copyright © 2009 pearson addison-wesley 7.5-1 7 applications of trigonometry and vectors

Copyright © 2009 Pearson Addison-Wesley 7.5-1

7Applications of Trigonometry and Vectors


7.1 Oblique Triangles and the Law of Sines

7.2 The Ambiguous Case of the Law of Sines

7.3 The Law of Cosines

7.4 Vectors, Operations, and the Dot Product

7.5 Applications of Vectors

7Applications of Trigonometry and Vectors

Copyright © 2009 Pearson Addison-Wesley

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Applications of Vectors7.5The Equilibrant ▪ Incline Applications ▪ Navigation Applications


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.


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Example 1 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.


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Example 1 FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.)

The magnitude of –v is the same as the magnitude of v.


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The required angle, , can be found by subtracting the measure of angle CAB from 180°. Use the law of sines to find the measure of angle CAB.


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Example 2 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.


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Example 2 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.


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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°.


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From right triangle ABC,

A force of approximately 17 lb will keep the wagon from sliding down the ramp.


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Example 3 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.


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Example 3 FINDING AN INCLINE ANGLE (cont.)

The hill makes an angle of about 23.6° with the horizontal.


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Example 4 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°.


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Example 4 APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)

Use the law of cosines to find |PE|.

The ship is about 10.9 miles from port.


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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.


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Example 5 APPLYING VECTORS TO A NAVIGATION PROBLEM

A plane with an airspeed of 192 mph is headed on a bearing of 121°. A north wind is blowing (from north to south) at 15.9 mph. Find the groundspeed and the actual bearing of the plane.

The groundspeed is represented by |x|.

We must find angle to determine the bearing, which will be 121° + .


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Find |x| using the law of cosines.

The plane’s groundspeed is about 201 mph.


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Find using the law of sines.

The plane’s groundspeed is about 201 mph on a bearing of 121° + 4° = 125°.

copyright © 2009 pearson addison-wesley 7.5-1 7 applications of trigonometry and vectors

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