Transcript No Slide Title

Chapter 2
Acute Angles and
Right Triangles
Copyright © 2005 Pearson Education, Inc.
2.1
Trigonometric Functions of
Acute Angles
Copyright © 2005 Pearson Education, Inc.
Development of Right Triangle Definitions of
B
Trigonometric Functions
c



Let ABC represent a right triangle with right A b C
angle at C and angles A and B as acute angles,
with side “a” opposite A, side “b” opposite B and
side “c” (hypotenuse) opposite C.
Place this triangle with either of the acute angles
in standard position (in this example “A”):
B
Notice that (b,a) is a point on the
b, a 
c
a
terminal side of A at a distance
A b C
“c” from the origin
Copyright © 2005 Pearson Education, Inc.
Slide 2-3
a
Development of Right Triangle Definitions of
Trigonometric Functions

Based on this diagram, each of the six
trigonometric functions for angle A would be
defined:
a
c
sin A 
csc A 
c
a
b
c
cos A 
sec A 
B
c
b
b, a 
c
a
b
a
tan A 
cot A 
b
a
A b C
Copyright © 2005 Pearson Education, Inc.
Slide 2-4
B
Right Triangle Definitions of
Trigonometric Functions
hypotenuse
c
A
a
side oppositeA
b C
side adjacent o
t A

The same ratios could have been obtained without placing an acute
angle in standard position by making the following definitions:
side oppositeA
sin A 
hypotenuse
side adjacent ot A
cos A 
hypotenuse
side oppositeA
tan A 
side adjacent ot A

hypotenuse
csc A 
side oppositeA
hypotenuse
sec A 
side adjacent ot A
side adjacent ot A
cot A 
side oppositeA
Standard “Right Triangle Definitions” of Trigonometric Functions
( MEMORIZE THESE!!!!!! )
May help to memorize"soh - cah - toa"
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.
sin A  side oppositeA  20  5
hypotenuse
52 13
side adjacent ot A
48 12
cos A 


hypotenuse
52 13
20 5
side oppositeA


tan A 
side adjacent ot A
48 12
Copyright © 2005 Pearson Education, Inc.
48
A
C
20
52
B
Slide 2-6
Development of Cofunction Identities

Given any right triangle, ABC, how does
the measure of B compare with A? c
o
B = 90  A


a
sin A   cos B  cos 90o  A
c
c
csc A   sec B  sec 90o  A
a
a
tan A   cot B  cot 90o  A
b
Copyright © 2005 Pearson Education, Inc.
a
B




A
b
C
Slide 2-7
Cofunction Identities


By similar reasoning other cofunction identities
can be verified:
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)
MEMORIZE THESE!!!
Copyright © 2005 Pearson Education, Inc.
Slide 2-8
Example: Write Functions in Terms of
Cofunctions

Write each function in
terms of its cofunction.

a) cos 38 =

b) sec 78 =
sin (90  38) =
csc (90  78) =
sin 52
csc 12
T hecofunctionidentitiescan be described as :
T hefunctionof an angle is equal to thecofunctionof its complement.
Copyright © 2005 Pearson Education, Inc.
Slide 2-9
Solving Trigonometric Equations Using
Cofunction Identities

Given a trigonometric equation that contains two
trigonometric functions that are cofunctions, it
may help to find solutions for unknowns by using
a cofunction identity to convert to an equation
containing only one trigonometric function as
shown in the following example
Copyright © 2005 Pearson Education, Inc.
Slide 2-10
Example: Solving Equations

Assuming that all angles are acute angles, find
one solution for the equation:
cot(4  8 )  tan(2  4 ).
cot 4  8o  cot 90o  2  4o





Anglesdon't have to be equal for cotangentsto be equal, but it is one way theycan be equal

4  8  90  2  4
4  8o  90o  2  4o
6  8o  86o
6  78o
  13o
o
Copyright © 2005 Pearson Education, Inc.
o
o

