MATH 125 Unit 2 1. B a

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1 MATH 15 Unit 1 Unit Law of Sines and Law of osines 1 Derive and identify the Law of Sines and the Law of osines 1 Derive and identify the Law of Sines. NOTE: See the objective overview for the derivation. sin A sin B Example: Solve each of the following, given = a b b= A= Try This: Solve each of the following, given sin sin B = c b c= B= Derive and identify the Law of osines. NOTE: See the objective overview for the derivation. Example: Solve each of the following, given the Law of osines b = a + c accos B a c = b = a + c + b bccos A abcos b= A= Try This: Solve each of the following, given the Law of osines b = a + c ac cos B a c = b = a + c + b bc cos A ab cos c= B= Dr. Paul Kennedy 8/17/015

2 MATH 15 Unit Apply the Law of Sines I: Two Angles 1 Solve triangles given two angles (AAS). A=5 B=37 = A=15 B= =100 a=4 b= c= a= b= c= 3 A= B=56 =75 A= B=5 =10 a= b= 6 c= a= b= c= 4 Solve triangles given two angles (ASA). A=0 B=55 = A=35 B= =80 a= b= c=10 a= b=14 c= A= B=56 =75 A= B=60 =105 a=15 b= c= a=17 b= c= Dr. Paul Kennedy 8/17/015

3 MATH 15 Unit 3 A 3 Solve applications with Law of Sines. Example: The height of an object above the ground can be determined by knowing the distance B and the angles of inclination from both B and. Determine h for each of the following. B=75ft. B=45, =0 B c h a b B=500 m. B=19, =9 Try This: B=3.5 mi. B=50, =13 B=4 km. B=9, =4 Dr. Paul Kennedy 8/17/015

4 MATH 15 Unit 4 3 Apply the Law of Sines II: Two Sides: The Ambiguous ase Two Sides: The Ambiguous ase 3.1 Solve the ambiguous case with two solutions. A=36 B= = A= B=55 = a=8 b=10 c= a=5 b= 3 c= A= 0 B= = A=50 B= = a=7 b= 10 c= a=13 b=15 c= 3. Solve the ambiguous case with one solution. A=65 B= = A= B=55 = a=7 b=5 c= a=4 b= 7 c= A= B=60 = A=50 B= = a= b= 6 c= a=13 b=11 c= Dr. Paul Kennedy 8/17/015

5 MATH 15 Unit Solve the ambiguous case with no solution. A= B=45 = A= B= =10 a=30 b=15 c= a= b= 30 c= 6 A= 50 B= = A= B= =100 a=0 b= 60 c= a=13 b= c= 5 4 Apply the Law of osines I 4.1 Solve for a missing side (SAS). A= B=4 = A= B= =10 a=30 b= c=10 a=4 b= 37 c= A= B= =78 A= B=3 = a=0 b= 60 c= a=1 b= c= 5 Dr. Paul Kennedy 8/17/015

6 MATH 15 Unit 6 4. Solve for a missing angle. (SSS) A= B= = A= B= = a=30 b=15 c=1 a=8 b= 30 c= 6 A= B= = A= B= = a=43 b= 6 c=5 a=17 b=19 c= Solve applications with Law of osines. Example: The distance, a=b, across a lake can be determined by knowing A, c, and b. Determine a for each of the following. A=85, c=45m, b=15 m A=15, c=5 mi, b=5 mi Try This: A=93, c=00 yd, b=350 yd A=115, c=3 km, b=5.5 km Dr. Paul Kennedy 8/17/015

7 MATH 15 Unit 7 5 Solve problems involving vectors 5.1 Given vectors compute its magnitude, direction, and resultant vectors. Example: For each of the following determine the magnitude, and direction in both degrees and radians of u and v. u= 3,6 v= -1,-4 3. Find the resultant vector for u and v, w = u + v, its magnitude and direction. Try This: For each of the following determine the magnitude, and direction in both degrees and radians. 1 u= 3,6 v= -1,-4 3. Find the resultant vector for u and v, w = u + v, its magnitude and direction. 5. Solve applications of vectors in the i j form. Example: Solve each problem An object of mass 00 kg is launched straight up into the air, and the magnitude of the force due to gravity acting on the object is 9.81 N. At the time of launch, wind blows from west to east, and the magnitude of the force acting on the object is 00 N. ompute the resultant force and the direction of the resultant force acting on the object, in radians, 0 θ < π, round to three decimal places. Two positive charges, charge 1 and charge, are separated by a certain distance. harge 1 is at the origin, and charge is at (,0). At the midpoint (1,0), the electric field due to charge 1 is: E 1 = N i, and the electric field due to charge is E = N j. What is the net electric field, E, at the midpoint? Try this: Solve each problem 1 An object of mass 5.00 kg is launched straight up into the air, and the magnitude of the force due to gravity acting on the object is 9.81 N. At the time of launch, wind blows from west to east, as well as up, and the magnitude of the force acting on the object is 3.00 N east, and 00 N straight up in the air. ompute the resultant force and the direction of the resultant force acting on the object, in radians, 0 θ < π, round to three decimal places. Two positive charges, charge 1 and charge, are separated by a certain distance. harge 1 is at the origin, and charge is at (5,0). At the midpoint (5,0), the electric field due to charge 1 is: E 1 = N i, and the electric field due to charge is E = N j. What is the net electric field, E, at the midpoint? i. ompute the dot product and angle between two vectors. Dr. Paul Kennedy 8/17/015

8 MATH 15 Unit 8 Example: Find the angle between the vectors u = 1,4 and v = 3, u = 1,4 and v = 3, Try This: Find the angle between the vectors u = 10, 1 and v = 3,5 u =,4 and v = 1, Dot Product Examples With a force of 30 N, a person pushes a box across the floor of 4.00m in the positive x direction at an angle of 45 below the horizontal. What is the work done on the box? Round to three decimal places if necessary. If the magnetic flux through a square loop of area 3.656m is equal to 0.518, and the magnetic field has a magnitude of T, what is the angle, in radians, between the magnetic field and the normal to the surface? Round to three decimal places if necessary. Dr. Paul Kennedy 8/17/015