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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.
Law of sines
Law of cosines
Heron’s formula
The formulas listed below will allow us to more easily deal with triangles that are not right triangles.
Formulas Law of sines
Law of cosines
Heron’s formula
)2)(2)(2(41
cPbPaPPA
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
cba sinsinsin
cos2222 accab cos2222 bccba
cos2222 abbac
oror
Law of Sines
cba sinsinsin
Law of Cosines
cos2222 accab
cos2222 bccba
cos2222 abbac
Use the Law of Sines to find the value of the side x.
continued on next slide
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.
Use the Law of Sines to find the value of the side x.
continued on next slide
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:
58angle
7052180angle
ACB
ACB
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.
Use the Law of Sines to find the value of the side x.
8097805.24
)58sin()52sin(7.26
)52sin(7.26)58sin(
52sin7.2658sin
x
x
xx
Use the Law of Cosines to find the value of the side x.
x
continued on next slide
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.
cos2222 bccba
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
Heron’s formula
)2)(2)(2(41
cPbPaPPA
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
cba sinsinsin
cos2222 accab cos2222 bccba
cos2222 abbac
oror
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.
88104091.28
114524.834
114524.834
39cos17642205
39cos17641764441
39cos)42)(21(24221
2
2
2
222
x
x
x
x
x
x
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?
continued on next slide
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.
26mph*3hr = 78 miles
harbor
ship 1 ship 2
140° 34mph*3hr = 102 miles
x
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
ship 1 ship 2
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°.
harbor
26mph*3hr = 78 miles
harbor
ship 1 ship 2
140° 34mph*3hr = 102 miles
x
cos2222 bccba
26mph*3hr = 78 miles
harbor
ship 1 ship 2
140° 34mph*3hr = 102 miles
x
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
ship 1 ship 2
140° 34mph*3hr = 102 miles
x
harbor
26mph*3hr = 78 miles
harbor
ship 1 ship 2
140° 34mph*3hr = 102 miles
x
3437309.169
29918.28677
29918.28677
140cos1591216488
140cos15912104046084
140cos)102)(78(210278
2
2
2
222
x
x
x
x
x
x
Since distance is positive, the ships are approximately 169.3437309 miles apart after 3 hours.
Approximating the area of a triangle
)2)(2)(2(41
cPbPaPPA
Heron’s Formula
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
OR
where P is the perimeter of the triangle and a,b, and c are the lengthsof the sides of the triangle.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-19
37.0 cm
48°20′
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-20
Vectors, Operations, and the Dot Product8.3Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors
8.3 Vectors A vector is an object that has a magnitude and a direction. Given two points P1: and P2: on the plane, a vector
v that connects the points from P1 to P2 is v = 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 = ai + bj. The magnitude of v is ||v|| = 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
22 ba
),( 11 yx ),( 22 yx
)( 12 xx )( 12 yy
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-22
Find the magnitude and direction angle for
Magnitude and Direction Angle
Magnitude:
Direction angle:
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-23
Vector v has magnitude 14.5 and direction angle 220°. Find the horizontal and vertical components.
Horizontal and Vertical Components
Horizontal component: –11.1
Vertical component: –9.3
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-24
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.
Example 4 Finding the Magnitude of a Resultant
because the adjacent angles of a parallelogram are supplementary.
Law of cosines
Find squareroot.
Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-25
Let u = 6, –3 and v = –14, 8. Find the following.
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:
jijiunit
length
53
537
53
532
53
7
53
2:
537)2( 22
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.
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.
Alternate Dot Product formula v · w = ||v||||w||cos(θ). The angle θ is the angle between the two vectors.
