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					                                  TRIGONOMETRY
                                SECTION 1.6: Pages 89-103
                                Right Triangle Trigonometry
                       ADDITIONAL EXPLANATIONS AND EXAMPLES

Section Objectives
Upon completion of this section, you should be able to

    1.        Define the inverse trigonometric functions for a given angle and
              relate these values to the values of the regular trigonometric functions.

    2.        Approximate values of the regular and inverse trigonometric functions
              using a calculator.

    3.        Determine the values of the six trigonometric functions for non-special
              right triangles.

    4.        Solve various right triangle problems.

    5.        Apply the concepts of right triangle trigonometry and the inverse
              trigonometric functions to real-world applications.


The Inverse Trigonometric Functions
Pages 89-91

Your text defines the inverse trigonometric functions as the functions whose
“input” is the general trigonometric value of an angle and whose “output” is the
angle. For example,

                                1                         1
                         sin-1   = 30 because sin (30) = .
                               2                          2


These functions are formally defined on page 90 of your text. Pay particular
attention to the domain of each inverse function. Right now we are only
interested in finding the inverse trigonometric values for acute angles. A more in-
depth discussion of the inverse trigonometric functions will take place when we
study section 4.1.

We will be using the inverse trigonometric functions later in this section in order
to find missing angles of right triangles. We can use our calculator to
approximate values of the inverse functions. On most calculators, the inverse
trigonometric functions are above the keys containing the regular trigonometric
functions. Hence, pressing the 2nd key on the calculator is required in order to



                                                                                     30
use the inverse functions. Refer to example 3 on page 91 of your text. For this
section, all angle values will be in degrees.


Right Triangles
Pages 91-92

                                    To the left is the figure of our unit circle
                                    reproduced from page 13 of your notes. We
                                    used this figure to obtain the following definitions
                   (cos , sin )
                                    of the trigonometric functions:
                    1
                                                                                 sin y
                           sin       sin  y         cos  x       tan          
                    cos
                                                                                 cos x

                                    In order to define these functions, we formed a
                                    right triangle in our circle (with the hypotenuse
                                    becoming a radius of our circle). Now we are
                                    going to remove the unit circle and re-define our
                                    trigonometric values for any right triangle.

We will do this by relating our angle  to the legs and
hypotenuse of the right triangle, as shown in the
figure at the right. From this new relationship, we can           Hypotenuse
form new definitions of our trigonometric functions:                              Opposite

               opposite               adjacent                        
     sin               and cos  
              hypotenuse             hypotenuse                       Adjacent


                     opposite 
                               
              sin  hypotenuse opposite
                               
       tan      
              cos  adjacent  adjacent
                    
                     hypotenuse
                                
                               

The reciprocal functions are defined in a similar manner. It is important to note
that for each definition above  must be an acute angle. Page 92 of your text
defines the six trigonometric functions in the following manner:


                             opp                 adj                   opp
                  sin                cos                 tan  
                             hyp                 hyp                   adj

                              hyp                hyp                   adj
                  csc                sec                 cot  
                              opp                adj                   opp


                                                                                          31
EXAMPLE 1 Similar to Example 4 on page 92

Use the information given for the right triangle to find the values of the six
trigonometric functions.

                               SOLUTION
                               First we need to find the value of the length of the
                               unknown side. We can accomplish this by using the
               6               Pythagorean Theorem:
                       5
                                     52 + b2 = 6 2
           
               b                     b=    36  25 =    11


Now that we know the lengths of all the sides, we can now find the values of the
trigonometric functions beginning with the sine function.

Therefore,

                    opp 5                       hyp 6
          sin                      csc        
                    hyp 6                       opp 5

                    adj   11                    hyp   6
          cos                      sec         
                    hyp   6                     adj   11

                    opp   5                     adj   11
          tan                      cot         
                    adj   11                    opp   5



Identifying Sides and Angles of Right Triangles
Page 93

For the remainder of this section, we will label the legs
of our right triangle using lowercase letters a and b,
and will use c to identify the hypotenuse. Anlges will
                                                                                
be labeled using the Greek letters alpha (), beta (),                 c
and gamma (). For our right triangles,  = 90. See                                a
the figure at the right. For the problems that follow,
these angles will ALWAYS be opposite the sides                             
as shown. Now we will practice “solving right                           b
triangles”.




