Transcript Document

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.
Right-triangle Based Definitions of
Trigonometric Functions

For any acute angle A in standard position.
y side opposite
sin A  
r
hypotenuse
x side adjacent
cos A  
r
hypotenuse
y side opposite
tan A  
x side adjacent
x side adjacent
cot A  
.
y side opposite
Copyright © 2005 Pearson Education, Inc.
r
hypotenuse
csc A  
y side opposite
r
hypotenuse
sec A  
x side adjacent
Slide 2-4
Example: Finding Trig Functions of
Acute Angles


Find the values of sin A, cos A, and tan A in the
right triangle shown.
side opposite 20
sin A 

hypotenuse
52
side adjacent 48
cos A 

hypotenuse 52
side opposite 20
tan A 

side adjacent 48
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48
A
C
20
52
B
Slide 2-5
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)
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Slide 2-6
Example: Write Functions in Terms of
Cofunctions


Write each function in
terms of its cofunction.
a) cos 38

b) sec 78
sec 78 = csc (90  78)
= csc 12
cos 38 = sin (90  38)
= sin 52
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Slide 2-7
Example: Solving Equations

Find one solution for the equation
cot(4  8 )  tan(2  4 ) .
o
o
Assume all angles are acute angles.
cot(4  8o)  tan(2  4o)
(4  8o)  (2  4o)  90o
6  12o  90o
6  78o
  13o
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Slide 2-8
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.
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Slide 2-9
Special Triangles

30-60-90 Triangle
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
45-45-90 Triangle
Slide 2-10
Function Values of Special Angles

sin 
30
1
2
3
2
3
3
3
2 3
3
2
45
2
2
2
2
1
1
2
2
60
3
2
1
2
3
3
3
2
2 3
3
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cos 
tan 
cot 
sec 
csc 
Slide 2-11
2.2
Trigonometric Functions of
Non-Acute Angles
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Answers to homework
1.
2.
3.
4.
5.
6.
7.
8.
Sin
21/29
45/53
n/p
k/z
C
H
B
G
Copyright © 2005 Pearson Education, Inc.
cos
20/29
28/53
m/p
y/z
tan
21/20
45/28
n/m
k/y
E
10. A
9.
Slide 2-13
Homework answers
11.
Sin
Cos
Tan
Csc
Sec
cot
12/13
5/13
12/5
13/12
13/5
5/12
12
5/3
13
14
3/5
6/7
7/12
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7/6
12/7
Slide 2-14
Reference Angles

A reference angle for an angle  is the positive
acute angle made by the terminal side of angle 
and the x-axis.
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Slide 2-15
Example: Find the reference angle for
each angle.


a) 218
Positive acute angle
made by the terminal side
of the angle and the xaxis is 218  180 = 38.



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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
Slide 2-16
Example: Finding Trigonometric Function
Values of a Quadrant Angle


Find the values of the
trigonometric functions for
210.
Reference angle:
210 – 180 = 30
Choose 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 2-17
Example: Finding Trigonometric Function
Values of a Quadrant Angle continued

The coordinates of P are

x = 3
y = 1

r=2
1
sin 210  
2
3
cos 210  
2
csc 210  2
sec 210  
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
3, 1
2 3
3
3
tan 210 
3
cot 210  3
Slide 2-18
Finding Trigonometric Function Values
for Any Nonquadrantal Angle 

Step 1


Step 2
Step 3

Step 4
If  > 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.
Find the reference angle '.
Find the trigonometric function values for
reference angle '.
Determine 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 .
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Slide 2-19
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.
Copyright © 2005 Pearson Education, Inc.
cos(240 )  cos120
1
  cos 60  
2
Slide 2-20
Homework
Pg 59-60 # 1-6, 10-13

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
Slide 2-21
Example: Evaluating an Expression
with Function Values of Special Angles


Evaluate cos 120 + 2 sin2 60  tan2 30.
Since
1
3
3
cos 120   , sin 60 
, and tan 30 
,
2
2
3
cos 120 + 2 sin2 60  tan2 30 =
2
 3  3
1
  + 2
 

2
2
3

 

1
3 3
   2  
2
4 9
2

3
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2
Slide 2-22
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Slide 2-23
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Slide 2-24
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
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Slide 2-25
2.3
Finding Trigonometric Function
Values Using a Calculator
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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 2-27
Example: Finding Function Values with
a Calculator



a) sin 38 24
Convert 38 24 to decimal
degrees.
24
 38.4
60
sin 38 24  sin 38.4
38 24  38



b) cot 68.4832 
Use the identity
1
cot  
.
tan 
cot 68.4832 
 .3942492
 .6211477
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Slide 2-28
Angle Measures Using a Calculator


