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Laws of Sines and Cosines
Sections 8.1 and 8.2
Objectives





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
Formulas

Law of sines

Law of cosines
sin  sin  sin 


a
b
c
a 2  b 2  c 2  2bc cos  or

Heron’s formula
b 2  a 2  c 2  2ac cos   or
c 2  a 2  b 2  2ab cos 
1
A
P (P  2a )(P  2b )(P  2c )
4
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
Law of Sines
sin  sin  sin 


a
b
c
Law of Cosines
a 2  b 2  c 2  2bc cos 
b 2  a 2  c 2  2ac cos  
c 2  a 2  b 2  2ab 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 ACB  180  52  70
angle ACB  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 sin 52

26.7
x
x sin( 58)  26.7 sin( 52)
26.7 sin( 52)
sin( 58)
x  24.8097805
x 
Use the Law of Cosines to find the
value of the side x.
x
a 2  b 2  c 2  2bc 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
Laws of Sines and Cosines
Sections 8.1 and 8.2
Objectives





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.
Formulas

Law of sines

Law of cosines
sin  sin  sin 


a
b
c
a 2  b 2  c 2  2bc cos  or

Heron’s formula
b 2  a 2  c 2  2ac cos   or
c 2  a 2  b 2  2ab cos 
1
A
P (P  2a )(P  2b )(P  2c )
4
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 2  212  422  2(21)(42) cos 39
x 2  441  1764  1764 cos 39
x 2  2205  1764 cos 39
x 2  834.114524
x   834.114524
x  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?
harbor
26mph*3hr = 78 miles
ship 1
140°
34mph*3hr = 102 miles
x
ship 2
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  2bc 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?
harbor
26mph*3hr = 78 miles
140°
34mph*3hr = 102 miles
x
ship 1
ship 2
x 2  782  1022  2(78)(102) cos 140 
x 2  6084  10404  15912 cos 140 
x  16488  15912 cos 140 
2
x 2  28677.29918
x   28677.29918
x  169.3437309
Since distance is positive, the
ships are approximately
169.3437309 miles apart after
3 hours.
Approximating the area of a triangle
Heron’s Formula
1
A
P (P  2a )(P  2b )(P  2c )
4 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
48°20′
37.0 cm
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
7-20
8.3 Vectors









A vector is an object that has a magnitude and a direction.
Given two points P1: ( x1 , y1 ) and P2: ( x2 , y2 )on the plane, a vector
v that connects the points from P1 to P2 is v =
( x2  x1 ) i + ( y 2  y1 ) 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 = ai + bj.
The magnitude of v is ||v|| =
a 2  b2
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.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
7-24
u = 6, –3 and v = –14, 8. Find the
following.
Let
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:
2
2
length  (2)  7  53
2
7
 2 53
7 53
unit :
i
j
i
j
53
53
53
53
8.1-8.3 Review Answers


8.4 Dot Product
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.
8.4 Vector Operations
Scalar multiplication: A vector can be multiplied by any scalar (or number).
Example: Let v = 5i + 4j, k = 7. Then kv = 7(5i + 4j) = 35i + 28j.
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 = 5i + 4j, w = –2i + 3j.
v + w = (5i + 4j) + (–2i + 3j) = (5 – 2)i + (4 + 3)j = 3i + 7j.
3v – 2w = 3(5i + 4j) – 2(–2i + 3j) = (15i + 12j) + (4i – 6j) = 19i + 6j.
Definition of Dot Product

The dot product of 2 vectors is the sum of the
products of their horizontal components and
their vertical components
v  a1i  b1 j , w  a2i  b2 j
v  w  a1a 2 b1b2
Example: Let v = 5i + 4j, w = –2i + 3j.
v · w = (5)(–2) + (4)(3) = –10 + 12 = 2.
Find the dot product of v&w if v=3i+j
and w= -2i - j
1.
2.
3.
4.
7
-5
-7
-4
Properties of Dot Product

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 |
5)(cu )  v  c(u  v)  u  (cv)
2
Angle between vectors, v and w
vw
cos  
v w
Find the angle between a=<4,-3> and
b=<1,2>.
Parallel Vectors


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 = ∏.
Perpendicular Vectors


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 compba: If a and b are nonzero
vectors, then compba = a • b
||b||
If c = 10i+4j and d = 3i-2j, find compdc and
compcd 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.
Trigonometric Form
of Complex Numbers
Lesson 8.5
Graphical Representation of a
Complex Number

Graph in coordinate plane

Called the complex plane

Horizontal axis
is the real axis

Vertical axis is
the imaginary
axis
3 + 4i
•
-2 + 3i
•
•
-5i
43
Absolute Value of a Complex Number

Defined as the length of the line segment



From the origin
To the point
Calculated by
using Pythagorean
Theorem
3 + 4i
•
3  4i  32  42  25  5
44
Find That Value, Absolutely

Try these


Graph the complex number
Find the absolute value
z  4  4i
z  5
z  5  6i
45
Trig Form of Complex Number