                                                                                        32
EXAMPLE 2 Similar to Example 5 on page 93

Solve the right triangle below by finding all sides and angles rounded to the
nearest tenth.

                             SOLUTION
                             We have several approaches to obtaining the correct
                             values for the missing lengths. Let’s start by finding
                 
            c                side c. Using our given angle, we can determine that
                         a   our known side-length 2 is adjacent to the angle while
                             side c is the hypotenuse. Therefore, we must use the
        72                  cosine function as shown below. Begin by using our
          2                  cosine equation from page 31.


                     2
       cos(72) 
                     c

To solve for c, we must first multiply both sides of the equation by c to eliminate
our denominator.

        ccos(72 )  2

Now solve our equation for c.

               2
       c
            cos( 72 )

If necessary, use your calculator to obtain an approximation for cos (72). Use
your approximation to find the value c. Round your approximation to the nearest
tenth.

              2
       c          6.472 or 6.5
            0.309

We can find a using the tangent function and a similar approach.

                     a
       tan(72)        2 tan(72)  a  a  2(3.0777) or 6.2
                     2

We finish by finding angle : 180 – 90 – 72 = 18

Therefore, a  6.2, c  6.5, and  = 18.




                                                                                  33
EXAMPLE 3 Similar to Example 6 on pages 93-94

Solve the right triangle below by finding all sides and angles rounded to the
nearest tenth.

                                  SOLUTION
                                  We begin by finding the value of angle . From our
                                 diagram, we know the length of the side opposite 
              c                   (which is 9) and the side adjacent  (which is 4).
                           9      Therefore, we need to use the tangent function as
                                  shown below.
          
              4



                      9
          tan( ) 
                      4

Now we can use the inverse tangent function to solve for .

                          9                   9
          If tan( )       , then   tan 1  
                          4                   4

Now we can use our calculator to approximate .

            tan 1 2.25   66.0

Now that we know , we can easily find : 180 – 90 – 66 = 24

Now we can find c.

              9          9             9
sin(66)        c            c         9.9
              c      sin( 66 )     0.9135

Therefore,   24.0,   66.0, and c  9.9.


Angle of Depression and Angle of Elevation
Page 94

Page 94 of your text displays two figures contrasting an angle of elevation
(beginning from the ground) and an angle of depression (beginning from the top
of an object).


                                                                                       34
It is important to note a horizontal line must be drawn from the high point of
an object in order to obtain the proper angle of depression. This is usually
not a problem with an anlge of elevation because we usually assume the ground
is a flat surface. These sepcal types of angles will be used frequently to solve
word problems and applications involving right triangle trigonometry. An example
follows.



EXAMPLE 4 Similar to Example 7 on page 95

A building casts a shadow along the ground
that is 37 feet long. The angle of elevation of
the sun is 56. How tall is the building?

SOLUTION
Let’s use the tangent function to find the
height of the building.

             height
tan(56)            37 tan(56)  h                   56
              37
                                                              37
               h  54.8548 feet

Therefore, the height of the building is approximately 54.9 feet high.



ASSIGNMENT            DUE AT THE NEXT CLASS MEETING
For each problem, please show all your work and any calculations.

Pages 98-103:         problems 10-16 even
                      problems 17, 18, 20, 23
                      problems 25-30
                      problems 33, 35, 37, 40, 44

For problems 33-44, be sure to consult examples 7, 8, and 9 in the text.


Please provide a brief 2-sentence reflection of what you accomplished during this
section. Feel free to refer to the objectives listed at the beginning of this section
and provide examples of your explanations. Also, please be sure to note any
difficulties you encountered with any of the section material.
Thank you!




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