Bearing

Complex Polar Coordinates

Binomial Theorem (Terms)

Trig Equations

Law of Cosines

Binomial Theorem (Basic)

Vectors

SSA Triangles

Comments

Please report any errors ASAP by email to sakim@fjuhsd.net or IM at kimtroymath.

Problems may be more difficult on test. Consult homework assignment. Not all topics covered.

Ones in read are the ones that have been completed. Remember, some material is on other powerpoints.

Green are always changing.

1) Figure out angles

2) Make vectors

3) Add vectors

4) Find magnitude

5) Find Bearing

A plane is traveling 400 miles per hour west. A wind from a direction of N 60o W is 10 mph. Find the ground speed and bearing of the plane. (I will round to hundredths)

)jsin180i180(cos400: Plane

180o

i400

Wind is coming FROM this direction, which is different from where it is heading.

60o

So it’s really heading E 30o S.

330o

)jsin330i330(cos10: Wind j5i67.8

Distribute the magnitude.

j5i33.391

Remember, magnitude is the same as ground speed.

22 baMagnitude

mph36.391

)5()33.391( 22

Remember inverse tangent is either quadrant I or IV. Make a sketch of the vector to see what quadrant the angle is supposed to be in.

QI Nothing QII Add 180 QIII Add 180 QIV Add 360

Then find the bearing afterwards. componentx

componenty

1tan

73.

33.391

5tan 1 QIII 73.180180

The angle for the bearing is .73. You can either subtract 180, or logically deduce it, or whatever you may need. You don’t always subtract 180. It depends on what quadrant it’s in.

SW 73

Read problems carefully, whether the wind is coming FROM a direction or is HEADING IN A direction. Heading in a direction is straight forward, coming from a direction is trickier.

22 )()( bar

iz 3 iw 2222 Putting into complex polar coordinate form.

1) Find radius

2) Find argument (angle)

a) Inverse Tangent

b) Figure out angle

i) QII, QIII add 180

ii) QIV, add 360

3 1

24 r

1tan

3 1

30

IIIQuadrant

210180

)240sin240(cos2 i

Convert the other complex number into complex polar form. Next click will give answer.

13518045

451tan22

22tan

42222

11

22

IIQuadrant

r

)135sin135(cos4 i

212

1

2

1

212121

222

111

)(

)(

if Remember,

cisr

r

z

z

cisrrzz

cisrz

cisrz

w

z zw

Find

360. subtractedthen

360, and 0between t isn'

argumentyour if Remember,

)345(8

135210)4)(2(

cis

ciszw

)75(2

1

))135210((4

2

w

z

cis

cis

Binomial Theorem

n

j

jjnn axj

nax

0

)()()(Common errors:

1) Parenthesis, you need them. Otherwise your powers will be messed up. (Math kryptonite)

2) Set up the bottom factorial carefully.

3) Keep sign.

)!(!

!

jnj

n

j

nC jn

3)32( yx

3

0

3 )3()2(3

j

jj yxj

)3)(2(3

3)3)(2(

2

3)3)(2(

1

3)3)(2(

0

3yxyxyxyx

3 0 2 1 21 30

Notice:

1) First term starts with exponent, goes down by 1.

2) Second term starts with 0, goes up by 1.

3) Bottom number matches up with second term exponent.

4) Exponents add up to n

5) Term number is ONE MORE than bottom number.

3223

3223

2754368

)27()9)(2)(3()3)(4)(3(8

yxyyxx

yyxyxx

1st term 2nd term 3rd term 4th term

62

62

62

112

)()2(6

8

6

862

yx

yx

isnumberBottom

becauseymeansx

62

26

2

112

)2()(2

8

2,

,)(

yx

xy

jx

xajn

nxwithtermax jjnjn

Binomial Theorem (Terms) Methods

1) Be safe, list them all, pick the one you need

2) Logic

3) Formula

List

Clear

Logic

Clear

Formula

Clear

6th term y with Term with x termFind

)2(42

8yx

80716253

4435261708

)()2(8

8)()2(

7

8)()2(

6

8)()2(

5

8

)()2(4

8)()2(

3

8)()2(

2

8)()2(

1

8)()2(

0

8

yxyxyxyx

yxyxyxyxyx

62

62

112

)()2)(28(

yx

yx

44

44

1120

)()2)(70(

yx

yx

53

53

448

)()2)(56(

yx

yx

1) Bottom number matches up with second term exponent.