Slide 2-11
Comparing the relative values of
trigonometric functions


Sometimes it may be useful to determine the
relative value between trigonometric functions of
angles without knowing the exact value of either
one
To do so, it often helps to draw a simple diagram
of two right triangles each having the same
hypotenuse and then to compare side ratios
Copyright © 2005 Pearson Education, Inc.
Slide 2-12
Example: Comparing Function Values

Tell whether the statement is true or false.
y
y
sin 31 > sin 29
sin 31o  31 and sin 29 o  29
r
r
y31
y29
Referring to drawing, which is bigger,
or
?
r
r
r
sin 31  sin 29
o


o
r
31o 29o
is T RUE!
x31
y31
y29
x29
Generalizing, in the interval from 0 to 90, as the angle
increases, so does the sine of the angle
Similar diagrams and comparisons can be done for the
other trig functions
Copyright © 2005 Pearson Education, Inc.
Slide 2-13
Equilateral Triangles



Triangles that have three equal side lengths are
equilateral
Equilateral triangles also have three equal
angles each measuring 60o
All equilateral triangles are similar
(corresponding sides are proportional)
60o
2
o
h
o
30
2 3
2
o
60
60
2
Copyright © 2005 Pearson Education, Inc.
o
60
1
h 2  12  2 2
h2  1  4
h2  3
h 3
Slide 2-14
Using 30-60-90 Triangle to Find Exact
Trigonometric Function Values

30-60-90 Triangle

Find each of these:
1
cos60 
2
0
tan600 
3
1
sin 30 
2
2
2 3
0

sec 30 
3
3
0
MEMORIZE THIS!
Copyright © 2005 Pearson Education, Inc.
Slide 2-15
Isosceles Right Triangles



Right triangles that have two legs of equal length
Also have two angles of measure 45o
All such triangles are similar
1 1  c
c
2
45o
1
2
2
2
2c
o
45
2
2 c
1
Copyright © 2005 Pearson Education, Inc.
Slide 2-16
Using 45-45-90 Triangle to Find Exact
Trigonometric Function Values

45-45-90 Triangle

Find each of these:
cos45 
0
tan450 
sin 45 
0
sec 450 
1

2
1
2
2
1

2
2
2
2
MEMORIZE THIS!
Copyright © 2005 Pearson Education, Inc.
Slide 2-17
Function Values of Special Angles
Are used a lot in T rigonometry and can quickly
be determinedfrommemorizedtriangles.

sin 
cos 
tan 
cot 
sec 
csc 
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
T hischartcan also be quickly completedby
memorizingthefirst columnand using identities.
Copyright © 2005 Pearson Education, Inc.
Slide 2-18
Usefulness of Knowing Trigonometric
Functions of Special Anlges: 30o, 45o, 60o


The trigonometric function values derived from
knowing the side ratios of the 30-60-90 and
45-45-90 triangles are “exact” numbers, not
decimal approximations as could be obtained
from using a calculator
You will often be asked to find exact trig function
values for angles other than 30o, 45o and 60o
angles that are somehow related to trig function
values of these angles
Copyright © 2005 Pearson Education, Inc.
Slide 2-19
Homework

2.1 Page 51
All: 1 – 14, 16 – 21, 23 – 26, 29 – 32, 35 – 42

MyMathLab Assignment 2.1 for practice

MyMathLab Homework Quiz 2.1 will be due for a
grade on the date of our next class meeting

Copyright © 2005 Pearson Education, Inc.
Slide 2-20
2.2
Trigonometric Functions of
Non-Acute Angles
Copyright © 2005 Pearson Education, Inc.
Reference Angles

A reference angle for an angle  is the positive
acute angle made by the terminal side of angle 
and the x-axis. (Shown below in red)
'
'
'
Referenceangle for is indicatedby  '
Copyright © 2005 Pearson Education, Inc.
Slide 2-22
Example: Find the reference angle for
each angle.



