Trigonometry. Week 1 Right Triangle Trigonometry


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1 Trigonometry Introduction Trigonometry is the study of triangle measurement, but it has expanded far beyond that. It is not an independent subject of mathematics. In fact, it depends on your knowledge of geometry and algebra. Specific topics or skills needed will be mentioned as they are used. I will be using and referring to a graphing calculator, when one is needed. I am a firm believer in students understanding what is behind what they do. It actually involves less work in the long run. But the more you understand how things work, the less you will need to memorize. The more you practice, the easier it will be. Therefore, I am going to focus on the main topics or trouble spots and reason through them. I will note very important ideas that we use several times in red letters. Make a list of these as we go through the lesson. When possible, specific websites for additional help will be given. This will save you a lot of time in searching. After going through each lesson or topic, look in the index of your textbook for the specific topic to read more and get more problems for practice. At any time you can also ask questions. Overview Our goal in this weekly lesson in 4 parts is to: Week 1 Right Triangle Trigonometry Define the six trigonometric functions for an acute angle in a right triangle and practice finding the functions from given information Introduce identities and their use Apply what we know to a few word problems (using the calculator to find specific values). Geometry terms and knowledge needed: Right triangle, hypotenuse Acute angle, right angle, complementary angles, degree measure, alternate interior angles Pythagorean Theorem (a red letter idea!). Algebra skills needed: Solving an equation by taking the square root of both sides and other methods Rationalizing a denominator (which uses multiplying by 1 in some form). Laws of exponents
2 Trigonometric Functions Defined in a Right Triangle The right triangle is important to the expansion of the trigonometric functions beyond it. You will see its presence in a number of situations that will be coming up in the next few weeks. This is why I have chosen to start with it. Textbook authors will vary on the order of content, so keep that in mind when looking for more help and problems in the book you use. We need a right triangle for reference. We can use capital letters to name the vertices, where C is usually reserved for the right angle. To use words like opposite, we need to designate an angle, so I will use angle A. We will be measuring or determining an acute angle by way of the sides. With the following diagram we can now define the trigonometric function for an acute angle A. We will abbreviate opposite side, adjacent side, and hypotenuse. sine: cosecant: cosine: secant: tangent: cotangent: The red letter definitions are the ones we use the most often. The others can also be remembered by an identity that will come later. Example 1: Let s find the six functions for angle A, given the 3 sides, in the following triangle. W e could find the functions for angle B as well. We will always keep answers exact if we can. We can certainly do that in this example. At the moment, we are not concerned about the particular measures of acute angles A and B. We will get to that later.
3 Example 2: Now, let s take an example where only 2 sides are given in a right triangle. We will use the side opposite to A as a, the side opposite to B as b, and the side opposite to C as c, just as in the Pythagorean Theorem. Find the six trigonometric functions of angle A. We see that a = 2 and b = 3. We need to find c first. From the Pythagorean Theorem we know that. So Although there are two square roots,, only the positive square root is appropriate here. We will talk more about this later. We have one more thing to do here. Simplest form does not have an irrational number (radical) in the denominator, so we must rationalize the denominator. Answers in math books will usually be in simplest form. From our definition of tangent, Examples 3: Suppose we had been asked to find the trigonometric functions for angle A given that. What would we do? We would draw a right triangle and label the angle A and, using the definition of we would label the opposite side with 2 and the adjacent side with 3. This is just another way to state the same problem as example 2. Do this to make sure you understand. á Now work the problem set for this section.
4 Problems for Trigonometric Functions Defined in a Right Triangle 1. In example 1 in this section, we found the six functions for angle A. Find the six functions for angle B. Fill in the functions for angle A and put your final answers for B in the following table. Sin Cos Tan Csc Sec Cot A B a. What is the relationship between angle A and angle B? b. What do you notice about the functions of angle B compared to those of angle A? 2. Find the six trig functions for angle A in the following right triangle. 3. Find sin, cos, and tan using the triangle below. Write answers exactly. 4. In the triangle below,. Find sin and cos exactly.
5 Answers to Problems from Trigonometric Functions Defined in a Right Triangle 1. Sin Cos Tan Csc Sec Cot A B a. Angle A and Angle B are complementary angles. (Their sum is 90.) b. Sin B = cos A, cos B = sin A, tan B = cot A, csc B = sec A, cot B = tan B b 2 = 169 b 2 = 144 b = 12 Using the definitions:. 3. From the drawing, we know that a = 2 and b = 5, so we need to find c. 4. Now let the calculator do work., so From the definitions, and.
