Sections 8.1 and 8.2





 Apply the law of sines to determine the lengths of side and measures of angle of a triangle.
Solve word problems requiring the law of sines.
Apply the law of cosines to determine the lengths of side and measures of angle of a triangle.
Solve word problems requiring the law of cosines.
Solve a word problem requiring Heron's formula.

The formulas listed below will allow us to more easily deal with triangles that are not right triangles.
 Law of sines
 Law of cosines

Heron’s formula



Law of sines sin a
Law of cosines

 sin b

 sin c
 a 2

 b 2  c 2  2 bc cos   or b
Heron’s formula
2
A 
1
4
 a 2
P
 c 2
( P


2 ac
2 a cos
)( P c 2
 or
 a 2 
2 b )( P b 2

 2 ab
2 c ) cos
a, b, and c are the lengths of the sides of the triangle
P is the perimeter of the triangle
A is the area of a triangle

sin  a
 sin b

 sin c


a 2  b 2  c 2  2 bc cos b 2  a 2  c 2  2 ac cos c 2  a 2  b 2  2 ab cos

Use the Law of Sines to find the value of the side x.
We are told to use the law of sines to find x. In order to use the law of sines, we need to have the lengths of two sides and the measures of the angle opposite those sides. In this case we have one side and the side we are looking for. We have the measure of the angle opposite the side we are looking for, but are missing the measure of the angle opposite the side we have.
continued on next slide

Use the Law of Sines to find the value of the side x.
Since we have the measures of two of the three angles, we can use the fact that the sum of the measures of the angles of a triangle add up to
180 degrees. This will give us: angle angle
ACB
ACB
 180  52  70
 58
Now that we have the measure of the angle opposite the side AB, we can apply the law of sines to find the value of x.
continued on next slide

Use the Law of Sines to find the value of the side x.
sin 58 x
26 .
7 sin(

58 ) sin

52 x
26 .
7 sin( 52 ) x x


26 .
7 sin( 52 sin( 58 )
)
24 .
8097805

Use the Law of Cosines to find the value of the side x.
x a 2  b 2  c 2  2 bc cos  
In order to use the law of cosines, we need the lengths of two sides and the measure of the angle between them. We have that here. We can let side a be x and angle α be the
39 degree angle. Sides b and c are the lengths 21 and 42.
continued on next slide

Sections 8.1 and 8.2





 Apply the law of sines to determine the lengths of side and measures of angle of a triangle.
Solve word problems requiring the law of sines.
Apply the law of cosines to determine the lengths of side and measures of angle of a triangle.
Solve word problems requiring the law of cosines.
Solve a word problem requiring Heron's formula.

a 2


Law of sines sin a
Law of cosines

 sin b

 sin c


 b 2  c 2 
Heron’s formula
A
2 bc
 cos  
1
4
P or
( P b

2 
2 a a 2 
)( P c 2

 2 ac
2 b cos
)( P  c 2 or
2 c

) a 2  b 2  2 ab cos
a, b, and c are the lengths of the sides of the triangle
P is the perimeter of the triangle
A is the area of a triangle

Use the Law of Cosines to find the value of the side x.
x
Now we plug into the law of cosine formula to find x.
x x x 2 x 2 x 2 x 2
 21 2  42 2  2 ( 21 )( 42 ) cos



441  1764  1764 cos
2205  1764 cos
834 .
114524


 834 .
114524
 28 .
88104091
Since length is positive, x is approximately 28.88104097

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them. If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours?
For this problem, the first thing that we should do is draw a picture.
Once we have the picture, we may be able to see which formula we can use to solve the problem.
continued on next slide

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them. If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours?
140° 34mph*3hr = 102 miles x
Looking at the labeled picture above, we can see that the have the lengths of two sides and the measure of the angle between them.
We are looking for the length of the third side of the triangle. In order to find this, we will need the law of cosines. x will be side a.
Sides b and c will be 78 and 102. Angle α will be 140°.
a 2  b 2  c 2  2 bc cos continued on next slide

Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them. If the first ship travels at 26 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 3 hours?
x x x 2 x 2 x 2 x 2



78 2 
6084
16488
102 2

 2 ( 78 )( 102 )
10404  cos
15912 cos
 140 
 140 
 15912 cos

140

 28677 .
29918


 28677 .
29918
 169 .
3437309
Since distance is positive, the ships are approximately
169.3437309 miles apart after
3 hours.

