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Chapter 2. Acute Angles and Right Triangles. 2.1. Trigonometric Functions of Acute Angles. Right-triangle Based Definitions of Trigonometric Functions. For any acute angle A in standard position. 48. C. A. 20. 52. B. Example: Finding Trig Functions of Acute Angles. - PowerPoint PPT Presentation
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Copyright © 2005 Pearson Education, Inc.
Copyright © 2005 Pearson Education, Inc.
Chapter 2
Acute Angles and
Right Triangles
Copyright © 2005 Pearson Education, Inc.
2.1
Trigonometric Functions of Acute Angles
Copyright © 2005 Pearson Education, Inc. Slide 2-4
Right-triangle Based Definitions of Trigonometric Functions For any acute angle A in standard position.
side opposite hypotenusesin csc
hypotenuse side opposite
side adjacent hypotenusecos sec
hypotenuse side adjacent
side oppositetan
side adjacent
side adjacentcot .
side opposite
y rA A
r y
x rA A
r x
yA
x
xA
y
Copyright © 2005 Pearson Education, Inc. Slide 2-5
Example: Finding Trig Functions of Acute Angles Find the values of sin A, cos A, and tan A in the
right triangle shown.
A C
B
52
48
20
side oppositesin
hypotenuse
side adjacentcos
hypotenuse
side opposite
52
5
tanside a
20
20
4
djac
8
8ent 4
2
A
A
A
Copyright © 2005 Pearson Education, Inc. Slide 2-6
Cofunction Identities
For any acute angle A,
sin A = cos(90 A) csc A = sec(90 A)
tan A = cot(90 A) cos A = sin(90 A)
sec A = csc(90 A) cot A = tan(90 A)
Copyright © 2005 Pearson Education, Inc. Slide 2-7
Example: Write Functions in Terms of Cofunctions Write each function in
terms of its cofunction.
a) cos 38
cos 38 = sin (90 38) = sin 52
b) sec 78
sec 78 = csc (90 78) = csc 12
Copyright © 2005 Pearson Education, Inc. Slide 2-8
Example: Solving Equations
Find one solution for the equation
Assume all angles are acute angles.
cot(4 8o) tan(2 4o)
(4 8o) (2 4o) 90o
6 12o 90o
6 78o
13o
cot(4 8o) tan(2 4o) .
Copyright © 2005 Pearson Education, Inc. Slide 2-9
Example: Comparing Function Values
Tell whether the statement is true or false.
sin 31 > sin 29
In the interval from 0 to 90, as the angle increases, so does the sine of the angle, which makes sin 31 > sin 29 a true statement.
Copyright © 2005 Pearson Education, Inc. Slide 2-10
Special Triangles
30-60-90 Triangle 45-45-90 Triangle
Copyright © 2005 Pearson Education, Inc. Slide 2-11
Function Values of Special Angles
260
1145
230
csc sec cot tan cos sin
1
23
2
3
3
3 2 3
3
2
2
2
2
2 2
3
2
1
23 3
3
2 3
3
Copyright © 2005 Pearson Education, Inc.
2.2
Trigonometric Functions of Non-Acute Angles
Copyright © 2005 Pearson Education, Inc. Slide 2-13
Answers to homework
Sin cos tan
1.21/29 20/29 21/20
2.45/53 28/53 45/28
3.n/p m/p n/m
4.k/z y/z k/y
5.C
6.H 9. E
7.B 10. A
8.G
Copyright © 2005 Pearson Education, Inc. Slide 2-14
Homework answers
Copyright © 2005 Pearson Education, Inc. Slide 2-15
Reference Angles
A reference angle for an angle is the positive acute angle made by the terminal side of angle and the x-axis.
Copyright © 2005 Pearson Education, Inc. Slide 2-16
Example: Find the reference angle for each angle.
a) 218 Positive acute angle
made by the terminal side of the angle and the x-axis is 218 180 = 38.
1387 Divide 1387 by 360 to get
a quotient of about 3.9. Begin by subtracting 360 three times. 1387 – 3(360) = 307.
The reference angle for 307 is 360 – 307 = 53
Copyright © 2005 Pearson Education, Inc. Slide 2-17
Example: Finding Trigonometric Function Values of a Quadrant Angle
Find the values of the trigonometric functions for 210.
Reference angle:
210 – 180 = 30Choose point P on the terminal side of the angle so the distance from the origin to P is 2.
Copyright © 2005 Pearson Education, Inc. Slide 2-18
Example: Finding Trigonometric Function Values of a Quadrant Angle continued
The coordinates of P are x = y = 1 r = 2
3, 1
3
1 3 3sin 210 cos210 tan 210
2 2 3
2 3csc210 2 sec210 cot 210 3
3
Copyright © 2005 Pearson Education, Inc. Slide 2-19
Finding Trigonometric Function Values for Any Nonquadrantal Angle Step 1If > 360, or if < 0, then find a
coterminal angle by adding or subtracting 360 as many times as needed to get an
angle greater than 0 but less than 360. Step 2Find the reference angle '. Step 3Find the trigonometric function values for
reference angle '. Step 4Determine the correct signs for the values
found in Step 3. (Use the table of signs in section 5.2, if necessary.) This gives the
values of the trigonometric functions for angle .