θw
v
Definition of Dot Product The dot product of 2 vectors is the sum of the
products of their horizontal components and their vertical components
2121
2211 ,
bbaawv
jbiawjbiav
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. 7
2. -5
3. -7
4. -4
Properties of Dot Product If u,v, & w are vectors and c is scalar, then
)()())(5
||)4
00)3
)()2
)1
2
cvuvucvcu
vvv
v
wuvuwvu
uvvu
Angle between vectors, v and w
wv
wvcos
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 BThe 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
43
Graphical Representation of a Complex Number Graph in coordinate plane
Called the complex plane
Horizontal axisis the real axis
Vertical axis is the imaginaryaxis
3 + 4i•-2 + 3i
•
• -5i
44
Absolute Value of a Complex Number Defined as the length of the line segment
From the origin To the point
Calculated byusing PythagoreanTheorem
3 + 4i•
2 23 4 3 4 25 5i
45
Find That Value, Absolutely Try these
Graph the complex number Find the absolute value
4 4z i 5z
5 6z i
46
Trig Form of Complex Number Consider the graphical representation
We note that a righttriangle is formed
a + bi•
θ
2 2
cos sin
cos sin
where
a b
r ra r b r
r z a b
b
a
r
How do we determine θ?
How do we determine θ?
1tanb
a
47
Trig Form of Complex Number Now we use
and substitute into z = a + bi
Result is
Abbreviation is often
cos sina r b r
cos sinz r i r
cisz r
48
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 = ? b = ?
49
Product of Complex Numbers in Trig Form Given
It can be shown that the product is
Multiply the absolute values Add the θ's
1 1 1 1 2 2 2 2cos sin cos sinz r i z r i
1 2 1 2 1 2cisz z r r
50
Quotient of Complex Numbers in Trig Form Given
It can be shown that the quotient is
1 1 1 1 2 2 2 2cos sin cos sinz r i z r i
1 11 2
2 2
cisz r
z r
51
Try It Out Try the following operations using trig form
Convert answers to standard form
4 cis120 6 cis315 15cis240
3cis135
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
29)5(252 22 i
Expressing complex numbers in polar form z = a + bi
)sin(cos
sin,cos
22
irz
bar
rbra
Express z = -5 + 3i in complex form
)54.sin()54.(cos(34
54.,5
3tan
343)5( 22
iz
r
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 Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
)5.322sin5.322(cos8
)5.262sin5.262(cos8
)5.202sin5.202(cos8
)5.142sin5.142(cos8
)5.82sin5.82(cos8
)5.22sin5.22(cos8)6
135sin
6
135(cos8
)135sin135(cos22,135,2
2tan
606
360,222)2(
126
126
126
126
126
1266
22
iz
iz
iz
iz
iz
iiz
iz
r
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 Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
)5.322sin5.322(cos8
)5.262sin5.262(cos8
)5.202sin5.202(cos8
)5.142sin5.142(cos8
)5.82sin5.82(cos8
)5.22sin5.22(cos8)6
135sin
6
135(cos8
)135sin135(cos22,135,2
2tan
606
360,222)2(
126
126
126
126
126
1266
22
iz
iz
iz
iz
iz
iiz
iz
r
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 Express in polar form, find the 1st root, then add 60
degrees successively to find the other 5 roots.
)5.322sin5.322(cos8
)5.262sin5.262(cos8
)5.202sin5.202(cos8
)5.142sin5.142(cos8
)5.82sin5.82(cos8
)5.22sin5.22(cos8)6
135sin
6
135(cos8
)135sin135(cos22,135,2
2tan
606
360,222)2(
126
126
126
126
126
1266
22
iz
iz
iz
iz
iz
iiz
iz
r
Formulas for index cards: Law of Sines: Sin A = Sin B = Sin C
a b c Law of Cosines:
a2 = b2 + c2 – 2bc Cos Aa2-b2-c2 = Cos A
-2bc Area of a Triangle
A = ½ ab Sin CA = √(s(s-a)(s-b)(s-c)); s = ½ (a+b+c)
||a|| = √(a12 + a2
2)
a1 = ||a||cos θ a2 = ||a||sin θ Unit Vector • a
cos θ = a•b angle between vectors
||a||||b|| compba = a•b
||b||
||||
1
a
Products and Quotients of Complex #’sz1z2 = 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
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