Approximating the area of a triangle
A 
Heron’s Formula
1
4
P ( P  2 a )( P  2 b )( P  2 c ) where P is the perimeter of the triangle and a , b , and c are the lengths of the sides of the triangle.
OR
The area of a triangle equals one-half the product of the lengths of any two sides and the sine of the angle between them.
i.e.
1/2ab sinγ = A

Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
48
° 20′
37.0 cm
7-19

8.3
Vectors, Operations, and the
Dot Product
Basic Terminology ▪ Algebraic Interpretation of Vectors ▪
Operations with Vectors ▪ Dot Product and the Angle Between
Vectors
7-20
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.










A vector is an object that has a magnitude and a direction.
x
1 y
1 x
2
( v that connects the points from P1 to P2 is v = x

2 x
1
) i + j .
Unit vectors are vectors of length 1.
i is the unit vector in the x direction.
j is the unit vector in the y direction.
A unit vector in the direction of v is v /|| v ||
A vector v can be represented in component form by v = a i + b j .
The magnitude of v is || v || = a
2  b
2
Using the angle that the vector makes with x-axis in standard position and the vector’s magnitude, component form can be written as v = || v
||cos(θ) i + || v
||sin(θ) j

Magnitude and Direction Angle
 Find the magnitude and direction angle for
Magnitude:
Direction angle:
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
7-22

Horizontal and Vertical Components
 Vector v has magnitude 14.5 and direction angle
220
°
. Find the horizontal and vertical components.
Horizontal component: –11.1
Vertical component: –9.3
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
7-23

Example 4 Finding the Magnitude of a Resultant
 Two forces of 32 and 48 newtons act on a point in the plane. If the angle between the forces is 76
°
, find the magnitude of the resultant vector.
because the adjacent angles of a parallelogram are supplementary.
Law of cosines
Find square root.
7-24
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

 Let u =

6, –3
 and v =
 –14, 8 
. Find the following.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
7-25

Unit Vector has the same direction as a given vector, but is 1 unit long
 Unit vector = (original vector)/length of vector
 Simply involves scalar multiplication once the length of the vector is determined (recall the length = length of hypotenuse if legs have lengths = a & b)
 Given vector, v = -2i + 7j, find the unit vector: length

(

2 )
2 
7
2 
53 unit :

2
53 i

7
53 j


2
53
53 i

7 53
53 j




Objectives
 Find dot product of 2 vectors
 Find angle between 2 vectors
 Use dot product to determine if 2 vectors are orthogonal
 Find projection of a vector onto another vector
 Express a vector as the sum of 2 orthogonal vectors
 Compute work.

Scalar multiplication: A vector can be multiplied by any scalar (or number).
Example: Let v = 5 i + 4 j , k = 7. Then kv = 7(5 i + 4 j ) = 35 i + 28 j .
Alternate Dot Product formula v · w = || v |||| w ||cos(θ). The angle θ is the angle between the two vectors. v θ w
Addition/subtraction of vectors: Add/subtract same components.
Example Let v = 5 i + 4 j , w = –2 i + 3 j . v + w = (5 i + 4 j ) + ( –2 i + 3 j ) = (5 – 2) i + (4 + 3) j = 3 i + 7 j .
3 v – 2 w = 3(5 i + 4 j ) – 2(–2 i + 3 j ) = (15 i + 12 j ) + (4 i – 6 j ) = 19 i + 6 j .

 The dot product of 2 vectors is the sum of the products of their horizontal components and their vertical components v v

 w a
1 i

 b
1 j , w
 a
1 a
2
 b
1 b
2 a
2 i
 b
2 j
Example: Let v = 5 i + 4 j , w = –2 i + 3 j .
v
· w = (5)(–2) + (4)(3) = –10 + 12 = 2.

3.
4.
1.
2.
Find the dot product of v&w if v=3i+j and w= -2i - j
7
-5
-7
-4

 If u,v, & w are vectors and c is scalar, then
1 ) u
 v
 v
 u
2 ) u

( v
 w )
 u
 v
 u
 w
3 ) 0
 v

0
4 ) v
 v

| v |
2
5 )( cu )
 v
 c ( u
 v )
 u

( cv )

cos
  v
 w v w

Find the angle between a=<4,-3> and b=<1,2>.