Copyright © 2005 Pearson Education, Inc. Slide 2-20
Example: Finding Trig Function Values Using Reference Angles Find the exact value of
each expression. cos (240) Since an angle of 240
is coterminal with and angle of 240 + 360 = 120, the reference angles is 180 120 = 60, as shown.
cos( 240 ) cos120
1cos60
2
Copyright © 2005 Pearson Education, Inc. Slide 2-21
HomeworkPg 59-60 # 1-6, 10-13
Copyright © 2005 Pearson Education, Inc. Slide 2-22
Example: Evaluating an Expression with Function Values of Special Angles Evaluate cos 120 + 2 sin2 60 tan2 30.
Since
cos 120 + 2 sin2 60 tan2 30 =
1 3 3cos 120 , sin 60 , and tan 30 ,
2 2 3
2 2
+ 2
1 3 32
2
1
4
3
3
2
9
3
2 3
2
Copyright © 2005 Pearson Education, Inc. Slide 2-23
Copyright © 2005 Pearson Education, Inc. Slide 2-24
Copyright © 2005 Pearson Education, Inc. Slide 2-25
Example: Using Coterminal Angles
Evaluate each function by first expressing the function in terms of an angle between 0 and 360.
cos 780 cos 780 = cos (780 2(360) = cos 60
= 1
2
Copyright © 2005 Pearson Education, Inc.
2.3
Finding Trigonometric Function Values Using a Calculator
Copyright © 2005 Pearson Education, Inc. Slide 2-27
Function Values Using a Calculator
Calculators are capable of finding trigonometric function values.
When evaluating trigonometric functions of angles given in degrees, remember that the calculator must be set in degree mode.
Remember that most calculator values of trigonometric functions are approximations.
Copyright © 2005 Pearson Education, Inc. Slide 2-28
Example: Finding Function Values with a Calculator a) Convert 38 to decimal
degrees.
b) cot 68.4832 Use the identity
cot 68.4832 .3942492
38 38
sin 38 24 si
38.4
38.4
2424
6
n
.6211477
0
1cot .
tan
sin 38 24
24
Copyright © 2005 Pearson Education, Inc. Slide 2-29
Angle Measures Using a Calculator
Graphing calculators have three inverse functions.
If x is an appropriate number, then
gives the measure of an angle whose sine, cosine, or tangent is x.
1 1sin ,cos , x x-1 or tan x
Copyright © 2005 Pearson Education, Inc. Slide 2-30
Example: Using Inverse Trigonometric Functions to Find Angles Use a calculator to find an angle in the
interval that satisfies each condition.
Using the degree mode and the inverse sine
function, we find that an angle having sine value .8535508 is 58.6 .
We write the result as
[0 ,90 ]
sin .8535508
1sin .8535508 58.6
Copyright © 2005 Pearson Education, Inc. Slide 2-31
Example: Using Inverse Trigonometric Functions to Find Angles continued
Use the identity Find the
reciprocal of 2.48679 to get
Now find using the inverse cosine function. The result is 66.289824
sec 2.486879
1cos .
sec
cos .4021104.
Copyright © 2005 Pearson Education, Inc. Slide 2-32
Page 64 # 22-29
22. 57.99717206
23. 55.84549629
24. 81.166807334
25. 16.16664145
26. 30.50274845
27. 38.49157974
28. 46.17358205
29. 68.6732406
Copyright © 2005 Pearson Education, Inc.
2.4
Solving Right Triangles
Copyright © 2005 Pearson Education, Inc. Slide 2-34
Significant Digits for Angles
A significant digit is a digit obtained by actual measurement. Your answer is no more accurate then the least accurate
number in your calculation.
Tenth of a minute, or nearest thousandth of a degree
5
Minute, or nearest hundredth of a degree4
Ten minutes, or nearest tenth of a degree3
Degree2
Angle Measure to Nearest:Number of Significant Digits
Copyright © 2005 Pearson Education, Inc. Slide 2-35
Example: Solving a Right Triangle, Given an Angle and a Side Solve right triangle ABC, if A = 42 30' and c = 18.4. B = 90 42 30'
B = 47 30'
AC
B
c = 18.4
4230'
sin A a
c
sin 42o30' a
18.4
a 18.4sin 42o30'
a 18.4(.675590207)
a 12.43
cos A b
c
cos42o30' b
18.4
b 18.4cos42o30'
b 13.57
Copyright © 2005 Pearson Education, Inc. Slide 2-36
Example: Solving a Right Triangle Given Two Sides Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm.
