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Method #3: Law of Sines and Cosines Also referred to as the analytical method.

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Presentation on theme: "Method #3: Law of Sines and Cosines Also referred to as the analytical method."— Presentation transcript:

1 Method #3: Law of Sines and Cosines Also referred to as the analytical method.

2 Steps  Draw a rough sketch of the vectors  The resultant is determined using:  Algebra  Trigonometry  Geometry

3 Any These Laws Work for Any Triangle. a c b C BA A + B + C = 180° Law of sines: a = b = c sin A sin B sin C Law of cosines: c 2 = a 2 + b 2 –2abcos C

4 Example 2: using method 3 Stan is trying to rescue Kyle from drowning. Stan gets in a boat and travels at 6 m/s at 20 o N of E, but there is a current of 4 m/s in the direction of 20 o E of N. Find the velocity of the boat.

5 Example (using same problem) Stan is trying to rescue Kyle from drowning. Stan gets in a boat and travels at 6 m/s at 20 o N of E, but there is a current of 4 m/s in the direction of 20 o E of N. Find the velocity of the boat.

6 Calculating: Magnitude: c 2 = a 2 + b 2 – 2abcosC = (6m/s) 2 + (4m/s) 2 – 2(6m/s)(4m/s)cos130° = 82.85 c = 9.10 m/s Direction: sin C = sin B c b sin 130°=sin B 9.10 4 sin B = 0.337B = 19.67° R = 19.67° + 20° = 39.67° R = 9.1 m/s @ 39.7° N of E

7 Use the Law of:  Sines when you know: 2 angles and an opposite side 2 sides and an opposite angle  Cosines when you know: 2 sides and the angle between them

8 Advantages and Disadvantages of the Analytical Method  Does not require drawing to scale.  More precise answers are calculated.  Works for any type of triangle if appropriate laws are used.  Can only add 2 vectors at a time.  Must know many mathematical formulas.  Can be quite time consuming.

9 This completes Method Three! Keep up the good work! This is our last time in class to learn these. problems #5, 6 due tomorrow

10 Another Problem Paul is on a railroad flat car which is moving east at 20.0 m/s (V cg = velocity of the car relative to the ground). Paul walks on the flat car at 5.0 m/s @ 40.0 o N of E as shown (V pc = velocity of Paul relative to the car). What is Paul’s velocity relative to the ground (V pg = velocity of Paul relative to the ground)? V pg = 24 m/s @ 7.7 o (or 7.7 o N of E)


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