AP Physics 1 Summer Packet Review of Trigonometry used in Physics

Transcription

1 AP Physics 1 Summer Packet Review of Trigonometry used in Physics For some of you this material will seem pretty familiar and you will complete it quickly. For others, you may not have had much or any of this material yet in your math classes. In either case, the concepts covered in this packet are some of the biggest stumbling blocks to progress in a physics class and it is important that we have as much of an understanding of it as possible to begin the year. Please review this materially thoroughly and take it seriously if you do, it will make your life in AP Physics 1 much easier. Please note, in addition to specific exercises, there are some places in the notes of this material where you need to fill in blanks don t just skip these! You will need to print this document and write your answers on it. It is to be turned in on the first day of class. This packet is designed to guide you through it without any need for additional resources. However, if you have questions you can reach me by (jmarshall@sahs1.org) or my cell phone ( ). If I do not answer a phone call directly, leave a message so I know it was a student. Marshall 1. Introduction Consider the following situation An airplane flies at a speed of 100 m/s toward the east (on an x- y graph this would be in a direction of 0 degrees). At the same time a strong wind is blowing toward the north at 20 m/s (on an x- y graph this would be in a direction of 90 degrees). This situation could be represented on a graph by considering the plane to be at the origin. The speed of the plane toward the east could be represented by arrow toward the east with a length of 100 units, while the speed of the wind could be represented by an arrow toward the north with a length of 20 units: Hopefully it makes sense that because of the wind, some of the speed of the wind would be added to the speed of the airplane and to a person observing the plane from the ground, the actual speed of the plane would be a little more than 100 m/s and the direction of the plane would no longer be directly east, but a direction that is a little north of east. But, how much is the speed of the plane changed by the wind? Is all of the wind speed added to the plane speed, or is only part of the wind speed added because the direction of plane and the wind are different? What is the actual direction the plane is now travelling in? Looking at the above diagram, we will find out that if we draw a line from the tip of each arrow in the direction of the other arrow to the point where the lines intersect, we will create a parallelogram (see diagram on next page): 1

2 Further, if we draw a line from the origin to the point where these two lines intersect, the length of this line will represent the actual speed of the plane: Additionally, the angle between the x- axis (the horizontal direction or east ) and this new line will represent the direction the plane is actually now flying: Is there some way we can actually determine how long this new line is and what the angle above the x axis it is? Notice that the line we drew from the tip of the plane speed arrow is the same length as the wind speed arrow. If we move the actual wind speed arrow to this location on the diagram we can see that the two arrows and line we originally drew form a right triangle: Is there some way that you know of that you can now find the length of the 2

3 unknown line? Of course there is! You should remember that we call the side of a right triangle that is opposite the right angle, the hypotenuse. The Pythagorean theorem tells us that for a right triangle, the square of the length of the hypotenuse (which we call side c ) is equal to the sum of the squares of the other sides of the triangle (sides a and b ): c 2 = a 2 + b 2 Or, in this case: (actual speed) 2 = ( plane speed) 2 + (wind speed) 2 If we substitute real values in this gives us: (actual speed) 2 = (100 m / s) 2 + (20 m / s) 2 If we take the square root of both sides of the equation we get: (actual speed) 2 = (100 m / s) 2 + (20 m / s) 2 or actual speed = (100 m / s) 2 + (20 m / s) 2 = 102 m / s So, with the wind blowing as stated, the plane is actually moving 102 m/s that s the easy part. You will use similar Pythagorean theorem calculations dozens (if not hundreds) of time throughout the AP Physics 1 course get used to it! However, we do not want to just know the actual speed of the plane we also want to know the direction the plane is now flying because of the wind that is, what is the angle between the original direction of the of the plane (originally the plane was flying along the x axis, or east, or at 0 degrees), and the new direction of the plane. Being able to determine this requires a basic knowledge or trigonometry, which some of you have not yet had in math classes. Even if you have, the review will be really helpful. 3

