4.4 Right Triangle Trigonometry

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1 4.4 Right Triangle Trigonometry Trigonometry is introduced to students in two different forms, as functions on the unit circle and as functions on a right triangle. The unit circle approach is the most natural setting for the trig functions since they are not just functions of angles between 0 and 180 but instead have as domain the set of all real numbers. The unit circle explains identities such as (cos ) 2 + (sin ) 2 = 1 and cos() = sin( + π 2 ). However, we would also like to apply trigonometry to right triangles which have a hypotenuse of length different than one. We may do this by using similar triangles Similar Triangles Previously, to examine our trig functions, we displayed a typical triangle on the unit circle with central angle, hypotenuse 1 and point. We can expand that triangle by the ratio r to get a triangle in which the hypotenuse has length r and the point is on a circle of radius r. When this happens, cos() will not be x but x/r. Similarly, sin() will be y/r. r to repeat an idea from section 4.2 of these notes we could imagine a triangle with hypotenuse of length H, opposite side of length, adjacent side of length. The point sits on the circle of radius H and sin() =. ll of these triangles are similar and so the trig functions of the angle, as ratios of side lengths, are unchanged. 1 y x r y H x Figure 17. Three similar triangles we can use when we examine trig functions 167

2 If we are willing to draw a right triangle and use the Pythagorean theorem, then we can solve any right triangle problem in which we are given a side and another angle. Here are some examples. Some worked problems using similar triangles. 1. Find sec() if sin() = 3 5. Solution. Draw a right triangle which has an angle with sin() = 3 5. ( triangle will do.) Then compute the secant of the angle. The secant is the reciprocal of cosine and so sec() = hyp adj = H. The answer is Find tan(α) if cos(α) = 2 5. Solution. Draw a right triangle which has an angle with cos() = 2 5. (The most obvious triangle will have a hypotenuse of length 5 and an side adjacent to with length 2.) The opposite side will then have length 21, by the Pythagorean Theorem. Then compute the tangent of the angle. (Tangent is the ratio opp adj.) The answer is Suppose sec = 7 2 and tan is negative. Identify the angle and then find all six trig functions of the angle. Solution. Since the tangent is negative and cosine (= 2 7 ) is positive, then the angle must be in the fourth quadrant. Draw a line segment of length 7 from the origin into the fourth quadrant, to a point P (2, y). By the Pythagorean theorem, the absolute value of y is = 45 = 9 5 = 3 5. So y = 3 5 and our line segment ends at the point P (2, 3 5). Now read off the values of the various trig functions: The reciprocals of these are cos = 2 7, sin = 3 5 7, tan = sec = 7 2, csc = 7 3 = , cot = 2 3 = pplications with right triangles nytime we have a right triangle, then, if we can measure one of the acute angles and also know the length of a side, then we know everything about the triangle. Consider the figure below. H Figure 18. typical right triangle 168