 Parallel: the angle between the vectors is either 0 (the vectors on top of each other) or
180 (vectors are in opposite directions).
 a and b are parallel if θ = 0 or θ = ∏.

 Two vectors v and w are orthogonal
(perpendicular) iff v
· w = 0
 Two vectors v and w are orthogonal
(perpendicular) if the angle between them is
θ=∏/2.

Let a = 1/2i – 3j and b = -2i+12j.
Show that a and b are parallel.

Show that the pair of vectors is orthogonal. 2i+3j; 6i-4j

 Formula for comp b a: If a and b are nonzero vectors, then comp b a = a • b
||b||

If c = 10i+4j and d = 3i-2j, find comp d c and comp c d and illustrate.

Work done by a force F moving an object from A to B
The work W done by a constant force a as its point of application moves along a vector b is W = a • b.

Lesson 8.5

Graphical Representation of a
Complex Number
 Graph in coordinate plane
 Called the complex plane
-2 + 3i
•
 Horizontal axis is the real axis
 Vertical axis is the imaginary axis
• -5i
3 + 4i
•
43

Absolute Value of a Complex Number


Defined as the length of the line segment


From the origin
To the point
3 + 4i
•
Calculated by using Pythagorean
Theorem
 i

3
2 
4
2 
25

5
44

 Try these


Graph the complex number
Find the absolute value z
 
5 z
  i z
   i
45

 Consider the graphical representation
 We note that a right triangle is formed cos
 a
 r
 a cos r
 sin
 b
 r
 b sin r
 r
θ a a + bi
• b
How do we determine θ ?
where r
 z
 a
2  b
2
  tan

1 b a
46

 Now we use a
 r cos
 b
 r sin
 and substitute into z = a + b i z r cos

 Result is sin
 z r cis

 Abbreviation is often
47

 Given the complex number -5 + 6 i



Write in trigonometric form r = ?
θ = ?
 Given z = 3 cis 315°




Write in standanrd form r = ?
a = ?
b = ?
48

Product of Complex Numbers in Trig
Form
 Given z
1
 r
1
 cos

1 sin

1
 z
2
 r
2
 cos

2 sin

2

 It can be shown that the product is
   
1 2 1 2 cis
  
1

2



Multiply the absolute values
Add the θ's
49

Quotient of Complex Numbers in Trig
Form
 Given z
1
 r
1
 cos

1 sin

1
 z
2
 r
2
 cos

2 sin

2

 It can be shown that the quotient is z
1
  z r
2 2 cis
  
1

2

50

 Try the following operations using trig form

4 cis120
   15cis240
3cis135
 Convert answers to standard form
51


8.6 Complex Numbers in Polar Form:
DeMoivre’s Theorem
Objectives







Plot complex numbers in the complex plane
Fine absolute value of a complex #
Write complex # in polar form
Convert a complex # from polar to rectangular form
Find products & quotients of complex numbers in polar form
Find powers of complex # in polar form
Find roots of complex # in polar form


a is a real number
 bi is an imaginary number
 Together, the sum, a+bi is a COMPLEX #


Complex plane has a real axis (horizontal) and an imaginary axis (vertical)
2 – 5i is found in the 4 th quadrant of the complex plane (horiz = 2, vert = -5)
 Absolute value of 2 – 5i refers to the distance this pt. is from the origin (continued)

 Since the horizontal component = 2 and vertical = -5, we can consider the distance to that point as the same as the length of the hypotenuse of a right triangle with those respective legs
2

5 i

2
2 
(

5 )
2 
29

Expressing complex numbers in polar form
 z = a + bi a
 r cos

, b
 r sin
 r z

 a
2  b
2 r (cos
  i sin

)

Express z = -5 + 3i in complex form r

(

5 )
2 
3
2 
34 tan
 z


3

5
,
  
.
54
34 (cos(

.
54 )
 i sin(

.
54 )

 [r (cos θ + i sin θ)] n = r n (cos n•θ + i sin n•θ)
Ex: Change (1 + i) 20 to the form a + bi

Taking a root
(DeMoivre’s Theorem)
 Taking the nth root can be considered as raising to the (1/n)th power
 Now finding the nth root of a complex # can be expressed easily in polar form


HOWEVER, there are n nth roots for any complex number & they are spaced evenly around the circle.
Once you find the 1 st root, to find the others, add 2pi/n to theta until you complete the circle

DeMoivre’s Theorem and nth roots:
W k
= n √r [cos (θ+2∏k ) + i sin (θ + 2∏k) n where k = 0,1,2,…n-1.
n

Find the fourth roots of -8-8√3i.