4 2. Lets define some terminology. As we have already stated: hypotenuse the hypotenuse of a right triangle is the side of the triangle opposite the right angle: There are two ways to identify the hypotenuse of a right triangle. 1) It will always be the longest side of the right triangle 2) Identify the right angle of the triangle it will be the one which is Then look across the triangle the hypotenuse will be the side of the triangle that is across from the right angle. Until you become thoroughly familiar with these concepts and they become intuitive (automatic) to you, whenever you have a right triangle I would get in the habit of immediately labeling the hypotenuse of a right triangle with a small letter h. There are still two other sides to the right triangle and we need a way to label or reference them. How we do this will depend on which angle (other than the right angle) we are talking about. In physics, we usually use the small Greek letter, θ (theta), to reference an angle. For the triangle we are talking about, we could label the two remaining angles, θ 1 and θ 2 : The angle we were concerned about finding for the direction of the airplane was θ 1. With reference to this angle, we can now label the sides. The side of the triangle that is next to the angle is called the adjacent side, because it is adjacent to the angle (yes, the hypotenuse is also adjacent to the angle, but we are already calling it the hypotenuse). The side of the triangle that is on the side opposite the angle is called the opposite side. As with the hypotenuse, whenever you have a right triangle I 4

5 would get in the habit of immediately labeling the adjacent side of a right triangle with a small letter a, and the opposite side with a small letter o: Recognize again, what you label as adjacent and opposite will depend on the angle you are interested in. We are interested in angle θ 1, but if we were interested in angle θ 2 instead, the adjacent and opposite sides would be switched: On the diagram of our airplane problem shown below, label the angle we wanted to find as θ, and also label the hypotenuse (h), adjacent side (a), and opposite side (o). 5

6 3. For any right triangle, for a given angle, the ratios of the lengths of the sides of the triangle will always be the same (huh?) the meaning of sine, cosine and tangent. As you become more and more experienced with physics involving right triangles, you will become aware that there are five features or pieces of information we want access to regarding the triangle. We already know that one of the angles is a 90 degree angle. To completely characterize the triangle we will want to know the number of degrees of the other two angles, and the lengths of each of the sides. You will find out, if you don t know this already, that if we have a right triangle, and are given information that provides us with (a) one of the non- right angles and the length of one of the sides, or (b) the length of two sides, we will be able to determine the other three features. The ability to do this depends on the concept that for any right triangle, for a given angle (that is not the right angle), the ratios of lengths of the sides will always be the same. Lets see if we can use our airplane triangle to make sense of this. Below is our triangle notice I have labeled the angle we are interested in as θ, and I have labeled the hypotenuse (h), the adjacent side (a), and the opposite side (o). Notice also the opposite side has a value of 20 and the adjacent side has a value of 100. What is the ratio of the value of the opposite side compared to the value of the adjacent side? You can see that it is o = 20 = 0.2. Stated another way, the length of a 100 the opposite side is 1/5 of the length of the adjacent side. What if we were to make this triangle five times bigger (see diagram on next page)? You can see from simply inspecting the two triangles that while the lengths of all the sides are multiplied by 5, all of the angles are 6

7 exactly the same in both triangles. Also, and most importantly, the ratio of the length of the opposite side to the length of the adjacent side remains the same. o = 20 = 100 = 0.2 a Because all of the angles of the right triangle have remained the same, both the smaller and larger triangle are stated to be similar they have the same shape. This means that even though the lengths of the sides have increased, the ratios of the lengths of each side must be the same. That is, even though we increased the value of the length of opposite side from 20 to 100, the ratio of the length of the opposite side to the adjacent side must still be the same. The opposite side must still have a length that is 1/5 of the length of the adjacent side. Therefore, the length of the adjacent side must now have a value of 500. Based on this reasoning, it should make sense that for this particular angle, θ, the ratio of the length of the opposite side to the length of the adjacent side, that is, o, will always be the same, no matter what the specific lengths a are. For this angle of θ, if the opposite side has a length of 2, the adjacent side must have a length of 10. If the opposite side has a length of 7, the adjacent side must have a length of 35. If the opposite side has a length of 15, the adjacent side must have a length of 75. If the opposite side has a length of 2.5, the adjacent side must have a length of 12.5 and so on. For this angle of θ, the ratio o will always equal 0.2. a 7

