Chapter 7 - Trigonometry

Transcription

1 Chapter 7 Notes Lessons Geometry 1 Chapter 7 - Trigonometry Table of Contents (you can click on the links to go directly to the lesson you want). Lesson Pages 7.1 and Trigonometry asics Pages Sine and Cosine Relationship Page asic Trigonometry word problems Pages dvanced trig word problems Pages Notes created by Mike Patterson edited by Mr. rentsen

2 Chapter 7 Notes Lessons Geometry 2 Lessons 7.1 and 7.2 asics of Trigonometry In lesson 6.6 we discussed that all right isosceles triangles have the same relationships of its sides no matter how large or small the triangle is. These ratio relationships are unchanging due to the similarity of the triangles. Similarity preserves the angle measures and maintains the sides proportionality. 7 cm cm 45 7 cm It is this proportionality of all that allowed us to generalie these relationships. The ratio of leg to hypotenuse: leg hypotenuse 2 This will be true for LL triangles. 5 cm The ratios within the triangle such as the leg to the hypotenuse or the leg to the leg are fied values because they would remain proportional no matter the sie of the triangle. If we calculated them we would find that: The ratio of leg to leg: leg leg This will be true for LL triangles. When looking at ratios within the triangle we need to name the sides so that we aren t confused by which side or leg that we are referring to. When we label sides it will always be based on a reference angle. In the previous eample we didn t have to discuss this because the two reference angles were the same. o 45 2 C 45 C If is our reference angle, then C is the opposite leg, is the adjacent leg, and C is the hypotenuse. The hypotenuse will always be opposite the right angle (the longest side). The opposite leg will always be the side of the triangle that does not form the reference angle. The adjacent leg will always be the non-hypotenuse side that forms the reference angle. What is the side opposite? What is the hypotenuse? What is the adjacent side to C? C Leg o C C is the opposite leg C is the hypotenuse. C is the adjacent leg 5 cm 5 2 cm Leg cm 9 3 cm 1 Leg 9 6 cm 45 Leg C If C is our reference angle, then is the opposite leg, C is the adjacent leg, and C is the hypotenuse. C

3 Chapter 7 Notes Lessons Geometry 3 So if we look at the ratios of the sides of a triangle we need to identify which angle we are referring to. In the below eample we will refer to the sides based on the reference angle of 30. Leg 30 Leg In this net eample, we will use the 60 angle as our reference angle. Notice how the and sides differ due to the new reference angle location. The hypotenuse will always stay stationary. Leg What we begin to recognie is that the 30 right triangle has three fied ratios: We also realie that all 60 right triangles have three fied ratios: We can also now etrapolate this concept to any reference angle in a right triangle. That there will be three ratios that eist for all of those as well. It is this concept that is the basis for TRIGONOMETRY. TRIGONOMETRY (Triangle Measure) allows us to link angle sie in right triangles to side proportionality. If we know the reference angle in a right triangle, then there are three fied ratios of its sides. The converse is also true, if we know the ratio of two sides of a right triangle then there eist one reference angle that has this ratio. Trigonometry connects angle measure to side measure, thus it is a very powerful piece of mathematics. These three ratios that we have established have names: 3 60 Leg 2 The Sine Ratio (sin) The Cosine Ratio (cos) The Tangent Ratio (tan) sin cos tan 60

4 Chapter 7 Notes Lessons Geometry 4 elow is a table of the three ratios (Sine, Cosine & Tangent) for all whole number value angles from 0 to 90 I highlighted the three reference angles that we just did. This table is a clear demonstration of how angles relate to side ratios and side ratios relate to angles. Degrees Sine (sin) Cosine (cos) Tangent (tan) Degrees Sine (sin) Cosine (cos) Tangent (tan)

5 Chapter 7 Notes Lessons Geometry 5 Lets look at how trigonometry helps us discover information about triangles. table is clipped to the right to demonstrate the power of these numbers. Knowing Sides. Get ngles Knowing ngles Get Sides θ 2.20 cm (adjacent) C tan 2.62 tan tan cm (opposite) The angle must be approimately 50 because its tangent ratio is opposite Without knowing anything about the opposite and the hypotenuse other than the reference angle, we are able to know that the value of ratio between them. sin 54 opposite hypotenuse hypotenuse 54 opposite hypotenuse Degrees Sine (sin) Cosine (cos) Tangent (tan) The chart allows us to find ratios of sides if we have angles or to find the reference angle if we have the ratio of sides. The chart is a nice way to learn the concept but our calculator is much more powerful because it can handle angle sies other than whole numbers. Using Trigonometry to determine the missing angle. a) b) c) θ 34 cm 13 cm 32 cm θ 25 cm θ 34 cm 22 cm tan 13 tan tan (0.3824) cos 25 cos cos ( ) sin 22 sin sin (0.6471) When we do the inverse of a function such as tan, we denote it by tan. What this means in terms of using the table is that we know the ratio and we want to know the angle, in the calcuator it is performing the inverse of that function, which is to determine the angle sie. In most calculators to use the inverse function you need to select the 2 nd button or INV button.

