Copyright © 2005 Pearson Education, Inc. Chapter 2 Acute Angles and Right Triangles.
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Transcript of Copyright © 2005 Pearson Education, Inc. Chapter 2 Acute Angles and Right Triangles.
Copyright © 2005 Pearson Education, Inc. Slide 2-3
Development of Right Triangle Definitions of Trigonometric Functions
Let ABC represent a right triangle with right 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”):
Notice that (b,a) is a point on the
terminal side of A at a distance
“c” from the origin b
ab,
A
B
C
ac
bA
B
C
ac
Copyright © 2005 Pearson Education, Inc. Slide 2-4
Development of Right Triangle Definitions of Trigonometric Functions
Based on this diagram, each of the six trigonometric functions for angle A would be defined:
b
ab,
A
B
C
ac
a
bAcot
b
aAtan
b
cA sec
c
bA cos
a
cA csc
c
aAsin
Copyright © 2005 Pearson Education, Inc. Slide 2-5
Right Triangle Definitions ofTrigonometric Functions
The same ratios could have been obtained without placing an acute angle in standard position by making the following definitions:
Standard “Right Triangle Definitions” of Trigonometric Functions( MEMORIZE THESE!!!!!! )
bA
B
C
acA opposite side
A oadjacent t side
hypotenuse
A opposite side
A oadjacent t sideAcot
A oadjacent t side
A opposite sideAtan
A oadjacent t side
hypotenuseA sec
hypotenuse
A oadjacent t sideA cos
A opposite side
hypotenuseA csc
hypotenuse
A opposite sideAsin
toa"-cah-soh" memorize tohelpMay
Copyright © 2005 Pearson Education, Inc. Slide 2-6
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
Atan
A cos
Asin
A oadjacent t side
A opposite side
hypotenuse
A oadjacent t side
hypotenuse
A opposite side
12
5
48
2013
12
52
48
13
5
52
20
Copyright © 2005 Pearson Education, Inc. Slide 2-7
Development of Cofunction Identities
Given any right triangle, ABC, how does
the measure of B compare with A?
B = B C
A
bc
aAsin
c
a B
A90 o
cos A90 ocos
A csc a
c sec A90 osec
A tan b
a cot A90 ocot
B
B
Copyright © 2005 Pearson Education, Inc. Slide 2-8
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)
!!THESE! MEMORIZE
Copyright © 2005 Pearson Education, Inc. Slide 2-9
Example: Write Functions in Terms of Cofunctions Write each function in
terms of its cofunction.
a) cos 38 =
sin (90 38) =
sin 52
b) sec 78 =
csc (90 78) =
csc 12
.complement its of cofunction the toequal is anglean offunction The
:as described becan identities cofunction The
Copyright © 2005 Pearson Education, Inc. Slide 2-10
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 2-11
Example: Solving Equations
Assuming that all angles are acute angles, find one solution for the equation:
cot(4 8 ) tan(2 4 ).
cot84cot o oo 4290
ooo 429084 equal becan way they one isit but equal, be tocotangentsfor equal be tohavet don' Angles
ooo 429084 oo 8686 o786 o13
Copyright © 2005 Pearson Education, Inc. Slide 2-12
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 2-13
Example: Comparing Function Values Tell whether the statement is true or false.