Graphing calculators have three inverse
functions.
If x is an appropriate number, then sin 1 x,cos1 x,
or tan -1 x gives the measure of an angle whose
sine, cosine, or tangent is x.
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Slide 2-29
Example: Using Inverse Trigonometric
Functions to Find Angles


Use a calculator to find an angle  in the
interval [0 ,90 ] that satisfies each condition.
sin   .8535508
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 sin 1 .8535508  58.6
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Slide 2-30
Example: Using Inverse Trigonometric
Functions to Find Angles continued

sec   2.486879
1
. Find the
Use the identity cos  
sec 
reciprocal of 2.48679 to get cos   .4021104.
Now find  using the inverse cosine function.
The result is   66.289824
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Slide 2-31
Page 64 # 22-29
22.
23.
24.
25.
26.
27.
28.
29.
57.99717206
55.84549629
81.166807334
16.16664145
30.50274845
38.49157974
46.17358205
68.6732406
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Slide 2-32
2.4
Solving Right Triangles
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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.
Number of
Significant Digits
Angle Measure to Nearest:
2
Degree
3
Ten minutes, or nearest tenth of a degree
4
Minute, or nearest hundredth of a degree
5
Tenth of a minute, or nearest thousandth of
a degree
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Slide 2-34
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
B = 47 30'
c = 18.4
a
sin A 
c
a
sin 42 30 ' 
18.4
a  18.4 sin 42o30 '
a  18.4(.675590207)
o
a  12.43
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4230'
C
A
b
cos A 
c
b
cos 42 30' 
18.4
b  18.4cos 42o30'
o
b  13.57
Slide 2-35
Example: Solving a Right Triangle
Given Two Sides

Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm.
side opposite
sin A 
hypotenuse
11.47

 .412293314
27.82
sin 1 A  24.35

B = 90  24.35
B = 65.65
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B
a = 11.47
c = 27.82
C
A
b2  c 2  a 2
b2  27.822  11.472
b  25.35
Slide 2-36
Solve the right triangle

A= 28.00o, and c = 17.4 ft
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Slide 2-37
Homework

Page 73 # 9-14
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Slide 2-38
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.
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
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.
Slide 2-39
Solving an Applied Trigonometry Problem

Step 1

Step 2

Step 3
Draw a sketch, and label it with the
given information. Label the quantity to
be found with a variable.
Use the sketch to write an equation
relating the given quantities to the
variable.
Solve the equation, and check that
your answer makes sense.
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Slide 2-40
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.
22.02
tan B 
28.34
1 22.02
B  tan
 37.85
28.34

22.02 m
B
28.34 m
The angle of elevation of the sun is 37.85.
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Slide 2-41
2.5
Further Applications of
Right Triangles
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Bearing

Other applications of right triangles involve
bearing, an important idea in navigation.
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Slide 2-43
Example

An airplane leave the airport flying at a bearing of N 32E
for 200 miles and lands. How far east of its starting point
is the plane?
e
e
sin 32 
200
e  200 sin 32 o
o
e  106

200
32º
The airplane is approximately 106 miles east of its
starting point.
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Slide 2-44
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  = 1 23'12" 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?
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Slide 2-45
Example: Using Trigonometry to
Measure a Distance continued

cot

2

d
b
2
b

d  cot
2
2

Let b = 2, change  to
decimal degrees.

b) Since  is 1 second
larger, use 1.386944.
d
1 23'12"  1.386667
2 1.386667
d  cot
 82.6341 cm
2
2
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
2
1.386944
cot
 82.6176 cm
2
2
The difference is .0170
cm.
Slide 2-46
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.
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Slide 2-47
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,


B
h
C
42.3
x
25.4
A
75 ft
D
h
tan 42.3 
or h  x tan 42.3 .
x
h
tan 25.4 
or h  (75  x) tan 25.4 .
75  x
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Slide 2-48
Example: Solving a Problem Involving
Angles of Elevation continued

Since each expression equals h, the
expressions must be equal to each other.
x tan 42.3  (75  x) tan 25.4
Solve for x.
x tan 42.3  75 tan 25.4  x tan 25.4
Distributive Property
x tan 42.3  x tan 25.4  75 tan 25.4
Get x-terms on one side.
x(tan 42.3  tan 25.4 )  75 tan 25.4
75 tan 25.4
x
tan 42.3  tan 25.4
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Factor out x.
Divide by the coefficient of x.
Slide 2-49
Example: Solving a Problem Involving
Angles of Elevation continued

We saw above that h  x tan 42.3 . Substituting for x.


75 tan 25.4
h
 tan 42.3 .
 tan 42.3  tan 25.4 

tan 42.3 = .9099299 and tan 25.4 = .4748349.
So, tan 42.3 - tan 25.4 = .9099299 - .4748349 =
.435095
and

 75 .4748349

 .435095

 .9099299  74.47796


The height of the Ferris wheel is approximately 74.48 ft.
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Slide 2-50