University:
Winston-Salem Forsyth County Schools
Course:
AP Calculus AB
Academic year:

2013
Views:

350

Pages:

2
Author:

Triston Herrera

Problem Solving Session #3: Law of Cosines Instructions: Solve the following problems with your group. Each member is responsible for writing on their board. Write down the name of the group member responsible for each board. Do all the work on the boards. Once finished, copy your final work from each board to this sheet. The person responsible for the plan board is the leader for that problem. The leader is responsible for communication and deciding on suggested ideas. Problem 1: Problem: A surveyor walks 380 meters from point A to point B, then turns 80 degrees and walks 240 meters to point C. Approximate the length AC. Round to two decimal places. Plan Board (Name): Gobil Sketch Board (Name): Garrett Solve Board (Name): Al Mathematical Work: b^2 = 240^2 + 380^2 - 2 * 240 * 380 * cos(80) b^2 = 200000 + 51673.43 b^2 = 251673.43 b = sqrt(251673.43) b = 483.4 meters Problem 2: Problem: A plane flies 810 miles from A to B with a bearing of N 75°E. Then it flies 648 miles from B to C with a bearing of N 32°E. Draw a diagram to represent the problem and find the straight line distance and bearing from C to A. Plan Board (Name): Austyn Sketch Board (Name): Gabriel Solve Board (Name): Gerrett Mathematical Work: b^2 = 648^2 + 810^2 - 2 * 648 * 810 * cos(157) b^2 = 1557.85^2 b = 1557.85 miles sin(157) / 1557.85 = sin(C) / 810 C = 24 degrees 180 - 75 - 24 = 81 degrees Bearing from C to A: S 56°W Problem 3: Problem: A baseball player in center field is playing approximately 330 feet from the television camera that is behind home plate. A batter hits the ball that goes to the wall 420 feet from the camera. Approximate the number of feet the center fielder has to run to make the catch if the camera turns 6 degrees to follow the play. Plan Board (Name): Garrett Sketch Board (Name): Austin Solve Board (Name): Gabriel Mathematical Work: b^2 = 420^2 + 330^2 - 2 * 420 * 330 * cos(6) b = 98.07 feet Problem 4: Problem: The lengths of the diagonals of a parallelogram are 10 feet and 16 feet. Find the lengths of the sides of the parallelogram if the diagonals intersect at an angle of 28 degrees. Plan Board (Name): Gabriel Sketch Board (Name): Gerrett Solve Board (Name): Austin Mathematical Work: b_1^2 = 8^2 + 5^2 - 2 * 8 * 5 * cos(28) b_1 = 4.28 feet b_2^2 = 8^2 + 5^2 - 2 * 8 * 5 * cos(152) b_2 = 12.63 feet Problem 5: Problem: Solve for c, d, and phi. The lengths of the diagonals are c and d. Given: a = 4, b = 8, and theta = 30 degrees Mathematical Work: d^2 = 4^2 + 8^2 - 2 * 4 * 8 * cos(70) d = 4.96 feet phi = 150 degrees c^2 = 4^2 + 8^2 - 2 * 4 * 8 * cos(50) c = 11.64 feet