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#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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Presentation on theme: "#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!"— Presentation transcript:

1 #3 NOTEBOOK PAGE 16 – 9/7-8/2010

2 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! Write a Summary!

3 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

4 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!

5 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.

6 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 ( 

7 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!

8 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 

9 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.

10 Vector Components & Trigonometry Any Vector can be broken down into x and y components. Example: a plane flies @ 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 

11 Given the vectors below Find the magnitude of the x and y components of each. A B C A.72 km @ 58 O NE B.48 km @ 42 O SE C.119km @ 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!

12 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

13 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 @ 58 O NE B.48 km @ 42 O SE C.119km @ 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 2 + 81km 2 ) 1/2 = 198km  = Tan -1 Ry / Rx = 24 O NE ON YOUR SLATE!

14 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 (-,+)

15 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.

16 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 = 30km @ 25 O East of North B = 30km @ 15 O North of East C = 30km @ 32 O South of East A = 30km @ 65 O North of East _______________! ____! ON YOUR SLATE!

17 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!

18 Page 16 & 17 17 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! !


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