6 Special Right Triangles Now, let s look at some very special right triangles those with 45 and those with 30 and 60. These are special because we can find the function values exactly. Since there are many right triangles with these angles, we will choose the simplest side lengths. (Would the ratios change if the triangles had longer sides? No. Remember similar triangles?) From the triangle, we see that.. Find the other three function values. In the triangle, we see that the other three function values for 30.. Find You will use these special triangles several times in this course, so you need to know them. Does that mean memorize the six functions for each triangle? No, the best way is to be able to make a quick sketch of these whenever you need them. To do that, you only need to remember a little bit. Angle Names The triangle has equal sides, so choose a side value of 1, and quickly calculate the hypotenuse using the Pythagorean Theorem. For the triangle, you only need to remember that. Use the definition of sine to sketch and label the angle, side, and hypotenuse. Complete the sketch with the 60 and use the Pythagorean Theorem to find the other side. This will get faster as you do it more often. I have used angle names A, B, and C. We will also use some other letters for the angle names. When we only need to refer to one angle, we might use the Greek letter, called theta. If we want to note both angles that are acute (and later on will be any size), then we might use the Greek letters α (called alpha) instead of A, and β (called beta) instead of B. á Now work the problem set for Special Right Triangles.
7 Problems for Special Right Triangles 1. In this section we found the sine, cosine, and tangent values for 30 and 45. Now find sin 60, cos 60, and tan 60. Sketch and label the triangle. 2. Determine if the following triangles are right triangles (Don t do it by looks or with a protractor). If a triangle is a right triangle, then pick an angle to call, and find at least one trigonometric function to determine if it is a special triangle( or ) and which one it is. 3. Find the following function values by first sketching and labeling a right triangle with the sides and angles. a. Cos 45 b. sin 60 c. tan 30
8 Answers to Problems for Special Right Triangles a. Since 2 is the longest side, it will be c in the Pythagorean Theorem. á á á, which is the longest side, so this is a right triangle. Using either acute angle as, This is a triangle. b. With 9, the longest side, as c, does No, So this is not a right triangle. c. With 8 as longest side, does so this is a right triangle. With as the lower acute angle, triangle., the sine of 30. This is a 3. a. You can sketch the following triangle if sides of 1 are easier to remember, or you can sketch a triangle with sides of and hypotenuse of 2, so that you do not need to always rationalize the denominator. Here b. You could use the answer from #1, but it is best to redo this as practice for now. part c:
9 The Beginning of Trigonometric Identities An identity is a statement (equation) that is true for all values substituted for the variable. For example, the equation is only true for x = 4. It isn t true for x = 3 or 10 or 5, etc. But the equation is true for any value you put in place of x. (Try some different values if you don t see that is true.) Let s develop the first identity. We need to refer to a general right triangle with as the angle and sides a, b, and c. We will start with the Pythagorean Theorem and use some algebra. Now divide both sides by. Writing without parentheses, this is Notice where the exponent is written. Do not write sin 2. That would mean is squared. By starting again and dividing by b 2 or a 2, we can find two more identities. These are called the Pythagorean Identities. Notice that the first one is the key identity. You can also get the second identity from this one by dividing both sides (all terms) by cos 2. You can get the third identity from the first by dividing both sides (all terms) by sin 2. To do that, we need a Quotient Identity to recognize what we get. This identity can be found by starting with. Divide both numerator and denominator by c ( the equivalent of multiplying by 1 in the form of ). This produces the complex fraction where we now see. We call the following the Quotient Identities.
10 The Reciprocal Identities are found by noticing the relationship between the functions as defined. In many situations that involve csc, sec, or cot, we tend to substitute for these functions and turn the problems into ones with sin, cos, and cot. We use these identities to find different parts of a triangle, verify another identity, or solve an equation. In this lesson we will only look at the first application. Example 4: Let be the acute angle in a right triangle such that sin =.42. Find cos and tan. (Since our accuracy here is only 2 significant digits, we will round off cos and tan to two decimal places.) Start with what you know and how it can get you to what you want to find. Since we know the value of sin and we want to find cos, we will go to the first Pythagorean identity. Substitute the value for sin. Use algebra to solve. We round to.91. Note: Again, there are two square roots, but only the positive one is appropriate here. (More on that later.) To find tan, we use the identity. To summarize, given that sin =.42, we found that cos =.91 and tan =.46. á Now work the problem set for The Beginning of Trigonometric Identities.
