Chapter 5 Trigonometric Functions Section 5.2 Trigonometric Functions of Acute Angles.
Chapter 4 Trigonometric Functions Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 4.8...
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Transcript of Chapter 4 Trigonometric Functions Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 4.8...
Chapter 4Trigonometric Functions
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1
4.8 Applications ofTrigonometric Functions
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2
Objectives:
•Solve a right triangle.•Solve problems involving bearings.•Model simple harmonic motion.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 3
Solving Right Triangles
Solving a right triangle means finding the missing lengths of its sides and the measurements of its angles. We will label right triangles so that side a is opposite angle A, side b is opposite angle B, and side c, the hypotenuse, is opposite right angle C.
When solving a right triangle,we will use the sine, cosine,and tangent functions, rather than their reciprocals.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4
Example: Solving a Right Triangle
Let A = 62.7° and a = 8.4. Solvethe right triangle, rounding lengthsto two decimal places.
90A B 90 90 62.7 27.3B A
tan bBa
tan 27.38.4b 8.4 tan 27.3 4.34b
sin aAc
8.4sin 62.7c
sin 62.7 8.4c
8.4 9.45sin 62.7
c
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5
Example: Finding a Side of a Right Triangle
From a point on level ground 80 feet from the base of the Eiffel Tower, the angle of elevation is 85.4°. Approximate the height of the Eiffel Tower to the nearest foot.
85.4
80 ft
tan85.480a
80tan85.4a 994 ft
The height of the Eiffel Tower isapproximately 994 feet.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6
Example: Finding an Angle of a Right Triangle
A guy wire is 13.8 yards long and is attached from the ground to a pole 6.7 yards above the ground. Find the angle, to the nearest tenth of a degree, that the wire makes with the ground.
13.8 yards6.7 yards
A
6.7sin13.8
A
1 6.7sin13.8
A
29.0
The wire makes an angle of approximately 29.0° with the ground.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7
Trigonometry and Bearings
In navigation and surveying problems, the term bearing is used to specify the location of one point relative to another. The bearing from point O to point P is the acute angle, measured in degrees, between ray OP and a north-south line. The north-south line and the east-west line intersect at right angles. Each bearing has three parts: a letter (N or S), the measure of an acute angle, and a letter (E or W).
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8
Trigonometry and Bearings (continued)
If the acute angle is measured on theeast side of the north-south line, thenwe write E last.
Second, we write the measure of theacute angle.
If the acute angle is measured fromthe north side of the north-south line,then we write N first.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9
Trigonometry and Bearings (continued)
If the acute angle is measured on thewest side of the north-south line, thenwe write W last.
Second, we write the measure of theacute angle.
If the acute angle is measured fromthe north side of the north-south line,then we write N first.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10
Trigonometry and Bearings (continued)
If the acute angle is measured on theeast side of the north-south line, thenwe write E last.
Second, we write the measure of theacute angle.
If the acute angle is measured fromthe south side of the north-south line,then we write S first.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11
Example: Understanding Bearings
Use the figure to find each of the following:a. the bearing from O to DPoint D is located to the south and to the east of the north-south line. The bearing from O to D is S25°E.b. the bearing from O to CPoint C is located to the south and tothe west of the north-south line.The bearing from O to C is S15°W.
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Simple Harmonic Motion
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Example: Finding an Equation for an Object in Simple Harmonic Motion
A ball on a spring is pulled 6 inches below its rest position and then released. The period for the motion is 4 seconds. Write the equation for the ball’s simple harmonic motion.When the ball is released (t = 0), the ball’s distance from its rest position is 6 inches down. Because it is down, d is negative. The greatest distance from rest position occurs at t = 0. Thus, we will us the equation with the cosine function,
0, 6t d
cos .d a t
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 14
Example: Finding an Equation for an Object in Simple Harmonic Motion
A ball on a spring is pulled 6 inches below its rest position and then released. The period for the motion is 4 seconds. Write the equation for the ball’s simple harmonic motion. is the maximum displacement. Because the ball is initially below rest position, a = –6The value of can be found using the formula for the period.
a
2period = 4
2 4 24 2
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 15
Example: Finding an Equation for an Object in Simple Harmonic Motion
A ball on a spring is pulled 6 inches below its rest position and then released. The period for the motion is 4 seconds. Write the equation for the ball’s simple harmonic motion.
We have found that a = –6 and The equation for the ball’s simple harmonic motion is
6cos .2d t
.2
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 16
Frequency of an Object in Simple Harmonic Motion
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Example: Analyzing Simple Harmonic Motion
An object moves in simple harmonic motion described by
where t is measured in seconds and d in
centimeters. Find the maximum displacement.The maximum displacement from the rest position is the amplitude. Because a = 12, the maximum displacement is 12 centimeters.
12cos ,4
d t
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 18
Example: Analyzing Simple Harmonic Motion
An object moves in simple harmonic motion described by
where t is measured in seconds and d in
centimeters. Find the frequency.
The frequency is cycle per second.
12cos ,4
d t
2f
4
2
14 2
18
18
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 19
Example: Analyzing Simple Harmonic Motion
An object moves in simple harmonic motion described by
where t is measured in seconds and d in
centimeters. Find the time required for one cycle.
The time required for one cycle is the period.
The time required for one cycle is 8 seconds.
12cos ,4
d t
2period =
2
4
2 4
8