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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. ccos(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! 35

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posted: | 11/21/2011 |

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