#3 NOTEBOOK PAGE 16 – 9/7-8/2010
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
#3
NOTEBOOK PAGE 16 – 9/7-8/2010
Page 16 & 17
1716Geometry & Trigonometry
P19 #2
P19 # 4
P20 #5
P20 # 7
Wed 9/8Tue 9/7 Problem Workbook.
Write questions!
Write a Summary!
Vector addition & Geometry• Pythagorean theorem • The square of the hypotenuse is equal to the sum of
the squares of the other two sides
•R2 = x2 + y2
•R = (x2 + y2)1/2
Given the vectors below Find the magnitude of the Resultant:
A
B
C
D1. A + C
2. A + B
3. B + H
Vector addition & Geometry
DRAW Graphical addition on your plastic slate! Show calculations to find magnitude of resultant.
ON YOUR SLATE!
Use the Pythagorean theorem
• When you can form a right triangle from the information given.
• You know the length of two sides of the triangle.
• The vectors are of the same unit of measure.
Vectors and Trigonometry• The Sine value of a right
triangle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse of the triangle.
• Cosine is a similar ratio
• The angle is called Theta (
sinOpp
Hyp
Example:
• If hyp = 6 and opp = 4• Sine = 4/6 = .66• This tells us that with the
same angle the opposite side is always 0.66 times the length of the hypotenuse!
• This is very useful!
When do you use Trigonometry?• When a Vector is described by it’s magnitude and
direction (angle • Since , and
• Then by algebra:– Opposite (or y component) = hyp x sin – Adjacent (or x component) = hyp x cos – We can always find the x and y components of a vector
it’s magnitude relative to the x axis!
sinOpp
Hyp
GIVEN: A, then AX = Acos and AY = Asin
Resolving a Vector Into ComponentsResolving a Vector Into Components
+x
+y
A
Ax
Ay
The horizontal, or x-component, of A is found by Ax = A cos
The vertical, ory-component, of A is found by Ay = A sin By the Pythagorean Theorem, Ax
2 + Ay2 = A2.
Every vector can be resolved using these formulas, such that A is the magnitude of A, and is the angle the vector makes with the x-axis.Each component must have the proper “sign”according to the quadrant the vector terminates in.
Vector Components & Trigonometry• Any Vector can be broken down into x and
y components. Example: a plane flies @ 100kph at 33O North of West. V = 100 kph = 33o
VX = V cos = 100 cos 33 = 83.9 kph VY = V sin = 100 sin 33 = 54.5 kph
ON YOUR SLATE!
GIVEN: A, then AX = Acos and AY = Asin
Given the vectors below Find the magnitude of the x and y
components of each.
A
B C A. 72 km @ 58O NE
B. 48 km @ 42O SE
C. 119km @ 26O NE
Vector Addition & Geometry
GIVEN: A, then AX = Acos and AY = Asin
Ax = 72cos(58) = 38km
Ay = 72sin(58) = 61km
Bx = 48cos(42) = 36km
By = 48sin(42) = -32km
Cx = 119cos(26) = 107km
Cy = 119sin(26) = 52km
ON YOUR SLATE!
Ax = A cos = Ay = A sin = Bx = B cos = By = B sin =Cx = C cos = Cy = C sin =
Rx = Ry =
Rx2 + Ry
2 = R2
4. Use the Pythagorean TheoremPythagorean Theorem to find the magnitude of the resultant vector.
3. Sum the y-components. This is the y-component of the resultant.
2. Sum the x-components. This is the x-component of the resultant.
1. Find the x- and y-components of each vector.Analytical Method of Vector AdditionAnalytical Method of Vector Addition
Given the vectors below Find the magnitude and direction of the sum of the following vectors by using x
and y component addition.
A
B C A. 72 km @ 58O NE
B. 48 km @ 42O SE
C. 119km @ 26O NE
Vector addition & Geometry
R= RX = 181km, RY = 81kmCy = 52km
Ax = 38km Ay = 61km
Bx = 36kmCx = 107km
By = -32km
R= (181km2 + 81km2)1/2 = 198km
= Tan-1 Ry/Rx = 24O NE
ON YOUR SLATE!
5. Find the reference angle by taking the inverse tangent of the absolute value of the y-component divided by the x-component.
= = TanTan-1-1 RRyy//RRxx
6. Use the “signs” of Rx and Ry to determine the quadrant.
NE(+,+)
NW(-,+)
SW
(-,-)SE
(-,+)
Sample Problem
A plane flies 65OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point.
A
B C
A plane flies 25OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point.
Vector addition & Geometry
• A = 30km @ 25O East of North
• B = 30km @ 15O North of East
• C = 30km @ 32O South of East
• A = 30km @ 65O North of East
_______________!
____!
ON YOUR SLATE!
A
B C
A plane flies 25OEast of North for 30 km, then 15O North of east for 42 km, then 32O South of east for 26 km. What is the reverse course and distance to the starting point.
Vector addition & Geometry
AX = 30km cos 65O NE = 12.7km
BX = 30km cos 15O NE =
CX = 30km cos -32O NE =
AY = 30km sin 65O NE =
BY = 30km sin 15O NE =
CY = 30km sin -32O NE =
ON YOUR SLATE!
Page 16 & 17
1716Geometry & Trigonometry Practice Problems
From Problem Workbook.
G.U.E.S.S.
P19 #2, 4
P20 #5, 7
Wed 9/8Tue 9/7
Write a Summary!
Write questions!
!
Related Documents