8 It should now further make sense that if the ratio of o is some value other a than 0.2, the angle θ would also be different. For example, if the opposite side had a value of 5 and the adjacent side had a value of 12, the ratio o a would be 5 =.416. As this ratio is larger than in the previous case, the angle 12 θ is also greater. Finally, it should make sense that when the ratio of the length of the opposite side to the length of adjacent side ( o ) is a certain value, we can measure a the angle that corresponds to this ratio and write that value down somewhere in a reference book, or that we can further write down the angles that correspond to all of the possible ratios of opposite side to adjacent side. But, it s going to get really annoying to keep saying ratio of the length of the opposite side to the length of the adjacent side, so we are going actually give this ratio a special name tangent. Lets not forget, we actually have three sides, so we can also have the ratio of the length of the opposite side to the length of the hypotenuse, or the ratio of the length of the adjacent side to the length of the hypotenuse. Of course, we give these ratios special names, too. sine (abbreviated sin) ratio of the length of the opposite side to the length of the hypotenuse: sin = o h cosine (abbreviated cos) ratio of the length of the adjacent side to the length of the hypotenuse: cos = a h tangent (abbreviated tan) ratio of the length of the opposite side to the length of the adjacent side: tan = o a 8

9 These ratios are also called trig functions (short for trigonometry). The point is, in a right triangle, for a given number of degrees for an angle θ, these ratios of the lengths of sides will always be the same no matter what the actual values are, and we have reference books that tell us what those ratios will be for any number of degrees. Conversely, if we have the ratio, we can find out what the corresponding number of degrees must be. Lets use an old fashioned illustration of this. Go to the following link on the internet: This page shows the reference I am talking about a trig table. In our first example we said that the ratio of the length of the opposite side to the length of the adjacent side was 20 =.2. Again, the ratio of 100 the opposite to the adjacent side is called the tangent o a = tan. So, look under the Tan column in this table and find the values.1944 and.2126 (.2 is between these two values). Notice that in the Deg column, the corresponding values for degrees are 11 and 12. That is, if the ratio of opposite to adjacent is.2, the corresponding angle must be somewhere between 11 and 12 degrees. 9

10 We also had a right triangle where the ratio was 5/12 or.416. Using the same trig table, between what two values must the angle θ be for this ratio? There are two ways to view this information. We can either know the angle θ and find out what the ratios sin, cos, and tan are. Or, we can know what the ratio is, and find out what the angle is that corresponds to it. However, we use calculators now, not trig tables (I had to use a trig table in high school because we did not have calculators). If you know the angle is a certain number of degrees, to find either sin, cos or tan on your calculator, simply hit the sin, cos, or tan button. This will print the name of the ratio on the screen and give you a left parenthesis. sin( cos( tan( Then type in the number of degrees and close the parentheses (always make a habit of closing parentheses). For now, make sure your calculator is set to degrees and not radians. When you find the sin of a certain angle, make sure you understand, you are finding the ratio of the length of the opposite side of a triangle to the length of the hypotenuse. (1) If the angle is small (<30 degrees), the opposite side will be short compared to the hypotenuse and the sin will be less than 0.5 (2) At an angle of 30 degrees, the opposite side will be one- half the length of the hypotenuse, and so the sin will be 0.5. (3) At an angle greater than 30 degrees, the length of the opposite side will approach the length of the hypotenuse and the sin will be greater than 0.5. For a right triangle, the length of the opposite side can never by greater than the hypotenuse so the sin can never be greater than 1. 10

11 When you find the cos of a certain angle, make sure you understand, you are finding the ratio of the length of the adjacent side of a triangle to the length of the hypotenuse. (1) If the angle is great (>60 degrees), the adjacent side will be short compared to the hypotenuse and the cos will be less than 1. (2) At an angle of 60 degrees, the adjacent side will be one- half the length of the hypotenuse, and so the cos will be 0.5. (3) At an angle less than 60 degrees, the length of the adjacent side will approach the length of the hypotenuse and the cos will be greater than 0.5. For a right triangle, the length of the adjacent side can never by greater than the hypotenuse so the cos can never be greater than 1. When you find the tan of a certain angle, make sure you understand, you are finding the ratio of the length of the opposite side of a triangle to the length of the adjacent side. (1) If the angle is great (>45 degrees), the opposite side will be longer than the adjacent side and the tan will be greater than 1. (2) At an angle of 45 degrees, the adjacent side will equal the length of the opposite side, so the tan will be 1. (3) At an angle less than 45 degrees, the length of the adjacent side will greater than the opposite side and the tan will be less than 1. 11

