Then
 1  i 

5
  2  cos 225  i sin 225  
5
 2   cos5(225)  i sin 5(225) 
5
 4 2  cos1125  i sin1125 
 2
2
 4 2
i

2
2


 4  4i
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Slide 10- 26
Roots of Complex Numbers

The nth roots of a complex number
r  cos  i sin   , r  0, are given by
r
1/ n
 

360 
360  
cos   k 
  i sin   k 
 ,
n 
n 
n
 n
where k = 0, 1, 2, …, n  1.
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Slide 10- 27
Example

Find the square roots of 1 

Trigonometric notation: 1 
 3  i.
 3  i  2 cos60  i sin 60
  60
360 
360  
 60
 2  cos60  i sin 60    2 cos 
k
k
  i sin 

2
2
2
2



 
1
2
1
2
 2 cos  30  k 180   i sin  30  k  180  


For k = 0, the root is 2  cos30  i sin30
For k = 1, the root is 2  cos 210  i sin 210
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Slide 10- 28
7.4
Polar Coordinates and Graphs

Graph points given their polar coordinates.

Convert from rectangular to polar coordinates and from
polar to rectangular coordinates.

Convert from rectangular to polar equations and from
polar to rectangular equations.

Graph polar equations.
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Polar Coordinates

Any point has rectangular
coordinates (x, y) and polar
coordinates (r, ).

To plot points on a polar graph:
 Locate the directed angle .
 Move a directed distance r
from the pole. If r > 0, move
along ray OP. If r < 0, move
in the opposite direction of
ray OP.
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Slide 10- 30
Example

Graph each of the following
points.
a)
A(3, 60)
b)
B(0, 10)
c)
C(5, 120)
d)
D(1, 60)
e)
 3 
E  2, 2 


f)
F  4, 
3




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Slide 10- 31
Conversion Formulas
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Slide 10- 32
Example

Convert (4, 2) to polar coordinates.
r  x2  y 2
2 1
tan   
4 2
  26.6
r  42  22
r  16  4
r  20  2 5


Thus (r, ) = 2 5,26.6

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Slide 10- 33
Another Example


 
Convert  5,  to rectangular coordinates.
 4
x  r cos 
y  r sin 


x  5cos
4
 2
x  5

2


5 2
x
2
y  5sin
4
 2
y  5

 2 
y
5 2
2
5 2 5 2 
,
 (x, y) = 

2 
 2
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Slide 10- 34
Polar and Rectangular Equations

Some curves have simpler equations in polar
coordinates than in rectangular coordinates. For others,
the reverse is true.

Convert x + 2y = 10 into a polar equation.
x + 2y = 10
r cos  2r sin   10
r (cos  2sin  )  10
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Slide 10- 35
Example


Convert r = 3 cos   sin  into a rectangular equation.
r  3cos  sin 
r 2  3r cos  r sin 
x 2  y 2  3x  y
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Slide 10- 36
Graphing Polar Equations

Graph r = 2 sin 

r

r

r
0
0
120
-1.732
240
1.732
15
-.5176
135
-1.414
255
1.932
30
-1
150
-1
270
2
45
-1.414
165
-.5176
285
1.932
60
-1.732
180
0
300
1.732
75
-1.932
195
.5176
315
1.414
90
-2
210
1
330
1
105
-1.932
225
1.414
345
.5176
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Slide 10- 37
Graphing with graphing calculator

Graph the equation
r + 2 = 2 sin 2.

Solution solve for r.
r  2sin 2  2
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Slide 10- 38
7.5
Vectors and Applications

Determine whether two vectors are equivalent.

Find the sum, or resultant, of two vectors.

Resolve a vector into its horizontal and vertical
components.

Solve applied problems involving vectors.
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Vector




A vector in the plane is a directed line segment. Two
vectors are equivalent if they have the same
magnitude and direction.
Consider a vector drawn from point A to point B AB :
 A is called the initial point
 B is called the terminal point
Equivalent vectors: vectors with the same length and
direction.
Magnitude: length of a vector, expressed as AB .
 
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Slide 10- 40
Example

Find the magnitude of the vector, v, with initial point
(1, 2) and terminal point (3, 2).
v  (3  (1)) 2  ( 2  2) 2
v  42  (4) 2
v  16  16
v  32  4 2
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Slide 10- 41
Vector Addition

The sum of two vectors is called the resultant.