11 Problems for The Beginning of Trigonometric Identities 1. You are going to develop a trigonometric identity. Start with the Pythagorean Identity. Divide both sides (all terms) of the equation by cos 2. Look at the first fraction and use a law of exponents to rewrite it  reverse. Reduce the second fraction. Use the same law of exponents to rewrite the third fraction. Now you should be able to rewrite the first fraction using a quotient identity and the third fraction by using a reciprocal identity. in 2. Let be an acute angle in a right triangle such that cos =.32. Find sin and tan to two decimal places. (Do not round off until the end.) 3. Let be an acute angle in a right triangle such that tan = Find sec to two decimal places. (Do not try to find yet. Use the identities in this section.)
12 Answers to Problems for The Beginning of Trigonometric Identities 1. which we write as tan = sec Using cos =.32,. to two decimal places. Then to two decimal places. 3. Use the identity from #1: to two decimal places.
13 Applications Involving Right Triangles Example 5: (This is more of a practice problem rather than a practical problem.) Solve the right triangle with c = 253 and angle B = (Label the figure with the given information.) Since A and B are complementary angles, A = = All angles are now known. Using the definition of cosine, we see that With calculator in degree mode, put in 253 sin 42.2 enter. To find b use the definition of sine. with 3 significant digits Now all three sides are known, so the triangle is solved. with 3 significant digits Important to note when solving: Make a sketch of the problem to orient what you know and to see what to do. Do not round off to the number of significant digits until the very last step. Rounding off too soon will cause some inaccuracy. Use algebra to do all the solving and let the calculator do the calculating in one step, as much as possible. The fewer numbers you write down from your calculator, the more accurate you will be and the less work you will actually have. For example, there was no need to write down the sin 42.2 out to 4 or 5 or even 9 digits. Just wait until you re ready to calculate and do it in one step. That is why we have a calculator to use to save us work. As much as possible, continue through the problem using the given information. Why? So that you do not possibly use inaccurate information that you found. In our example, we could have
14 used angle A, other functions, or even the Pythagorean Theorem to find the other side. Instead, I kept with what was given to maintain as much accuracy as possible. However, you can now use these other things to check your work, keeping in mind that numbers have been rounded off. Your check may not be exact, but it should be close. Example 6: A ladder must reach the roof of a building 47 feet high. The ladder leans against the building at an angle of 70 (for safety purposes). What ladder length is necessary? (Make a sketch.) We have been given the side opposite 70 and want to find the hypotenuse of the triangle. So we will use the sine. To calculate, in degree mode, put in 47 sin 70 enter. (two significant digits) How could you check your answer? One possibility is to find 47/50 on your calculator and see if that is approximately the same as sin 70 on your calculator. á Now do the problem set for Applications Involving Right Triangles. To make sure that you understand that these are functions of an angle that is, changing the angle will change the function, but changing sides and keeping the angle will keep the function value go to the following interactive sites by pressing control+click with the cursor over the address and then click on go to the student worksheet.
15 Problems for Applications Involving Right Triangles 1. Solve the right triangle with c = 145 and angle A = Draw a figure and label the given information and what you need to find (a, b, angle B). Find answers to three significant digits. 2. In order to find the distance from the beach to an offshore island, the students from a trigonometry class used the following procedure. Calling a tree on the island point T and a spot on the beach point B, the students walked from B 120 feet on a perpendicular to the line BT and marked point C. They found that angle BCT was 70. How far was it from the beach to the island? 3. From a window 152 feet above ground, the angle of depression (an angle from a horizontal line downward) of a park bench in a park across the street is How far is it from the base of the building to the park bench (to three significant digits)? (What is the relationship between the 56.5 and the angle PB?)
16 Answers to Applications Involving Right Triangles 1. With the given information labeled, we find that angle B = = All angles are known. To find a, To find b, All sides are known, and the triangle is solved. 2. We need to find distance d. 3. From geometry is the theorem that when two lines are cut by a transversal, alternate interior angles are equal. So we know that angle PB is Calling the distance from PB to the base of the building d, we can see that Notes: Label your sketches with meaningful letters to help you see the situation clearly. Remember to have your calculator in degree mode. Do all algebraic solving first and leave calculating to the last step. Take the calculation out at least one more place than needed for rounding off.
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