218
Positive acute angle made by
the terminal side of the angle
and the x-axis is:
218  180 = 38




Copyright © 2005 Pearson Education, Inc.
1387
First find coterminal angle
between 0o and 360o
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 2-23
Comparison of Trigonometric Functions of
Angles vs Functions of Reference Angles


Each angle below has the same reference angle
Choosing the same “r” for a point on the terminal
side of each (each circle same radius), you will
notice from similar triangles that all “x” and “y”
values are the same except for sign
x , y  



'
x , y' 

Copyright © 2005 Pearson Education, Inc.

x , y 


 '

'
x , y 


Slide 2-24
Comparison of Trigonometric Functions of
Angles vs Functions of Reference Angles




Based on the observations on the previous slide:
Trigonometric functions of any angle will be
the same value as trigonometric functions of
its reference angle, except for the sign of the
answer
The sign of the answer can be determined by
quadrant of the angle
Also, we previously learned that the
trigonometric functions of coterminal angles
always have equal values
Copyright © 2005 Pearson Education, Inc.
Slide 2-25
Finding Trigonometric Function Values
for Any Non-Acute 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. (Hint: All students take
calculus.) This gives the values of the
trigonometric functions for angle .
Copyright © 2005 Pearson Education, Inc.
Slide 2-26
Example: Finding Exact Trigonometric
Function Values of a Non-Acute Angle



Find the exact values of
the trigonometric
functions for 210. (No
Calculator!)
Reference angle:
210 – 180 = 30
Remember side ratios for
30-60-90 triangle.
Corresponding sides:
600
1,
1
2
3, 2
Copyright © 2005 Pearson Education, Inc.
300
3
Slide 2-27
2
Example Continued
300



600
1
3
Trig functions of any angle are equal to trig functions of
its reference angle except that sign is determined from
quadrant of angle
210o is in quadrant III where only tangent and cotangent
are positive
Based on these observations, the six trig functions of
210o are:
1
o
o
sin 210   sin 30  
csc 210o   csc 30o  2
2
3
2 3
o
o
o
o
cos 210   cos30  
sec 210   sec 30  
2
3
3
o
o
tan 210  tan30 
cot 210o  cot 30o  3
3
Copyright © 2005 Pearson Education, Inc.
Slide 2-28
Example: Finding Trig Function Values
Using Reference Angles

Find the exact value of:
cos (240)

Coterminal angle
between 0 and 360:
240 + 360 = 120
2
300

the reference angles is:
180  120 = 60
Copyright © 2005 Pearson Education, Inc.
600

1
3



cos  2400  cos 1200 
1
0
 cos 60  
2
 
Slide 2-29
Expressions Containing Powers of
Trigonometric Functions

An expression such as:
sin 
2

Has the meaning:

Example: Using your memory regarding side ratios of
30-60-90 and 45-45-90 triangles, simplify:
sin  
2
2
2
 2  3
2 3
sin 45  tan 30        4  9 
 2   3 
1 3
9 6
3
1
 
 

2 9 18 18 18
6
2
0
Copyright © 2005 Pearson Education, Inc.
2
0
Slide 2-30
Example: Evaluating an Expression
with Function Values of Special Angles

Evaluate cos 120 + 2 sin2 60  tan2 30.

Individual trig function values before evaluating are:
1
3
3
cos 120   , sin 60 
, and tan 30 
,
2
2
3

2
2
Substituting into the expression:
 3  3
1
2
2
cos 120 + 2 sin 60  tan 30   + 2 
 

2
2
3

 

1
3 3
   2  
2
4 9
2

3
Copyright © 2005 Pearson Education, Inc.
Slide 2-31
Finding Unknown Special Angles that Have
a Specific Trigonometric Function Value





0
0
Example: Find all values of  in the interval 0 , 360  given:
2
cos  
2
Use your knowledge of trigonometric function values of 30o, 45o and
60o angles* to find a reference angle that has the same absolute
value as the specified function value
Use your knowledge of signs of trigonometric functions in various
quadrants to find angles that have both the same absolute value and
sign as the specified function value
*NOTE: Later we will learn to use calculators to solve equations that
don’t necessarily have these special angles as reference angles
Copyright © 2005 Pearson Education, Inc.
Slide 2-32
Example: Finding Angle Measures Given an
Interval and a Function Value


