Copyright © 2017, 2013, 2009 Pearson Education, Inc. Applications of Trigonometry and Vectors Copyright © 2017, 2013, 2009 Pearson Education, Inc. 1

Geometrically Defined Vectors and Applications 7.4 Geometrically Defined Vectors and Applications Basic Terminology ▪ The Equilibrant ▪ Incline Applications ▪ Navigation Applications

Basic Terminology Scalar: The magnitude of a quantity. It can be represented by a real number. A vector in the plane is a directed line segment. Consider vector OP O is called the initial point. P is called the terminal point.

Basic Terminology Magnitude: length of a vector, expressed as |OP| Two vectors are equal if and only if they have the same magnitude and same direction. Vectors OP and PO have the same magnitude, but opposite directions. |OP| = |PO|

Basic Terminology A = B C = D A ≠ E A ≠ F

Two ways to represent the sum of two vectors The sum of two vectors is also a vector. The vector sum A + B is called the resultant. Two ways to represent the sum of two vectors

Sum of Two Vectors The sum of a vector v and its opposite –v has magnitude 0 and is called the zero vector. To subtract vector B from vector A, find the vector sum A + (–B).

Scalar Product of a Vector The scalar product of a real number k and a vector u is the vector k ∙ u, with magnitude |k| times the magnitude of u.

Properties of Parallelograms 1. A parallelogram is a quadrilateral whose opposite sides are parallel. 2. The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles of a parallelogram are supplementary. 3. The diagonals of a parallelogram bisect each other, but do not necessarily bisect the angles of the parallelogram.

Use the law of cosines with either triangle. Example 1 FINDING THE MAGNITUDE OF A RESULTANT Two forces of 15 and 22 newtons act on a point in the plane. (A newton is a unit of force that equals 0.225 lb.) If the angle between the forces is 100°, find the magnitude of the resultant vector. The angles of the parallelogram adjacent to P measure 80° because the adjacent angles of a parallelogram are supplementary. Use the law of cosines with either triangle.

The magnitude of the resultant vector is about 24 newtons. Example 1 FINDING THE MAGNITUDE OF A RESULTANT (continued) The magnitude of the resultant vector is about 24 newtons.

The Equilibrant Sometimes it is necessary to find a vector that will counterbalance a resultant. This opposite vector is called the equilibrant. The equilibrant of vector u is the vector –u.

Example 2 FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT Find the magnitude of the equilibrant of forces of 48 newtons and 60 newtons acting on a point A, if the angle between the forces is 50°. Then find the angle between the equilibrant and the 48-newton force. The equilibrant is –v.

The magnitude of –v is the same as the magnitude of v. Example 2 FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.) The magnitude of –v is the same as the magnitude of v.

Example 2 FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.) The required angle, , can be found by subtracting angle CAB from 180°. Use the law of sines to find angle CAB.

Example 2 FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.)

The vertical force BA represents the force of gravity. Example 3 FINDING A REQUIRED FORCE Find the force required to keep a 50-lb wagon from sliding down a ramp inclined at 20° to the horizontal. (Assume there is no friction.) The vertical force BA represents the force of gravity. BA = BC + (–AC) Vector BC represents the force with which the weight pushes against the ramp.

Vector BF represents the force that would pull the weight up the ramp. Example 3 FINDING A REQUIRED FORCE (cont.) Vector BF represents the force that would pull the weight up the ramp. Since vectors BF and AC are equal, |AC| gives the magnitude of the required force. Vectors BF and AC are parallel, so the measure of angle EBD equals the measure of angle A.

Example 3 FINDING A REQUIRED FORCE (cont.) Since angle BDE and angle C are right angles, triangles CBA and DEB have two corresponding angles that are equal and, thus, are similar triangles. Therefore, the measure of angle ABC equals the measure of angle E, which is 20°.

From right triangle ABC, Example 3 FINDING A REQUIRED FORCE (cont.) From right triangle ABC, A force of approximately 17 lb will keep the wagon from sliding down the ramp.

Example 4 FINDING AN INCLINE ANGLE A force of 16.0 lb is required to hold a 40.0-lb lawn mower on an incline. What angle does the incline make with the horizontal? Vector BE represents the force required to hold the mower on the incline. In right triangle ABC, the measure of angle B equals θ, the magnitude of vector BA represents the weight of the mower, and vector AC equals vector BE.

The hill makes an angle of about 23.6° with the horizontal. Example 4 FINDING AN INCLINE ANGLE (cont.) The hill makes an angle of about 23.6° with the horizontal.

Example 5 APPLYING VECTORS TO A NAVIGATION PROBLEM A ship leaves port on a bearing of 28.0° and travels 8.20 mi. The ship then turns due east and travels 4.30 mi. How far is the ship from port? What is its bearing from port? Vectors PA and AE represent the ship’s path. We are seeking the magnitude and bearing of PE. Triangle PNA is a right triangle, so the measure of angle NAP = 90° − 28.0° = 62.0°.

The ship is about 10.9 mi from port. Example 5 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued) Use the law of cosines to find |PE|. The ship is about 10.9 mi from port.

Now add 28.0° to 20.4° to find that the bearing is 48.4°. Example 5 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued) To find the bearing of the ship from port, first find the measure of angle APE. Use the law of sines. Now add 28.0° to 20.4° to find that the bearing is 48.4°.

Airspeed and Groundspeed The airspeed of a plane is its speed relative to the air. The groundspeed of a plane is its speed relative to the ground. The groundspeed of a plane is represented by the vector sum of the airspeed and windspeed vectors.

Example 6 APPLYING VECTORS TO A NAVIGATION PROBLEM An airplane that is following a bearing of 239º at an airspeed of 425 mph encounters a wind blowing at 36.0 mph from a direction of 115º. Find the resulting bearing and ground speed of the plane.

Example 6 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)

Example 6 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued) Vector c represents the airspeed and bearing of the plane, and vector a represents the speed and direction of the wind. Angle ABC has its measure the sum of angle ABN1 and angle N1BC. Angle SAB measures 239º – 180º = 59º. Because angle ABN1 is an alternate interior angle to it, ABN1 = 59º Angle E1BF measures 115º – 90º = 25º. Thus, angle CBW1 also measures 25º because it is a vertical angle. Angle N1BC is the complement of 25º, which is 90º – 25º = 65º.

By these results, angle ABC = 59º + 65º = 124º. Example 6 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued) By these results, angle ABC = 59º + 65º = 124º. To find │b│, we use the law of cosines.

Add 4º to 239º to find the resulting bearing of 243º. Example 6 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued) To find the resulting bearing of b, we must find the measure of angle α in the figure and then add it to 239º. Use the law of sines. Add 4º to 239º to find the resulting bearing of 243º.