Transcript Slide 1

Bearing
SSA Triangles
Complex Polar Coordinates
Binomial Theorem (Basic)
Binomial Theorem (Terms)
Vectors
Trig Equations
Law of Cosines
Comments
Please report any errors ASAP by email to [email protected] 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.
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)
1) Figure out angles
Plane : 400(cos 180î  sin180  ĵ)  400î
Wind : 10(cos 330î  sin330  ĵ)  8.67 î  5ĵ
3) Add vectors
2) Make vectors
4) Find magnitude
5) Find Bearing
 391.33î  5ĵ
(391.33) 2  (5) 2
60o
180o
391.36 mph
  tan 1
5
 .73 180  180.73
 391.33
QIII
330o
W 73 S
Remember
inverse
tangent
either
I orsubtract
IV. FROM
Make
aasketch
of the
Read
problems
carefully,
whether
thequadrant
windeither
is coming
direction
ordeduce
isvector
HEADING
The angle
for
the
bearing
is is
.73.
You
can
180,
or
logically
it,toorsee
what
quadrant
the
angle
to
be in.
IN
A direction.
Heading
asupposed
direction
is straight
forward,
fromon
a direction
is
whatever
you may
need.inisYou
don’t always
subtract
180. coming
It depends
what quadrant
trickier.
it’s
in. Nothing
QI

QII this
180
QIII
 Add
180 QIV
 where
Add 360
Remember,
isAdd
the
same as
ground
speed.
Wind
is coming
FROM
direction,
which
is different
from
it is heading.
Distribute
themagnitude
magnitude.
y  component
1
2
2   tan
Then
find
the
bearing
afterwards.
o
So it’s really heading E 30 S. Magnitude  a  b
x  component
z   3  1i
w  2 2  2 2i
r  ( a )2  ( b )2
r  Find
4 2
Putting into complex polar
coordinate form.
1) Find radius
2) Find argument (angle)
a) Inverse Tangent
z


zw
1
i) QII, QIII add 180
  30180  210w
tan 
2

2
ii) QIV, add 360
z 2
r   2 2  2 2   4  (cis (210  135))
zw  (2)( 4) cis III
210  135
w 4
Remember,
if complex
Quadrant
Convert
the other
number into complex polar
1
 8cis
(345
2) 2 
1
1

cis
(
75

)
z1  rNext
1 ) will give
 240 isintan240)1  245
form.
1 (cisclick
tan
2
(cos
Remember,
  2if 2your
 argument
answer.


z 2  r2 (cis  2 )
isn' t between 0 and 360,
Quadrant
z1 z 2  r1r2 cis 1   2 
then subtractedII360.
b) Figure out angle

  
 45  180  135
4(cos 135  i sin 135)
z1 r1
 cis 1   2 
z 2 r2
 n
n!
nCj  
 j   j!(n  j )!
 
Parenthesis, you need them.
 n  n j
( x  a)    ( x) (a) j
j 0  j 
Binomial Theorem
Common errors:
1)
Otherwise your powers will be
messed up. (Math kryptonite)
n
n
 3
(2 x  3 y)    (2 x)3 j (3 y ) j
j 0  j 
3
3
2) Set up the bottom factorial
carefully.
3) Keep sign.
2
1 3
3 3
0 3
3
0
1
2 3
 (2 x)( 3 y )   (2 x)( 3 y )   (2 x)( 3 y )   (2 x)( 3 y )
0
1
2
3
8 x  (3)( 4 x )( 3 y)  (3)(2 x)(9 y )  (27 y )
1st term
Notice:
3
2 term
2nd
3rd term
2
8 x  36 x y  54 xy  27 y
3
2
2
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.
4th term3
3
Binomial Theorem (Terms)
List
Logic
Methods
Formula
1) Be safe, list them all, pick the one you need
2) Logic
Clear
Clear
Clear
3) Formula
( 2 x  y )8
Find term with x 2
Term with y 4
6th term
6

8
8
 8  number
x 2 8means
number
4
8y because
0
7
1n Bottom
6
2 is 8
5
3Bottom


 (n2 x) ( y )   (2 xj ) ( y )   (2n x )j (j y )   (2 x) ( y )   (2 x) 4 ( y )is4 5
4
( x2 0a6) 8, term with
x is 4 3 
1 x  
 2
a power

 n  Other

8
 
3
5
j




(
2
x
)
(

y
)
Bottom
number
is
6
 58
8
8
8
3
5
2
6  88  4 1
0
4 7
) 









