Applications of Trigonometry and Vectors

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

7

Applications of Trigonometry and Vectors


7.1 Oblique Triangles and the Law of Sines

7.3 The Law of Cosines

7.4 Vectors, Operations, and the Dot Product

7.5Applications of Vectors

Applications of Trigonometry and Vectors7


Oblique Triangles and the Law of Sines

7.1

Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle


Solve triangle ABC if A = 28.8°, C = 102.6°, and c = 25.3 in.

7.1 Example 1 Using the Law of Sines to Solve a Triangle (SAA) (page 292)


Use the Law of Sines to find the lengths of the missing sides.

7.1 Example 1 Using the Law of Sines to Solve a Triangle (SAA) (cont.)


Kurt wishes to measure the distance across the Gasconade River. He determines that C = 117.2°, A = 28.8°, and b = 75.6 ft. Find the distance a across the river.

7.1 Example 2 Using the Law of Sines in an Application (ASA) (page 305)


7.1 Example 2 Using the Law of Sines in an Application (ASA) (cont.)

Use the Law of Sines to find the length of side a.

The distance across the river is about 65.1 ft.


The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. Find the distance between the ship and the lighthouse at each location.

7.1 Example 3 Using the Law of Sines in an Application (ASA) (page 293)

Let x = the distance to the lighthouse at bearing N 52° W andy = the distance to the lighthouse at bearing N 23° W.



The lighthouse is located at Z, and the ship is first located at Y and then at X.



The distance between the ship and the second location is about 9.4 km.

The distance between the ship and the first location is about 4.7 km.


7.1 Example 4 Finding the Area of a Triangle (SAS) (page 294)

Find the area of triangle DEF in the figure.


7.1 Example 5 Finding the Area of a Triangle (ASA) (page 294)

Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′.

We must find AC (side b) or AB (side c) in order to find the area of the triangle.


7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.)


7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.)

37.0 cm

48°20′


The Law of Cosines7.3Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle ▪ Derivation of Heron’s Formula


Two boats leave a harbor at the same time, traveling on courses that make an angle of 82°20′ between them. When the slower boat has traveled 62.5 km, the faster one has traveled 79.4 km. At that time, what is the distance between the boats?

7.3 Example 1 Using the Law of Cosines in an Application (SAS) (page 308)

The harbor is at C. The slower boat is at A, and the faster boat is at B.

We are seeking the length of AB.


Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.

7.3 Example 1 Using the Law of Cosines in an Application (SAS) (cont.)

The two boats are about 94.3 km apart.


Solve triangle ABC if B = 73.5°, a = 28.2 ft, and c = 46.7 ft.

7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) (page 308)

Use the law of cosines to find b because we know the lengths of two sides of the triangle and the measure of the included angle.


7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) (cont.)

47.2 ft

Now use the law of sines to find the measure of another angle.


Solve triangle ABC if a = 25.4 cm, b = 42.8 cm, and c = 59.3 cm.

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (page 309)

Use the law of cosines to find the measure of the largest angle, C.


Use either the law of sines or the law of cosines to find the measure of angle B.

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.)

118.6°


Alternative:

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.)

118.6°


7.3 Example 4 Designing a Roof Truss (SSS) (page 309)

Find the measure of angle C in the figure.


7.3 Example 5 Using Heron’s Formula to Find an Area (SSS)

(page 311) The distance “as the crow flies” from Chicago to St. Louis is 262 mi, from St. Louis to New Orleans is 599 mi, and from New Orleans to Chicago is 834 mi. What is the area of the triangular region having these three cities as vertices?

The semiperimeter s is


7.3 Example 5 Using Heron’s Formula to Find an Area (SSS)

(page 311)

The area of the triangular region is about 40,800 mi2.


Vectors, Operations, and the Dot Product7.4Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors


Find the magnitude and direction angle for

7.4 Example 1 Finding Magnitude and Direction Angle

(page 322)

Magnitude:

Direction angle:


Graphing calculator solution

7.4 Example 1 Finding Magnitude and Direction Angle (cont.)


Vector v has magnitude 14.5 and direction angle 220°. Find the horizontal and vertical components.