sin 31 > sin 29
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
o29o31 29y
31yrr
29x31x
r
y
r
y 29o31o 29sin and 31sin
? or bigger, is which drawing, toReferring 2931
r
y
r
y
TRUE! is 29sin 31sin oo
Copyright © 2005 Pearson Education, Inc. Slide 2-14
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)
o60
2
2
2o60
o60 2o30
o60
1
h
3
3
41
21
2
2
222
h
h
h
h
3
Copyright © 2005 Pearson Education, Inc. Slide 2-15
Using 30-60-90 Triangle to Find Exact Trigonometric Function Values
30-60-90 Triangle Find each of these:
THIS! MEMORIZE
060tan
030sin
030sec
060cos2
1
3
2
1
3
2
3
32
Copyright © 2005 Pearson Education, Inc. Slide 2-16
Isosceles Right Triangles
Right triangles that have two legs of equal length Also have two angles of measure 45o
All such triangles are similar
2
1
1
o45
o45c
c
c
c
2
2
112
222
Copyright © 2005 Pearson Education, Inc. Slide 2-17
Using 45-45-90 Triangle to Find Exact Trigonometric Function Values
45-45-90 Triangle Find each of these:
THIS! MEMORIZE
045tan
045sin
045sec
045cos 2
1
2
2
2
1
2
2
1
2
Copyright © 2005 Pearson Education, Inc. Slide 2-18
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
. trianglesmemorized from determined be
quicklycan andry Trigonometin lot a used Are
.identities using andcolumn first thememorizing
by completedquickly be alsocan chart This
Copyright © 2005 Pearson Education, Inc. Slide 2-19
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 2-20
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 2-22
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)
'
' '
'by indicated is for angle Reference
Copyright © 2005 Pearson Education, Inc. Slide 2-23
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
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
Copyright © 2005 Pearson Education, Inc. Slide 2-24
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
yx ,
yx , yx ,
yx , '
' '
'
Copyright © 2005 Pearson Education, Inc. Slide 2-25
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 2-26
Finding Trigonometric Function Values for Any Non-Acute 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. (Hint: All students take calculus.) This gives the values of the trigonometric functions for angle .
Copyright © 2005 Pearson Education, Inc. Slide 2-27
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:
2 ,3 ,1 12
3030
060
Copyright © 2005 Pearson Education, Inc. Slide 2-28
Example Continued
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:
3 30cot210cot 3
330tan210tan
3
3230sec210 sec
2
330cos210cos
230csc210 csc 2
130sin210sin
oooo
oooo
oooo
12
3030
060
Copyright © 2005 Pearson Education, Inc. Slide 2-29
Example: Finding Trig Function Values Using Reference Angles Find the exact value of:
cos (240)
Coterminal angle between 0 and 360: 240 + 360 = 120
the reference angles is:
180 120 = 60 0240cos 0120cos
060cos2
1
12
3030
060
Copyright © 2005 Pearson Education, Inc. Slide 2-30
Expressions Containing Powers of Trigonometric Functions
An expression such as:
Has the meaning:
Example: Using your memory regarding side ratios of 30-60-90 and 45-45-90 triangles, simplify:
2sin
2sin
0202 30tan45sin
22
3
3
2
2 9
3
4
2
9
3
2
1
18
6
18
9
18
3
6
1
Copyright © 2005 Pearson Education, Inc. Slide 2-31
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:
Substituting into the expression:
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 2-32
Finding Unknown Special Angles that Have a Specific Trigonometric Function Value
Example: Find all values of in the interval given:
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
00 360 ,0
2
2cos
Copyright © 2005 Pearson Education, Inc. Slide 2-33
Example: Finding Angle Measures Given an Interval and a Function Value
Find all values of in the interval given:
Which special angle has the same absolute value cosine as this angle?
In which quadrants is cosine negative? Putting 45o reference angles in quadrants II and
III, gives which two angles as answers?
00 360 ,0
2
2cos
045III and II
000 13545180 000 22545180
Copyright © 2005 Pearson Education, Inc. Slide 2-34
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.
2.3
Finding Trigonometric Function Values Using a Calculator
Copyright © 2005 Pearson Education, Inc. Slide 2-36
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 2-37
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:
This is the sec A value
Acos
1
Copyright © 2005 Pearson Education, Inc. Slide 2-38
Example: Finding Function Values with a Calculator
Convert 38 to decimal degrees and use sin key.
Find tan of the angle and use reciprocal key
cot 68.4832 .3942492
38 38
sin 38 24 si
38.4
38.4
2424
6
n
.6211477
0
sin 38 24
24
o4832.68cot
Copyright © 2005 Pearson Education, Inc. Slide 2-39
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
If x is an appropriate number, then 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
1 1sin ,cos , x x -1 or tan x
Copyright © 2005 Pearson Education, Inc. Slide 2-40
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 2-41
Example: Using Inverse Trigonometric Functions to Find Angles continued Find one value of given: Use reciprocal identities to get:
Now find using the inverse cosine function.