Find all values of  in the interval 00 , 3600  given:
2
cos  
2



Which special angle has the same absolute
0
value cosine as this angle? 45
In which quadrants is cosine negative? II and III
Putting 45o reference angles in quadrants II and
III, gives which two angles as answers?
1800  450  1350
Copyright © 2005 Pearson Education, Inc.
1800  450  2250
Slide 2-33
Homework

2.2 Page 59
All: 1 – 6, 10 – 17, 25 – 32, 36 – 37, 48 – 53,
61- 66

MyMathLab Assignment 2.2 for practice

MyMathLab Homework Quiz 2.2 will be due for a
grade on the date of our next class meeting

Copyright © 2005 Pearson Education, Inc.
Slide 2-34
2.3
Finding Trigonometric Function
Values Using a Calculator
Copyright © 2005 Pearson Education, Inc.
Function Values Using a Calculator




As previously mentioned, 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.
Also, angles measured in degrees, minutes and
seconds must be converted to decimal degrees
Remember that most calculator values of
trigonometric functions are approximations.
Copyright © 2005 Pearson Education, Inc.
Slide 2-36
Function Values Using a Calculator



Sine, Cosine and Tangent of a specific angle may be
found directly on the calculator by using the key labeled
with that function
Cosecant, Secant and Cotangent of a specific angle may
be found by first finding the corresponding reciprocal
function value of the angle and then using the reciprocal
key label x-1 or 1/x to get the desired function value
Example: To find sec A, find cos A, then use the
reciprocal key to find:
1
cos A
This is the sec A value
Copyright © 2005 Pearson Education, Inc.
Slide 2-37
Example: Finding Function Values with
a Calculator
sin 38 24

cot 68.4832o
Convert 38 24 to decimal
degrees and use sin key.
24
 38.4
60
sin 38 24  sin 38.4
38 24  38

Find tan of the angle and
use reciprocal key
cot 68.4832 
 .3942492
 .6211477
Copyright © 2005 Pearson Education, Inc.
Slide 2-38
Finding Angle Measures When a
Trigonometric Function of Angle is Known

When a trigonometric ratio is known, and the angle is unknown,
inverse function keys on a calculator can be used to find an angle*
that has that trigonometric ratio

Scientific calculators have three inverse functions each having an
“apparent exponent” of -1 written above the function name. This use
of the superscript -1 DOES NOT MEAN RECIPROCAL

1
1
-1
If x is an appropriate number, then sin x,cos x, or tan x
gives the measure of an angle* whose sine, cosine, or tangent is x.
* There are an infinite number of other angles, coterminal and other,
that have the same trigonometric value
Copyright © 2005 Pearson Education, Inc.
Slide 2-39
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
Copyright © 2005 Pearson Education, Inc.
Slide 2-40
Example: Using Inverse Trigonometric
Functions to Find Angles continued


Find one value of  given: sec   2.486879
Use reciprocal identities to get:
1
cos  
 .4021104
2.486879

Now find  using the inverse cosine function.
The result is:
  66.289824
Copyright © 2005 Pearson Education, Inc.
Slide 2-41
Homework

2.3 Page 64
All: 5 – 29, 55 – 62

MyMathLab Assignment 2.3 for practice

MyMathLab Homework Quiz 2.3 will be due for a
grade on the date of our next class meeting

Copyright © 2005 Pearson Education, Inc.
Slide 2-42
2.4
Solving Right Triangles
Copyright © 2005 Pearson Education, Inc.
Measurements Associated with Applications
of Trigonometric Functions




In practical applications of trigonometry, many of
the numbers that are used are obtained from
measurements
Such measurements many be obtained to
varying degrees of accuracy
The manner in which a measured number is
expressed should indicate the accuracy
This is accomplished by means of “significant
digits”
Copyright © 2005 Pearson Education, Inc.
Slide 2-44
Significant Digits


“Digits obtained from actual measurement”
All digits used to express a number are
considered “significant” (an indication of
accuracy) if the “number” includes a decimal



The number of significant digits in 583.104 is: 6
The number of significant digits in .0072 is: 4
When a decimal point is not included, then
trailing zeros are not “significant”