If u  AB and v = BC, then u + v = AB  BC  AC.
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Slide 10- 42
Parallelogram Law

Vector addition is
commutative.
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Slide 10- 43
Example

Forces of 10 newtons and 50 newtons act on an object at
right angles to each other. Find the magnitude of the
resultant and the angle of the resultant that it makes with the
larger force.
v  102  502
v  100  2500
v  2600
v  10 26  51

10
tan  
50
1  10 
  tan  
 50 
  11.3
v
10
10

50
The resultant vector, v, has magnitude 51 and makes an
angle of 11.3 with the larger force.
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Slide 10- 44
Components

Resolving a vector into its vector components—write a
vector, w, as a sum of two vectors, u and v, which are
the components.

Most often, the two components will be the horizontal
component and the vertical component of the vector.
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Slide 10- 45
Example

A vector w has a magnitude of 45 and rests on an
incline of 20. Resolve the vector into its horizontal and
vertical components.
u
cos 20 
45
u  45cos 20
u  42.3

v
sin 20 
45
v  45sin 20
v  15.4
45
v
20
u
The horizontal component of w is 42.3 right and the
vertical component of w is 15.4 up.
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Slide 10- 46
7.6
Vector Operations

Perform calculations with vectors in component form.

Express a vector as a linear combination of unit vectors.

Express a vector in terms of its magnitude and its
direction.

Solve applied problems involving forces in equilibrium.
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Position Vectors

Standard position—the initial point is the origin and
the terminal point is (a, b).

Notation: v = a, b
 a is the scalar horizontal component
 b is the scalar vertical component

Scalar—numerical quantity rather than a vector
quantity
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Slide 10- 48
Component Form of a Vector

The component form of AC with A = (x1, y1) and
C = (x2, y2) is AC  x  x , y  y .
2
1
2
1
Example: Find the component form of AB if
A = (3, 2) and B = (5, 6).

Solution: AB  5  3,  6  (2)  8, 4
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Slide 10- 49
Length of a Vector

The length, or magnitude, of a vector v =
is given by v  v12  v2 2 .
v1 , v2
Equivalent Vectors
Let u = u1 , u2 and v = v1 , v2 . Then
u1 , u2  v1 , v2 if and only if u1 = v1 and u2 = v2.
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Slide 10- 50
Operations on Vectors
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Operations on Vectors continued
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Slide 10- 52
Example

Let u = 4,10 and v = 5, 3 . Find each of the
following.
a) 4v
4 5, 3  20, 12
b) 2u + v
2 4,10  5, 3
 13,17
c) 2u  3v
2 4,10  3 5, 3
d) |u + 4v|
 4,10  4 5, 3
 24, 2
 (24) 2  (2) 2
 576  4
 580  24.1
 7, 29
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Slide 10- 53
Properties of Vector Addition and Scalar
Multiplication
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Slide 10- 54
Unit Vectors

A vector of magnitude 1.

If v is a vector and v  O, then
1
v
 v, or
,
v
v
is a unit vector in the direction of v.
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Slide 10- 55
Example


Find a unit vector that has the same direction
as w  5,12 .
w  (5) 2  (12) 2
w  25  144
w  169  13
1
u w
13
1

5,12
13
5 12

,
13 13
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Slide 10- 56
Linear Combinations

i = 1,0 is a vector parallel to the x-axis.
j = 0,1 is a vector parallel to the y-axis.

The linear combination, v, of unit vectors i and j, where

v  v1 , v2 , is v1i  v2 j.
Example: The linear combination of i and j for 2,3
is 2i + 3j.
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Slide 10- 57
Direction Angles

A vector u in standard position on the unit circle can be
written as
u x, y  cos ,sin   (cos )i  (sin  ) j
where  is called the direction angle. It is measured
counterclockwise from the positive x-axis.
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Slide 10- 58
Example

Determine the direction angle  of the vector
w = 3i + 4j.
Solution:
w  3i  4 j  3, 4
4
tan  
3
  53.1

 is a 2nd quadrant angle, so
 = 53.1 + 180 = 126.9.
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Slide 10- 59
Definitions
The dot product of two vectors u = u1 , u2
and v = v1 , v2 is u  v  u1v1  u2v2 .
(Note that u1v1 + u2v2 is a scalar, not a vector.)
Angle Between Two Vectors
If  is the angle between two nonzero vectors u and v,
uv
then
cos 
u v
.
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Slide 10- 60
Example

Find the angle between u =

Solution: Find u • v, |u|, |v|.
4, 2 and v = 3, 1
.
u • v = 4(3) + 2(1) =14
u  (4) 2  (2) 2
 16  4  20
v  (3)  (1)
2
2
 9  1  10
uv
14
cos 

u v
20  10
 14 
  cos 

 20  10 
  171.9
1
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