The result is:
66.289824
sec 2.486879
4021104.486879.2
1cos
Copyright © 2005 Pearson Education, Inc. Slide 2-42
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 2-44
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 2-45
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: The number of significant digits in .0072 is:
When a decimal point is not included, then trailing zeros are not “significant” The number of significant digits in 32,000 is: The number of significant digits in 50,700 is: 3
2
46
Copyright © 2005 Pearson Education, Inc. Slide 2-46
Significant Digits for Angles
The following conventions are used in expressing accuracy of measurement (significant digits) in angle measurements
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 2-47
Calculations Involving Significant Digits
An answer is no more accurate than the least accurate number in the calculation
Examples:
calculator toaccording 4.44444440072.
000,32
digitst significan two000,400,40072.
000,32
calculator toaccording 1586.2160458.42sin3200 0
digitst significan two2200458.42sin3200 0
Digits?t Significan
2
4
Digits?t Significan
2 5
Copyright © 2005 Pearson Education, Inc. Slide 2-48
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 2-49
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 AC
B
c = 18.4
4230'
c
aA sin
4.185.42sin 0 a
a05.42sin4.18
a675590207.4.18
a4.12
c? and a A, relatesfunction Which trigc? and b A, relatesfunction Which trig
c
bA cos
4.185.42cos 0 b
b05.42cos4.18
b6.13
a
b
Copyright © 2005 Pearson Education, Inc. Slide 2-50
Example: Solving a Right Triangle Given Two Sides Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm.
AC
B
c = 27.82a = 11.47
2 2 2
2 2 227.82 11.47
25.35
b c a
b
b
sides?given two theandA relatesfunction What trig
B? findyou wouldHow)digits"t significan" (Note
b? findyou wouldHow
b
82.27
47.11sin
hypotenuse
oppositeA
412293314.sin A
412293314.sin 1A035.24A
412293314.82.27
47.11cos B
01 65.65412293314.cos B
Copyright © 2005 Pearson Education, Inc. Slide 2-51
Angles of “Elevation” and “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.
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.
"depression of angle" and
elevation" of angle" involve problemsn applicatio Some
Copyright © 2005 Pearson Education, Inc. Slide 2-52
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 2-53
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.
30.5123
horizontal above pole ofheight xx
1230426tan 0 x
x0426tan123 0
x8.61
ground? from pole ofHeight
ft. 1.6730.58.61
Copyright © 2005 Pearson Education, Inc. Slide 2-54
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
Equation?
Copyright © 2005 Pearson Education, Inc. Slide 2-55
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 2-57
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 2-58
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
NN N N
032
0164
02290304
Copyright © 2005 Pearson Education, Inc. Slide 2-59
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 2-60
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.
061
0331
090
A B
C
7.3
NN
d029 061
formed!* is ngleRight triaangle? acutean and 3.7 , relatingfunction Trig d
7.329cos 0 d
d029cos7.3
dkm 2.3Cosines of Law using leany triang with done beCan *
Copyright © 2005 Pearson Education, Inc. Slide 2-61
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
Example: S 45o E would represent 45o rotation to the east of a south line
N
S
030
045
Copyright © 2005 Pearson Education, Inc. Slide 2-62
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?
The airplane is approximately 106 miles west of its starting point.
sin32200
200sin32
106
e
e
e
e
200
32
function? triginvolvingEquation
Copyright © 2005 Pearson Education, Inc. Slide 2-63
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.
2
cot2
cot2 2
b
d
bd
Copyright © 2005 Pearson Education, Inc. Slide 2-64
Example: Using Trigonometry to Measure a Distance Find d when =
and b = 2.0000 cm
Let b = 2, change to decimal degrees.
2
cot2
cot2 2
b
d
bd
1 23'12" 1.386667
2 1.386667co 8t 2.6341
2m
2
c
d
1 23'12"
cm 634.82
:Digitst Significan
d
Copyright © 2005 Pearson Education, Inc. Slide 2-65
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 2-66
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,
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
on.substitutiby equations of system thisSolve
Copyright © 2005 Pearson Education, Inc. Slide 2-67
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
Resulting Equation
Distributive Property
Factor out x.
Get x-terms on one side.
Divide by the coefficient of x.
Copyright © 2005 Pearson Education, Inc. Slide 2-68
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 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 .4748349.9099299 74.
.435095