2) Exponents add up to n

3) Term number is one more than bottom number.

44

44

1120

)()2(4

8

4

4

yx

yx

ispowerOther

isnumberBottom

53

53

448

)()2(5

8

5

yx

yx

isnumberBottom

The rest, you use logic to set it up so that you can use the formula. Refer to other slide or logic button for logical rules. You use logic to set up the x term.

||||||||cos

,, 2211

vu

vu

bavbauGiven

1212

22

11

yy,xx

:is vector The

)y,Q(xpoint terminaland

)y,P(xpoint initialGiven

jrsinircos

:by form

jbiain written becan vector the

,argument andr magnitudeGiven

)1,1(

)4,3(

Q

P

)7,2(

)5,3(

Q

P

54

41)3(1

,

,

125

57)3(2

,

,

left. on the vectors

twoebetween th angle theFind

||13||||41||

12,55,4cos

4113

80

4113

6020cos

04.16

2121

2211 ,,

bbaavu

bavbauGiven

21

21

11

||||

,

bau

bauGiven

formjbiacomponentinwrite

rwithvectorGiven

)ˆˆ(

200,4

ji ˆ200sin4ˆ200cos4 ji ˆ37.1ˆ76.3

Check the General Trig Powerpoint Ch 6 for good equation examples. If you want a specific hw problem done, e-mail me. I’ll try to fit 1 or 2 in here.

Law of Cosines – To be used with SSS or SAS triangle. There are no ambiguous cases for these triangles.

654 cba 7068 Cba

SSS – 3 sides given SAS – 2 sides given, name of angle is not the letter of the other sides. (or make a sketch)

Baccab

Abccba

Cabbac

cos2

cos2

cos2

222

222

222

Quick Check, small angle small side, middle angle middle side, big angle big side.

CHECK MODE OF CALCULATOR!

Ccos)5)(4(2546 222 Ccos404136

Ccos405

Ccos8

1

Cm81.82

Just showing LOC step. Use LOS to finish the problem.

A

B

Cb

ac

A

B

Cb

ac

70cos)6)(8(268 222 c8339.321002 c

1661.672 c20.8c

SSA triangle. Use law of sines. Rules

h = bsinA

a is the side opposite the angle.

b is side adjacent to angle

a < h, no triangle

a = h, one right triangle

a < b and h < a, 2 triangles

a ≥ b, one triangle

A

b a

A

A

A

a

a a

a

b

b

b

You can use common sense. Set up a law of sines. And then try to find the supplement. See if 0, 1, or both triangles work. Common sense is nice because even if you use the rules, if you notice there are 2 triangles, you will need to use the supplement anyways.m A = 40o; a = 4, b = 5

cCm

bBm

aAm

5

40o

40o 4

4

5

464.534

40sin5sin

5

sin

4

40sin

B

B

B

53.46o

cCm

bBm

aAm

cCm

bBm

aAm

cCm

bBm

aAm

Check supplement and the third angle.

180-53.46o = 126.54o

126.54o

86.54o

13.46o

Both triangles are ok, you can finish using law of sines.

triangles2

421.340sin5

54 :Rules Using

m C = 30o; c = 6, b = 4

630o

4

630o

4

47.196

30sin4sin

4

sin

6

30sin

B

B

B

19.47o

Check supplement and the third angle.

180-19.47o = 160.53o

160.53o

130.53o

-10.53o

2nd triangle is impossible, only need to solve for the top one.

triangle1

46 :Rules Using

9

10

11

12

13

Comments

Make sure you really understand graphing and solving equations.

Graphing, know the formulas, and how to find period, amplitude and shift.