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

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.



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.


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


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?


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



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


harbor

ship 1 ship 2

140° 34mph*3hr = 102 miles

x

cos2222 bccba


harbor

ship 1 ship 2

140° 34mph*3hr = 102 miles

x


harbor

ship 1 ship 2

140° 34mph*3hr = 102 miles

x

harbor


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′


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


Find the magnitude and direction angle for

Magnitude and Direction Angle

Magnitude:

Direction angle:


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


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.


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



)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



)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

Documents

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