The number of significant digits in 32,000 is: 2
The number of significant digits in 50,700 is: 3
Copyright © 2005 Pearson Education, Inc.
Slide 2-45
Significant Digits for Angles

The following conventions are used in expressing accuracy of
measurement (significant digits) in angle measurements
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
Copyright © 2005 Pearson Education, Inc.
Slide 2-46
Calculations Involving Significant Digits


2
4
An answer is no more accurate than the least
accurate number in the calculation
Examples:
32,000
 4444444 .4 according to calculator
.0072
Significant Digits?
32,000
 4,400 ,000 two significan t digits
.0072
2
5
3200sin 42.4580  2160.1586 accordingtocalculator
Significant Digits?
3200sin 42.4580  2200 two significant digits
Copyright © 2005 Pearson Education, Inc.
Slide 2-47
Solving a Right Triangle


To “solve” a right triangle is to find the measures
of all the sides and angles of the triangle
A right triangle can be solved if either of the
following is true:


One side and one acute angle are known
Any two sides are known
Copyright © 2005 Pearson Education, Inc.
Slide 2-48
Example: Solving a Right Triangle,
Given an Angle and a Side


Solve right triangle ABC, if
A = 42 30' and c = 18.4.
How would you find angle B?
B = 90  42 30'
B = 47 30‘ = 47.5
B
c = 18.4
a
4230'
C
A
b
Which trigfunctionrelatesA, a and c?
a
Which trigfunctionrelatesA, b and c?
sin A 
c
a
sin 42 .5 
18 .4
18.4 sin 42.50  a
0
18.4.675590207  a
12.4  a
Copyright © 2005 Pearson Education, Inc.
cos A 
b
c
b
cos 42.5 
18.4
18.4 cos 42.50  b
13.6  b
0
Slide 2-49
Example: Solving a Right Triangle
Given Two Sides

Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm.
What trigfunctionrelatesA and the two given sides?
opposite 11.47
B
sin A 

hypotenuse 27.82
c = 27.82
a = 11.47
sin A  .412293314
A
C
A  sin 1 .412293314
b
A  24.350 (Note"significant digits")
How would you find B?
How would youfind b?
11.47
2
2
2
b

c

a
cos B 
 .412293314
27.82
2
2
2
b

27.82

11.47
B  cos1 .412293314 65.650
b  25.35
Copyright © 2005 Pearson Education, Inc.
Slide 2-50
Angles of “Elevation” and “Depression”
Some application problemsinvolve" angle of elevation"
and " angle of depression"

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.
Copyright © 2005 Pearson Education, Inc.

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 2-51
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.
Copyright © 2005 Pearson Education, Inc.
Slide 2-52
Example: Application

Shelly McCarthy stands 123 ft from the base of a
flagpole, and the angle of elevation to the top of
the pole is 26o40’. If her eyes are 5.30 ft above
the ground, find the height of the pole.
x
x  height of pole above horizontal
x
tan 26 40 
123
5.30
123
0
123tan 260 40  x
61 .8  x
Copyright © 2005 Pearson Education, Inc.
Height of pole from ground?
61.8  5.30  67.1 ft.
Slide 2-53
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.
Equation?
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.
Copyright © 2005 Pearson Education, Inc.
Slide 2-54
Homework

2.4 Page 72
All: 11 – 14, 21 – 28, 35 – 36, 41 – 44, 48 – 49

MyMathLab Assignment 2.4 for practice

MyMathLab Homework Quiz 2.4 will be due for a
grade on the date of our next class meeting

Copyright © 2005 Pearson Education, Inc.
Slide 2-55
2.5
Further Applications of
Right Triangles
Copyright © 2005 Pearson Education, Inc.
Describing Direction by Bearing
(First Method)



Many applications of trigonometry involve
“direction” from one point to another
Directions may be described in terms of
“bearing” and there are two widely used
methods
The first method designates north as being 0o
and all other directions are described in terms of
clockwise rotation from north (in this context the
angle is considered “positive”, so east would be
bearing 90o)
Copyright © 2005 Pearson Education, Inc.
Slide 2-57
Describing Bearing Using First Method