6 Chapter 7 Notes Lessons Geometry 6 Using Trigonometry to determine the missing side. a) b) c) 12 cm cm 18 cm 55 C tan tan tan tan 38 (12) (0.7813)(12) 9.38 cm sin 8 sin16 sin sin16 8 (0.2756) cm cos cos cos (cos 55 )(18) (0.5736)(18) cm

7 Chapter 7 Notes Lessons Geometry 7 Lesson 7.3 Sine and Cosine Relationships The trigonometry table can reveal a number of patterns that sometimes get hidden in calculator only use. PTTERN #1 Identical Sine and Cosine Values If you look at the trigonometry table to the right you see that every sine value has a matching value for cosine. Why would that happen? Let s look at the pattern closer. Sin 5 = Cos 85 Sin 10 = Cos 80 Sin 23 = Cos 67 Sin 33 = Cos 57 Sin 45 = Cos 45 We see that if the two angles are complementary then this relationship works: ut why does that work? sin cos(90 ) Deg Sine (sin) Cosine (cos) Tangent (tan) Deg Sine (sin) Cosine (cos) Tangent (tan) UND The answer is quite a simple one. In a right triangle the two acute angles are always complementary and these two ratios are comparing the eact same sides of the triangle. Let me show you what I mean. Let us use the eample of a 23, 67 and 90 right triangle. 67 y y 23 sin sin 67 y cos cos67 cos cos 23 y sin sin 23 sin 67 cos 23 cos67 y sin 23 The two complementary angles are in the same triangle and they are actually references the same sides.

8 Chapter 7 Notes Lessons Geometry 8 Lesson 7.4 Trigonometry Word Problems In this objective we apply the principles that we have learned concerning similarity, trigonometry and the Pythagorean Theorem to solve real world problems. While we call them real world problems many of them have been greatly simplified so that the mathematics at this level can be applied. Real world problems have many variables and are often too comple to approach with this simple level of mathematics but as students progress through mathematics we add more detail to the problems making them more real. t this stage we want them to gain proficiency in the skill within a contet to see where these mathematical tools apply. We call these problems real world while students call them word problems. Yes that is often true Real world problems do come with contet and so yes they look and feel different than the standard symbol based problems that we often do. It also means that we need to gain proficiency with language as well as mathematics to succeed with these types of problems. Most of these notes will be discussing the contet of these problems and the translation of word to mathematics and not the solving of them. That skill was discussed in an earlier objective. Let me start with how we refer to angles. THE NGLE OF ELEVTION To elevate is the move in an upward direction. Thus the angle of elevation is an upward angle from the horion. The angle of elevation is the reference angle from the position in an upward direction from the horion. The stranded person on the island looks up to the plane at an NGLE OF ELEVTION. Very few students struggle with this concept The name and the placement seem to come easily. This is not true concerning the net angle, many students struggle with its placement and understanding. THE NGLE OF DEPRESSION To be depressed is to be down, thus the angle of depression is the reference angle from the position in a downward direction from the horion. Sound simple like the angle of elevation but it isn t quite as easy. Eample #1 Eample #2 Eample #3 In eample #1, the angle of depression (1) has been placed correctly, as the angle from the airplanes horion looking down. In eample #2, the most common error has taken place; the angle of depression (1) was PLCED IN THE WRONG LOCTION. Student s make this error because no horion has been drawn and so they see this as the only angle from that reference point. Finally, in eample #3, the angle of depression (1) has been correctly drawn and located - placing it from the horion down.

9 Chapter 7 Notes Lessons Geometry 9 I also want to discuss some of the CLSSIC environments of trigonometry and Pythagorean problems and their particular terminology. THE LDDER PROLEM The leaning ladder or pole is a common trigonometry problem because of the natural right angle formed between the wall and the ground. The ladder or pole becomes the hypotenuse of the triangle. word that gets used that may be new to the student is the FOOT of the ladder the foot of a ladder is the bottom of the ladder that is placed on the ground. The two complement angles are often referred to as (1) the angle formed between the ladder and the ground and (2) the angle formed between the ladder and the wall. THE KITE PROLEM The kite problem is another classic!! The length of the string of course is the hypotenuse. The perpendicular distance from the kite to the ground is the height or altitude of the kite and then other leg gets referred to as the horiontal or ground distance. The only angle I ever see used here is (1) the angle of elevation from the boy to the kite. THE GUY WIRE guy wire is a wire that is used as a brace to stabilie something. They are used on young trees to help stabilie them in windy or stormy conditions so that they don t pull their roots out of the ground. They are also used to stabilie antennas, satellite dishes, towers, etc. they are basically brace supports. In these problems the tree and the ground form the right angle, and the wire or brace becomes the hypotenuse. The other descriptions are similar to other problems already discussed.