7.4 Example 2 Finding Horizontal and Vertical Components

(page 323)

Horizontal component: –11.1

Vertical component: –9.3



7.4 Example 2 Finding Horizontal and Vertical Components

(cont.)


Write each vector in the form ⟨a, b .⟩

7.4 Example 3 Writing Vectors in the Form ⟨a, b ⟩ (page 323)

u: magnitude 8, direction angle 135°

v: magnitude 4, direction angle 270°

w: magnitude 10, direction angle 340°


Two forces of 32 and 48 newtons act on a point in the plane. If the angle between the forces is 76°, find the magnitude of the resultant vector.

7.4 Example 4 Finding the Magnitude of a Resultant (page 324)

because the adjacent angles of a parallelogram are supplementary.

Law of cosines

Find squareroot.


Let u = 6, ⟨ –3 and ⟩ v = ⟨–14, 8 . Find the following.⟩

7.4 Example 5 Performing Vector Operations (page 324)


u = 6, ⟨ –3 , ⟩ v = ⟨–14, 8⟩

7.4 Example 5 Performing Vector Operations (cont.)



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


Find the magnitude of the equilibrant of forces of 54 newtons and 42 newtons acting on a point A, if the angle between the forces is 98°. Then find the angle between the equilibrant and the 42-newton force.

7.5 Example 1 Finding Magnitude and Direction of an Equilibrant (page 332)

The equilibrant is –v.


7.5 Example 1 Finding Magnitude and Direction of an Equilibrant (cont.)

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

Law of cosines


7.5 Example 1 Finding Magnitude and Direction of an Equilibrant (cont.)

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


Find the force required to keep a 2500-lb car parked on a hill that makes a 12° angle with the horizontal.

7.5 Example 2 Finding a Required Force (page 332)

The vertical force BA represents the force of gravity.

BA = BC + (–AC)


7.5 Example 2 Finding a Required Force (cont.)

Vector BC represents the force with which the weight pushes against the hill.

Vector BF represents the force that would pull the car up the hill.

Since vectors BF and AC are equal, 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.

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 12°.



From right triangle ABC,

A force of approximately 520 lb will keep the car parked on the hill.


A force of 18.0 lb is required to hold a 74.0-lb crate on a ramp. What angle does the ramp make with the horizontal?

7.5 Example 3 Finding an Incline Angle (page 333)

Vector BF represents the force required to hold the crate on the incline.

In right triangle ABC, the measure of angle B equals θ, the magnitude of vector BA represents the weight of the crate, and vector AC equals vector BF.


7.5 Example 3 Finding an Incline Angle (cont.)

The ramp makes an angle of about 14.1° with the horizontal.


A ship leaves port on a bearing of 25.0° and travels 61.4 km. The ship then turns due east and travels 84.6 km. How far is the ship from port? What is its bearing from port?

7.5 Example 4 Applying Vectors to a Navigation Problem

(page 333)

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° − 25.0° = 65.0°.


(cont.)

The measure of angle PAE = 180° − 65.0 = 115.0°.

Law of cosines

The ship is about 124 km from port.


To find the bearing of the ship from port, first find the measure of angle APE.


(cont.)

Law of sines

Now add 38.3° to 25.0° to find that the bearing is 63.3°.


A small plane follows a bearing of 70° at an airspeed of 230 mph and encounters a wind blowing at 25.0 mph from a direction of 290°. Find the resulting bearing and ground speed of the plane.


(page 334)



(cont.) 1 70m ABS 1 290 02 0 27m W BD

1 20m E BC 1 790 2 00m S BC

1 1

70 70 140

m ABC m ABS m S BC

2 2 2

2 2

2 cos( )

25 230 2(25)(230)cos140

b a c ac ABC

Use the law of cosines to find AC = |b|

62,334 250b



(cont.)

The bearing is about 70° + 4° = 74°. The groundspeed is about 250 mph and the bearing is about 74 degrees.

Use the law of sines to find α, and then determine the bearing, 70° + α.

sin140 sin

250 25

25sin140sin 0.064278761

250

3.69 4

Applications of Trigonometry and Vectors

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