Consider the graphical representation
We note that a right
triangle is formed
a
cos  
r
a  r cos 
b
sin  
r
b  r sin 
where r  z  a 2  b 2
a + bi
•
r
b
θ
a
How do we
determine θ?
b
  tan
a
1
46
Trig Form of Complex Number

Now we use
a  r cos 
b  r sin 
and substitute into z = a + bi

Result is
z  r  cos  i  r  sin 
z  r  cis

Abbreviation is often
47
Try It Out

Given the complex number -5 + 6i




Write in trigonometric form
r=?
θ=?
Given z = 3 cis 315°



Write in standanrd form
r=?
a=?
48
Product of Complex Numbers in Trig
Form

Given
z1  r1  cos1  i  sin 1 

z2  r2  cos2  i  sin 2 
It can be shown that the product is
z1  z2  r1  r2  cis 1  2 


Multiply the absolute values
Add the θ's
49
Quotient of Complex Numbers in Trig
Form

Given
z1  r1  cos1  i  sin 1 

z2  r2  cos2  i  sin 2 
It can be shown that the quotient is
z1 r1
  cis 1   2 
z2 r2
50
Try It Out

Try the following operations using trig form
 4  cis120    6  cis315 

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
Complex number = z = a + bi






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 4th 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)
Find the absolute value

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  5i  2  (5)  29
2
2
Expressing complex numbers in polar
form

z = a + bi
a  r cos  , b  r sin 
r  a b
z  r (cos   i sin  )
2
2
Express z = -5 + 3i in complex form
r  (5)  3  34
2
2
3
tan  
,   .54
5
z  34 (cos( .54)  i sin( .54)
DeMoivre’s Theorem

[r (cos θ + i sin θ)]n = rn (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 1st root, to find the others,
add 2pi/n to theta until you complete the
circle
DeMoivre’s Theorem and nth roots:
Wk = n√r [cos (θ+2∏k ) + i sin (θ + 2∏k)
n
n
where k = 0,1,2,…n-1.
Example:
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 6th roots of z= -2 + 2i

r
Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
( 2) 2  2 2  2 2 ,
360
 60
6
2
,   135, z  2 2 (cos 135  i sin 135)
2
135
135
z  6 8 (cos
 i sin
)  12 8 (cos 22.5  i sin 22.5)
6
6
z  12 8 (cos 82.5  i sin 82.5)
tan  
6
6
6
z  12 8 (cos 142.5  i sin 142.5)
6
z  12 8 (cos 202.5  i sin 202.5)
6
z  12 8 (cos 262.5  i sin 262.5)
6
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 6th roots of z= -2 + 2i

r
Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
( 2) 2  2 2  2 2 ,
360
 60
6
2
,   135, z  2 2 (cos 135  i sin 135)
2
135
135
z  6 8 (cos
 i sin
)  12 8 (cos 22.5  i sin 22.5)
6
6
z  12 8 (cos 82.5  i sin 82.5)
tan  
6
6
6
z  12 8 (cos 142.5  i sin 142.5)
6
z  12 8 (cos 202.5  i sin 202.5)
6
z  12 8 (cos 262.5  i sin 262.5)
6
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 6th roots of z= -2 + 2i

r
Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
( 2) 2  2 2  2 2 ,
360
 60
6
2
,   135, z  2 2 (cos 135  i sin 135)
2
135
135
z  6 8 (cos
 i sin
)  12 8 (cos 22.5  i sin 22.5)
6
6
z  12 8 (cos 82.5  i sin 82.5)
tan  
6
6
6
z  12 8 (cos 142.5  i sin 142.5)
6
z  12 8 (cos 202.5  i sin 202.5)
6
z  12 8 (cos 262.5  i sin 262.5)
6
z  12 8 (cos 322.5  i sin 322.5)
Formulas for index cards:



Law of Sines: Sin A = Sin B = Sin C
a
b
c
Law of Cosines:
a2 = b2 + c2 – 2bc Cos A
a2-b2-c2 = Cos A
-2bc
Area of a Triangle
A = ½ ab Sin C
A = √(s(s-a)(s-b)(s-c)); s = ½ (a+b+c)

||a|| = √(a12 + a22)

a1 = ||a||cos θ
Unit Vector 1 • a



a2 = ||a||sin θ
|| a ||
cos θ = a•b
||a||||b||
compba = a•b
||b||
angle between vectors

Products and Quotients of Complex #’s
z1z2 = r1r2(cos(θ1+ θ2)+isin(θ1+ θ2))
z1 = r1 (cos(θ1- θ2)+isin(θ1- θ2))
z2 r2

DeMoivre’s Theorem
[r(cos θ + isin θ)]n = rn(cosnθ + isin nθ)

nth roots
Wk = n√r [cos (θ +2пk) + isin (θ +2пk)]
n
n
where k = 0,1,…n-1

Dot Product <a1,a2> • <b1,b2> = a1b1 + a2b2

Work Done = a • b (force * distance)
Review Test Answers