10 Chapter 7 Notes Lessons Geometry 10 THE SHDOW VIEWING SOMETHING First of all shadows are LWYS ON THE GROUND. Too often shadows are viewed by students as the hypotenuse that is wrong. We don t often calculate for the sun s rays because it isn t something that makes sense to measure, so we are usually finding the man s height or his shadow or (1) the angle formed by the sun s rays and the ground. The other angle (2) would be the angle formed by the sun s rays and the man. When viewing something we use the phrase, the line of sight, to refer to the distance from your position to the object that you are viewing. So in this case, the pilot spots the island and that distance is called his line of sight. lso in this case because the pilot is looking down from his horion, the angle (1) that he is using is the angle of DEPRESSION from the pilot/plane to the island. Planes and objects in the sky often give their height by stating the LTITUDE of the object. Solve the given word problem. a) b) c) 3.4 guy wire is attached to a tree 3 ft from the ground. What is the angle that is formed between the wire and the ground (to the nearest degree)? 15 ft ladder leans against a wall at 52. How far from the wall is the foot of the ladder (to the nearest ft)? The sun s ray strike the ground at 55, 21 m from the base of the tree. What is the height of the tree (to the nearest metre)? 15 ft 3.4 ft 3.0 ft θ 3 sin sin ( ) The angles is approimately cos52 15 cos 52 (15) 9.23 ft The distance is approimately 9 ft. tan tan 55 (21) m The height of the tree is approimately 30 m.

11 Chapter 7 Notes Lessons Geometry 11 d) little boy flies his kite. The string formes an angle of elevation of 37 and from where he stands to directly under the kite is 45 ft. How long is the kite string (to the nearest ft)? ft 45 cos37 cos cos ft The string is appro. 56 feet long.

12 Chapter 7 Notes Lessons Geometry 12 Lesson 7.5 Trigonometry Word Problems Trigonometry problems can escalate in difficulty in a number of ways. One of those ways is to use two triangles and their measurements to come up with a single answer. Really this doesn t increase the difficulty too much you just need to keep track of the relationship between the two triangles involved. it is mostly an increase of work and not so much difficulty. Eample Problem flagpole is at the top of a building. 300 ft from the base of the building, the angle of elevation of the top of the pole is 32 and the angle of elevation of the bottom of the pole is 30. Determine the length of the flagpole (to the nearest foot). This problem looks quite comple because of the overlapping angles of elevation but really if we think through the problem we can simplify it by solving two different triangles ft The way to get the height of just the flagpole would be to calculate the total height from ground to top of flag () and then calculate the distance from ground to the bottom of the flagpole and then subtract the two values ft tan (tan 32 )(300) ft ft y tan y (tan 30 )(300) y ft y y will give us the flagpole height, = ft SYSTEM OF EQUTIONS In algebra 1 we learned how to solve two variables if we had two equations. This process was called solving a system of equations. We learned a few different ways to solve them; (1) by graphing, (2) by substitution, and finally (3) by elimination. We will look a trigonometry problem that will give us two equations and two variables and we will use substitution to solve it. Eample Problem (on net page)

13 Chapter 7 Notes Lessons Geometry 13 Eample Problem Sally and Jonathan are on either side of the tree and 40 ft apart. Sally sees the top of the tree at 42 and Jill sees the top of the steeple at 36. How high is the tree (to the nearest foot)? 42 HEIGHT 36 The first step is to break up the 40 ft. Most students break the 40 ft into two groups of 20 but this can t be because if the distances were equal then so would there angles (the tree would be the perpendicular bisector of an isosceles triangle) but the angles aren t equal so neither will the distances. HEIGHT h tan 42 h (tan 42 )( ) h tan h (tan 36 )(40 ) (tan 42 )( ) (tan 36 )(40 ) (tan 42 )( ) (tan 36 )(40) (tan 36 )( ) (tan 42 )( ) (tan 36 )( ) (tan 36 )(40) h tan h (tan 42 )(17.83) h The tree is 16 ft tall. This is just one way to solve this problem, as mentioned earlier there are a number of algebraic techniques that could be applied here.