(
2
x
)
(

y
)

(
2
x
)
(

y
)

(
2
x
)
(

y
)

(
2
x
)
(

y


(
2
x
)
(

y
)
2










x ,8 (j25x) 22( y ) 6
3 5
6
8
 47 


448
x
y
 6
3
5
4
4
4
4
2
6

(
56
)(
2
x
)
(

y
)
The
rest,
you
use
logic
to
set
it
up
so
that



1120
x
y

(
70
)(
2
x
)
(

y
)
 8  (28)(2 266x) ( y2)
you can use the formula. Refer to other
slide
 112
(112
 yxx)2yy(62 x)
 448x 3 y 5
 1120 x 4 y 4
 2
 112 x 2 y 6
or logic button for logical rules. You use
logic to set up the x term.
1) Bottom number matches up with second term exponent.
2) Exponents add up to n
3) Term number is one more than bottom number.
Given initial point P(x 1 , y1 )
and terminal point Q(x 2 , y 2 )
The vector is :
x 2  x1 , y 2  y1
P (3,4)
P(3,5)
Q (1,1)
Q(2,7)
1  (3) ,  1  4
2  (3) ,  7  5
4,  5
5,  12

Given u  a1 , b1

v  a2 , b2
 
u v
cos    
|| u |||| v ||
Find the angle between th e two
vectors on the left.
4,5  5,12
cos  
|| 41 |||| 13 ||
20  60
80
cos  

13 41 13 41
  16.04


Given u  a1 , b1 v  a2 , b2
 
u
 v  a1a2  b1b2
the vector can be written in aî  bĵ

Given u  a1 , b1
form by :
Given vector with r  4,  200

2
2
|| u || a1   b1 
rcos î  rsin  ĵ write in component (aiˆ  bˆj ) form
Given magnitude r and argument  ,
4 cos 200iˆ  4 sin 200 ˆj
 3.76iˆ  1.37 ˆj
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.
SSS – 3 sides given
SAS – 2 sides given, name of angle is not
the letter of the other sides. (or make a
sketch)
a  8 b  6 C  70
a4 b5 c6
B
B
A
2 b
C
6  4  5  2(4)(5) cos C
36  41 40 cos C
 5  40 cos C
1
 cos C
8
82.81  mC
2
2
c 2  a 2  b 2  2ab cos C
a 2  b 2  c 2  2bc cos A
b 2  a 2  c 2  2ac cos B
a
c
a
c
A
b
C
c 2  82  62  2(8)(6) cos 70
c 2  100  32.8339
2
c  67.1661
c  8.20
Just showing LOC step. Use LOS to finish the problem.
Quick Check, small angle small side, middle angle
middle side, big angle big side.
CHECK MODE OF CALCULATOR!
Rules
SSA triangle. Use law of sines.
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
Check
sin 40 sin B

supplement and
4
5
the third angle.
5 sin 40
sin B 
180-53.46o =
4
126.54o
B  53.464
mC = 30o; c = 6, b = 4
sin 30 sin B Check

supplement and
6
4
the third angle.
4 sin 30
sin B 
180-19.47o =
6
160.53o
B  19.47
Both triangles are ok, you can finish
using law of sines.
2nd triangle is impossible, only need
to solve for the top one.
Using Rules : 4  5
Using Rules : 6  4
5 sin 40  3.21  4
1 triangle
h = bsinA
a is the side opposite the
angle.
b is side adjacent to angle
a < h, no triangle
b
A
a = h, one right triangle
b
mA 
mB 
53.46o
a 4
b5
mC  86.54o c 
mA  40o
a4
mB 126.54o b  5
mC 13.46o c 
mA 
130.53o
mB 
19.47o
mC  30o
a
A
a < b and h < a, 2 triangles
2 triangles
40o
a
b
a
b 4
c6
mA -10.53o a 
mB 160.53o b  4
mC  30o
c 6
a
a
A
a ≥ b, one triangle
b
A
a
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.