Note: All directions can be described as an
angle in the interval: [ 0o, 360 )
Show bearings: 32o, 164o, 229o and 304o
N
N
N
N
320
1640
2290
Copyright © 2005 Pearson Education, Inc.
3040
Slide 2-58
Hints on Solving Problems Using Bearing



Draw a fairly accurate figure showing the
situation described in the problem
Look at the figure to see if there is a triangular
relationship involving the unknown and a
trigonometric function
Write an equation and solve the problem
Copyright © 2005 Pearson Education, Inc.
Slide 2-59
Example

Radar stations A and B are on an east-west line
3.7 km apart. Station A detects a plane at C on
a bearing of 61o, while station B simultaneously
detects the same plane on a bearing of 331o.
Find the distance from A to C.
N
N
Right triangle is formed!*
T rigfunctionrelatingd , 3.7 and an acuteangle?
C
610d
d
0
0
cos 29 
90
3. 7
290
610
0
3
.
7
cos
29
d
B
A
3 .7
3.2 km  d
3310
* Can be done with any triang
le using Law of CosinesSlide 2-60
Copyright © 2005 Pearson Education, Inc.
Describing Direction by Bearing
(Second Method)


The second method of defining bearing is to indicate
degrees of rotation east or west of a north line or east or
west of a south line
Example: N 30o W would represent 30o rotation to the
west of a north line
N
300

Example: S 45o E would represent 45o rotation to the
east of a south line
450
S
Copyright © 2005 Pearson Education, Inc.
Slide 2-61
Example: Using Bearing

An airplane leaves the airport flying at a bearing of
N 32 W for 200 miles and lands. How far west of its
starting point is the plane?
e
Equationinvolvingtrigfunction?

e
200
sin 32 
200
32
e  200sin 32
e  106
The airplane is approximately 106 miles west of its
starting point.
Copyright © 2005 Pearson Education, Inc.
Slide 2-62
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.
cot

2

d
b
2
b

d  cot
2
2
Copyright © 2005 Pearson Education, Inc.
Slide 2-63
Example: Using Trigonometry to
Measure a Distance

Find d when  = 1 23'12"
and b = 2.0000 cm
 d
cot  b
2 2
b

d  cot
2
2

Let b = 2, change  to
decimal degrees.
1 23'12"  1.386667
Significant Digits :
d  82.634 cm
2
1.386667
d  cot
 82.6341 cm
2
2
Copyright © 2005 Pearson Education, Inc.
Slide 2-64
Example: Solving a Problem Involving
Angles of Elevation

Sean wants to know the height of a Ferris wheel.
He doesn’t know his distance from the base of
the wheel, but, from a given point on the ground,
he finds the angle of elevation to the top of the
Ferris wheel is 42.3o . He then moves back 75 ft.
From the second point, the angle of elevation to
the top of the Ferris wheel is 25.4o. Find the
height of the Ferris wheel.
Copyright © 2005 Pearson Education, Inc.
Slide 2-65
Example: Solving a Problem Involving
Angles of Elevation continued



The figure shows two
unknowns: x and h.
Use the two triangles, to
write two trig function
equations involving the
two unknowns:
In triangle ABC,
B
h
C
42.3
x
25.4
A
75 ft
D
h
tan 42.3 
or h  x tan 42.3 .
x

In triangle BCD,
h
tan 25.4 
or h  (75  x) tan 25.4 .
75  x
Solve thissystemof equationsby substitution.
Copyright © 2005 Pearson Education, Inc.
Slide 2-66
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
Resulting Equation
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
Copyright © 2005 Pearson Education, Inc.
Factor out x.
Divide by the coefficient of x.
Slide 2-67
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  

 .9099299  74.
 .435095 
The height of the Ferris wheel is approximately 74 ft.
Copyright © 2005 Pearson Education, Inc.
Slide 2-68
Homework

2.5 Page 81
All: 11 – 16, 23 – 28

MyMathLab Assignment 2.5 for practice

MyMathLab Homework Quiz 2.5 will be due for a
grade on the date of our next class meeting

Copyright © 2005 Pearson Education, Inc.
Slide 2-69