12 What if we are given the ratio (that is, the sin, cos, or tan) and want to know the angle. The process of finding the angle given the ratio is called finding the inverse. If you are given the sin and want to know the angle, you find the inverse of the sin (designated sin - 1 ) (this is also known as the arcsin ). If you are given the cos and want to know the angle, you find the inverse of the cos (designated cos - 1 ) (this is also known as the arccos). If you are given the tan and want to know the angle, you find the inverse of the tan (designated tan - 1 ) (this is also known as the arctan). To accomplish this on the calculator, hit the 2 nd button, and then the ratio you want to use (sin, cos, or tan). On the screen will then be printed the name of the ratio, raised to the negative 1 power (this means inverse of ), followed by a left parenthesis. sin - 1 ( cos - 1 ( tan - 1 ( Then type in the value of the ratio and close the parentheses. The answer will then be the number of degrees. Lets test this. In our above example we said that when the ratio of the opposite side to the adjacent side was 0.2, the corresponding angle was between 11 and 12 degrees according to the trig table. So, find the inverse tangent of 0.2 on your calculator. You should get , confirming what our trig table said. The relationship between the ratios of the different sides of a right triangle and the corresponding angles is so important that we have a mnemonic to help us remember the names of the ratios and the corresponding sides: SOH- CAH- TOA (pronounced sew- caw- toe- a) 12

13 Lets make sure you have a good understanding of how to find values for sin, cos and tan given an angle, and how to find angles given a value for sin, cos and tan. Using the trig table ( given the following angles, which are not exact whole number angles, state the values between which the actual sin, cos and tan must be (notice that the table headings for values of trig functions between 0 and 45 degrees are at the top of the table, while the table headings for values of trig functions between 45 and 90 degrees are listed at the bottom of the table). (round to the nearest thousandth). I have done the first one for you Angle sin is between cos is between tan is between and and and and and and and and and and and and and and and Now, using your calculator, find the exact values of the trig functions for these same angles (round to nearest thousandth). Angle sin cos tan

14 Given the following values for sin, use the trig table ( trig/tables.htm) to determine the angles between which the actual angle must be. Then use your calculator to determine the exact value (round to nearest hundredth of a degree) Value of sin angle is between exact value (nearest hundredth).6092 and (think this means that the length of opposite side is.6092 times the length of the hypotenuse what does that mean for the size of the angle don t.0035 and (think this means that the length of opposite side is.0035 times the length of the hypotenuse what does that mean for the size of the angle don t.1987 and (think this means that the length of opposite side is.1987 times the length of the hypotenuse what does that mean for the size of the angle don t.8888 and (think this means that the length of opposite side is.8888 times the length of the hypotenuse what does that mean for the size of the angle don t.3443 and (think this means that the length of opposite side is.3443 times the length of the hypotenuse what does that mean for the size of the angle don t Given the following values for cos, use the trig table( trig/tables.htm) to determine the angles between which the actual angle must be. Then use your calculator to determine the exact value (round to nearest hundredth of a degree) Value of cos angle is between exact value (nearest hundredth).6092 and (think this means that the length of adjacent side is.6092 times the length of the hypotenuse what does that mean for the size of the angle don t.0035 and (think this means that the length of adjacent side is.0035 times the length of the hypotenuse what does that mean for the size of the angle don t.1987 and (think this means that the length of adjacent side is.1987 times the length of the hypotenuse what does that mean for the size of the angle don t 14