B = 90 24.35B = 65.65
AC
B
c = 27.82a = 11.47
1
side oppositesin
hypotenuse
11.47.412293314
27.82
sin 24.35
A
A
2 2 2
2 2 227.82 11.47
25.35
b c a
b
b
Copyright © 2005 Pearson Education, Inc. Slide 2-37
Solve the right triangle
A= 28.00o, and c = 17.4 ft
Copyright © 2005 Pearson Education, Inc. Slide 2-38
Homework
Page 73 # 9-14
Copyright © 2005 Pearson Education, Inc. Slide 2-39
Definitions
Angle of Elevation: from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint X.
Angle of Depression: from point X to point Y (below) is the acute angle formed by ray XY and a horizontal ray with endpoint X.
Copyright © 2005 Pearson Education, Inc. Slide 2-40
Solving an Applied Trigonometry Problem
Step 1 Draw a sketch, and label it with the given information. Label the quantity
to be found with a variable.
Step 2 Use the sketch to write an equation relating the given quantities to the variable.
Step 3 Solve the equation, and check that your answer makes sense.
Copyright © 2005 Pearson Education, Inc. Slide 2-41
Example: Application
The length of the shadow of a tree 22.02 m tall is 28.34 m. Find the angle of elevation of the sun.
Draw a sketch.
The angle of elevation of the sun is 37.85.
22.02 m
28.34 mB
1
22.02tan
28.3422.02
tan 37.8528.34
B
B
Copyright © 2005 Pearson Education, Inc.
2.5
Further Applications of Right Triangles
Copyright © 2005 Pearson Education, Inc. Slide 2-43
Bearing
Other applications of right triangles involve bearing, an important idea in navigation.
Copyright © 2005 Pearson Education, Inc. Slide 2-44
Example
An airplane leave the airport flying at a bearing of N 32E for 200 miles and lands. How far east of its starting point is the plane?
The airplane is approximately 106 miles east of its starting point.
sin 32oe
200
e 200sin 32o
e 106
e
200
32º
Copyright © 2005 Pearson Education, Inc. Slide 2-45
Example: Using Trigonometry to Measure a Distance A method that surveyors use to determine a small
distance d between two points P and Q is called the subtense bar method. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q. Angle is measured, then the distance d can be determined.
a) Find d when = and b = 2.0000 cm b) Angle usually cannot be measured more accurately
than to the nearest 1 second. How much change would there be in the value of d if were measured 1 second larger?
1 23'12"
Copyright © 2005 Pearson Education, Inc. Slide 2-46
Example: Using Trigonometry to Measure a Distance continued
Let b = 2, change to decimal degrees.
b) Since is 1 second larger, use 1.386944.
The difference is .0170 cm.
2
cot2
cot2 2
b
d
bd
1 23'12" 1.386667
2 1.386667co 8t 2.6341
2m
2
c
d
2 1.38694482co .61
26t
27 m c
d
Copyright © 2005 Pearson Education, Inc. Slide 2-47
Example: Solving a Problem Involving Angles of Elevation Sean wants to know the height of a Ferris wheel.
From a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3 . He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4 . Find the height of the Ferris wheel.
Copyright © 2005 Pearson Education, Inc. Slide 2-48
Example: Solving a Problem Involving Angles of Elevation continued The figure shows two
unknowns: x and h. Since nothing is given about
the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse, (the tangent ratio).
In triangle ABC,
In triangle BCD,
xC
B
h
DA 75 ft
42.3 25.4
tan 42.3 or tan 4 2.3 .h
hx
x
tan 25.4 or (75 ) tan 25 5
.4 .7
xh
hx
Copyright © 2005 Pearson Education, Inc. Slide 2-49
Example: Solving a Problem Involving Angles of Elevation continued Since each expression equals h, the
expressions must be equal to each other.
tan 42.3 75 tan 25.4 tan 25.4
tan 42.3 tan 25.4 75 tan 25.4
(tan 42.3 tan 25.4 ) 75 tan 25.4
75 tan 25.4
tan 42.3 tan 25
tan 42.3 (75
.4
) tan 25.4x x
x x
x x
x
x
Solve for x.
Distributive Property
Factor out x.
Get x-terms on one side.
Divide by the coefficient of x.
Copyright © 2005 Pearson Education, Inc. Slide 2-50
We saw above that Substituting for x.
tan 42.3 = .9099299 and tan 25.4 = .4748349. So, tan 42.3 - tan 25.4 = .9099299 - .4748349 = .435095 and
The height of the Ferris wheel is approximately 74.48 ft.
75 tan 25.4tan 42.3 .
tan 42.3 tan 25.4h
Example: Solving a Problem Involving Angles of Elevation continued
tan 42.3 .h x
75 .4748349 .435095
.9099299 74.47796
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