#3 NOTEBOOK PAGE 16 – 9/7-8/2010. Page 16 & 17 17 16 Geometry & Trigonometry P19 #2 P19 # 4 P20 #5 P20 # 7 Wed 9/8 Tue 9/7 Problem Workbook. Write questions!

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#3 NOTEBOOK PAGE 16 – 9/7-8/2010

Page 16 & Geometry & Trigonometry P19 #2 P19 # 4 P20 #5 P20 # 7 Wed 9/8 Tue 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 R 2 = x 2 + y 2 R = (x 2 + y 2 ) 1/2

Given the vectors below Find the magnitude of the Resultant: A B C D 1.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 ( 

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! GIVEN: A, then A X = Acos  and A Y = Asin 

Resolving a Vector Into Components +x +y A AxAx AyAy  The horizontal, or x-component, of A is found by A x = A cos  The vertical, or y-component, of A is found by A y = A sin  By the Pythagorean Theorem, A x 2 + A y 2 = A 2. 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 100kph at 33 O North of West.  V = 100 kph   = 33 o  V X = V cos   = 100 cos 33  = 83.9 kph  V Y = V sin   = 100 sin 33  = 54.5 kph ON YOUR SLATE! GIVEN: A, then A X = Acos  and A Y = Asin 

Given the vectors below Find the magnitude of the x and y components of each. A B C A O NE B O SE 26 O NE Vector Addition & Geometry GIVEN: A, then A X = Acos  and A Y = 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!

A x = A cos  =A y = A sin  = B x = B cos  = B y = B sin  = C x = C cos  =C y = C sin  = Rx =Rx =R y = R x 2 + R y 2 = R 2 Pythagorean Theorem 4. Use the Pythagorean 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 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 O NE B O SE 26 O NE Vector addition & Geometry R= R X = 181km, R Y = 81km Cy = 52km Ax = 38kmAy = 61km Bx = 36km Cx = 107km By = -32km R= (181km km 2 ) 1/2 = 198km  = Tan -1 Ry / Rx = 24 O 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.  = Tan -1 R y /R x 6. Use the “signs” of R x and R y to determine the quadrant. NE (+,+) NW (-,+) SW (-,-) SE (-,+)

Sample Problem A plane flies 65 O East of North for 30 km, then 15 O North of east for 42 km, then 32 O South of east for 26 km. What is the reverse course and distance to the starting point.

A B C A plane flies 25 O East of North for 30 km, then 15 O North of east for 42 km, then 32 O South of east for 26 km. What is the reverse course and distance to the starting point. Vector addition & Geometry A = 25 O East of North B = 15 O North of East C = 32 O South of East A = 65 O North of East _______________! ____! ON YOUR SLATE!

A B C A plane flies 25 O East of North for 30 km, then 15 O North of east for 42 km, then 32 O South of east for 26 km. What is the reverse course and distance to the starting point. Vector addition & Geometry A X = 30km cos 65 O NE = 12.7km B X = 30km cos 15 O NE = C X = 30km cos -32 O NE = A Y = 30km sin 65 O NE = B Y = 30km sin 15 O NE = C Y = 30km sin -32 O NE = ON YOUR SLATE!

Page 16 & Geometry & Trigonometry Practice Problems From Problem Workbook. G.U.E.S.S. P19 #2, 4 P20 #5, 7 Wed 9/8 Tue 9/7 Write a Summary! Write questions! !