3 In most of the applications in this section, we will be able to draw a triangle like that in the figure above. (The right triangle may have a different orientation; the right angle may be a the top of our figure or on the leftside... but we should be able to draw a right triangle when we start these problems!) We will then find one of the acute angles of the triangle (such as, drawn in red) and we will also be able to find the length of one of the sides. nce we have this information, the lengths of the other two sides can be computed using our trig functions. Here is are some examples involving practical applications. 1. radio tower is stands on a flat field. I walk 1000 feet away from the base of the radio tower and look up at the top of the tower. I measure a 71 angle between the horizon and the top of the tower. How tall is the tower? Solution. We begin by drawing a picture much like Figure 18, above. The radio tower is a vertical straight line perpendicular to the ground. We want to know the length of this vertical line segment. (Call its length.) Draw the ground as a horizontal line and mark the length of that horizontal line as = 1000 feet. The hypotenuse of this right triangle makes an angle = 71 with the ground. The tangent of 71 is tan 71 = = So the radio tower is 2904 feet tall Solving for we have = 1000 tan 71 2, 904 feet. 2. I am flying a kite on the beach. The kite is attached to 3000 feet of string. t the time that the string plays out and I have used all the string, the kite makes an angle with the horizon of 38. ssuming that the 3000 feet of string is a straight line, how high is the kite? Solution. gain, as before, begin by drawing a picture! (See Figure 18.) The kite and the person holding the kite and the ground directly below the kite form the three vertices of a right triangle with the right angle being at the point on the ground directly below the kite. The hypotenuse of this right triangle is H = 3000 feet long and the kite is feet above the ground. The sine of = 38 is sin 38 = H = 3000 and so the height is equal to = 3000 sin(38 ) 1847 feet. So the kite is at heigth 1847 feet (which is quite high!!) Parallax stronomers use simple right triangles to find the distance to nearby stars. Here is how they do this. s the earth revolves around the sun, it marks out a circle 2 of radius 93 million miles. (This is a fact worth remembering: the distance from the earth to the sun is about 93,000,000 miles or about 8 light minutes.) ver the course of the year, a nearby star should appear to move back and forth in the night sky as the earth revolves around the sun and so one should be able to measure that angle of apparent motion and use a right triangle (with one side equal to 93,000,000 miles) to compute that distance. In the figure below, we let the sun from a right angle vertex of a triangle and we set the earth and star as the other two vertices. If the star does not move, it would appear to form a right angle with the earth in this figure. But if the star is nearby then the line of sight to the star forms an angle α with the anticipated line of sight. The star appears to have moved. 2 To be precise, the earth s orbit is really an ellipse but the shape of this ellipse fortunately for us! is very close to the shape of a circle. 169

4 α α Earth Sun Earth Figure 19. Parallax. By a standard result from geometry, this angle α (the apparent motion of the star) is also the acute angle of the triangle at the vertex given by the star. If is the distance from the star to the sun and = 93, 000, 000 = miles then the cotangent of α is and so = cot α. The ancient Greeks thought of this idea and attempted to measure parallax. But when they did this, the stars didn t seem to move!! So either this picture was wrong (maybe the earth was the center of the universe?) or the stars must be billions of miles away! Convinced that the universe could not be billions of miles in size, most Greeks agreed with ristotle s belief that the earth was the center of the universe and that the sun revolved around the earth. Now we know better and indeed, with modern equipement, we have been able to measure the parallax of some stars. worked problem. The closest star to us, Proxima Centauri 3 has a parallax of 0.77, that is, 0.77 arcseconds. How far away is Proxima Centauri? Solution. minute of arc is one-sixtieth of a degree; a second of arc is one-sixtieth of a minute. So an arcsecond is 1 60 = degrees. The angle 0.77 arcseconds is equal to 60 = degrees. The tangent of this angle is tan( ) The cotangent of is the reciprocal of this, approximately So the distance to Proxima Centauri is = ( ) ( ) miles or 3 Proxima means close 170

5 about miles! The fastest rocket ever made reaches speeds of 25,000 miles per hour. rocket traveling at that speed would take over one hundred thousands years to reach Proxima Centauri. Light travels about miles in a year, so this distance is about 4.24 light years. nd this is our closest star... It is no wonder that the Greeks could not measure parallax; the universe is indeed unbelievably large! ther resources on the trigonometry of right triangles In the free textbook, Precalculus, by Stitz and Zeager (version 3, July 2011, available at stitz-zeager.com) this material is covered in section 10.3 In the free textbook, Precalculus, n Investigation of Functions, by Lippman and Rassmussen (Edition 1.3, available at this material is covered in section 5.5. In the textbook by Ratti & McWaters, Precalculus, Unit Circle pproach, 2nd ed., c much of this material appears in section 6.1 In the textbook by Stewart, Precalculus, Mathematics for Calculus, 6th ed., c (here at mazon.com) this material appears in chapter 5. Many online sources introduce trigonometry by beginning with a right triangle. has a number of videos on this topic. Khan cademy Homework. s class homework, please complete Worksheet 4.4, Trig Functions and Right Triangles available through the class webpage. 171

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