If you’re working with degrees add 360/n to the angle measure to complete the circle.
 Example: Find the 6 th roots of z= -2 + 2i
6
6
6
6
 r

Express in polar form, find the 1 st root, then add 60 degrees successively to find the other 5 roots.
(

2 )
2 
2
2 
2 2 ,
360

6

60

6
6 tan z z


6


2
8
,
 
135

,
2
(cos
135

6
 i z
 sin
2 2 (cos 135

135

)
6

12

8 (cos i sin 135

)
22 .
5
  i sin

12 8 (cos 82 .
5
  i sin 82 .
5

)
22 .
5

) z

12 8 (cos 142 .
5
  i sin 142 .
5

) z

12 8 (cos 202 .
5
  i sin 202 .
5

) z

12 8 (cos 262 .
5
  i sin 262 .
5

) z

12 8 (cos 322 .
5
  i sin 322 .
5

)

If you’re working with degrees add 360/n to the angle measure to complete the circle.
 Example: Find the 6 th roots of z= -2 + 2i
6
6
6
6
 r

Express in polar form, find the 1 st root, then add 60 degrees successively to find the other 5 roots.
(

2 )
2 
2
2 
2 2 ,
360

6

60

6
6 tan z z


6


2
8
,
 
135

,
2
(cos
135

6
 i z
 sin
2 2 (cos 135

135

)
6

12

8 (cos i sin 135

)
22 .
5
  i sin

12 8 (cos 82 .
5
  i sin 82 .
5

)
22 .
5

) z

12 8 (cos 142 .
5
  i sin 142 .
5

) z

12 8 (cos 202 .
5
  i sin 202 .
5

) z

12 8 (cos 262 .
5
  i sin 262 .
5

) z

12 8 (cos 322 .
5
  i sin 322 .
5

)

If you’re working with degrees add 360/n to the angle measure to complete the circle.
 Example: Find the 6 th roots of z= -2 + 2i
6
6
6
6
 r

Express in polar form, find the 1 st root, then add 60 degrees successively to find the other 5 roots.
(

2 )
2 
2
2 
2 2 ,
360

6

60

6
6 tan z z


6


2
8
,
 
135

,
2
(cos
135

6
 i z
 sin
2 2 (cos 135

135

)
6

12

8 (cos i sin 135

)
22 .
5
  i sin

12 8 (cos 82 .
5
  i sin 82 .
5

)
22 .
5

) z

12 8 (cos 142 .
5
  i sin 142 .
5

) z

12 8 (cos 202 .
5
  i sin 202 .
5

) z

12 8 (cos 262 .
5
  i sin 262 .
5

) z

12 8 (cos 322 .
5
  i sin 322 .
5

)




Law of Sines: Sin A = Sin B = Sin C c
Law of Cosines: a 2 = b 2 + c 2 – 2bc Cos A a 2 -b 2 -c 2 = Cos A
-2bc a b
Area of a Triangle
A = ½ ab Sin C
A = √(s(s-a)(s-b)(s-c)); s = ½ (a+b+c)


||a|| = √(a
1
2 + a
2
2 )

 a
1
= ||a||cos θ
Unit Vector • a
|| a ||

 cos θ = a•b
||a||||b|| comp b a = a•b
||b|| a
2
= ||a||sin θ angle between vectors


Products and Quotients of Complex #’s z
1 z
2
= r
1 r
2
(cos(θ
1
+ θ
2
)+isin(θ
1
+ θ
2
))
 z
1 z
2
= r r
1
2
(cos(θ
1
- θ
2
)+isin(θ
1
DeMoivre’s Theorem
- θ
2
))
[r(cos θ + isin θ)] n = r n (cosnθ + isin nθ)
 nth roots
W k
= n √r [cos (θ +2пk) + isin (θ +2пk)] n where k = 0,1,…n-1 n

 Dot Product <a
1
,a
2
> • <b
1
,b
2
> = a
1 b
1
+ a
2 b
2

Work Done = a • b (force * distance)