15 .8888 and (think this means that the length of adjacent side is.8888 times the length of the hypotenuse what does that mean for the size of the angle don t.3443 and (think this means that the length of adjacent side is.3443 times the length of the hypotenuse what does that mean for the size of the angle don t Here s something interesting you may have already noticed it add up the values of the exact angles for each value of sin and cos as follows: angle for sin - 1 (.6092) + angle for cos - 1 (.6092) = angle for sin - 1 (.0035) + angle for cos - 1 (.0035) = angle for sin - 1 (.1987) + angle for cos - 1 (.1987) = angle for sin - 1 (.8888) + angle for cos - 1 (.8888) = angle for sin - 1 (.3443) + angle for cos - 1 (.3443) = sin and cos are said to be reciprocal functions this means that the sin of one angle will equal the cos of the complement of that angle (remember that two complementary angles add up to 90 degrees) so, sin(35 0 ) = cos(55 0 ). This means that if you know the sin of an angle, you also automatically know the cos of its complement. Given the following values for tan, use the trig table( trig/tables.htm) to determine the angles between which the actual angle must be. Then use your calculator to determine the exact value (round to nearest hundredth of a degree) Value of tan angle is between exact value (nearest hundredth).0178 and (think this means that the length of opposite side is.0178 times the length of the adjacent side what does that mean for the size of the angle don t and (think this means that the length of opposite side is times the length of the adjacent side what does that mean for the size of the angle don t and (think this means that the length of opposite side is times the length of the adjacent side what does that mean for the size of the angle don t.8888 and 15

16 (think this means that the length of opposite side is.8888 times the length of the adjacent side what does that mean for the size of the angle don t and (think this means that the length of opposite side is times the length of the adjacent side what does that mean for the size of the angle don t 16

17 4. How to use sin, cos, and tan (trig functions) to find unknown sides/angles of right triangles. In the introductory problem we saw that in solving some types of physics problems, we might have a critical need to be able to determine the value of unknown lengths of sides and angles of a right triangle. Underlying this skill is the understanding that for a specific angle in a right triangle, the ratio of the length of any of the sides to one of the other sides is a constant value, no matter what the actual lengths are. For a particular angle, we can either look up these ratios (sin, cos, and tan) on a trig table, or we can find them using a calculator. Knowing these ratios for a particular angle, if you know the length of one of the sides you can determine the length of the other side. Or, if you know the length of two sides, you can determine the angle associated with those two sides. The following exercises will guide you through and give you practice in the strategy we need to follow to develop these necessary skills. - - Given the right triangle illustrated to the right - - Given the angle designated θ 1. Label the sides of the triangle as: h for hypotenuse a for adjacent o for opposite θ 2. The sin of θ is defined as the ratio of the length of the side to the length of the. 3. The cos of θ is defined as the ratio of the length of the side to the length of the. 4. The tan of θ is defined as the ratio of the length of the side to the length of the side. (Recognize that if we had called the other non- right angle θ, the sides that we called opposite and adjacent would be reversed, although the hypotenuse would still be the same side.) 17

18 5. We commonly use the abbreviations listed in question 1 for the hypotenuse, adjacent and opposite sides of a right triangle as they relate the angle designated as θ. Using those abbreviations, complete the following equations: sin θ = cos θ = tan θ = 6. Remember that we have created a mnemonic to help us remember these ratios Write out that mnemonic: 7. For the following right triangles (starting on the next page), you have been given all three sides of the triangle. For example, for triangle 1 you have been given that one side has a value of 14.7, another has a value of 3.31, and the last has a value of Get used to the idea that we will commonly label sides of a triangle with variables that mean something to us and their corresponding value. In the first triangle, V, VA, and VB are referring to velocity, velocity A and velocity B. It will not only be important for you to recognize the hypotenuse, adjacent side and opposite side of a right triangle, but that these sides correspond to specific values in a physics problem and you can substitute these values in for h, o, and a to help you solve problems. Recognizing this will allow you determine which trig functions you need to use to solve problems. For each triangle, for each angle listed (θ 1 and θ 2): (a) Write out the general formula for the trig function asked for using the standard abbreviations, h for hypotenuse, o for opposite, and a for adjacent (b) Then write that formula in terms of the labels on the triangle; (c) Then plug the correct values in and solve Recognize that what the adjacent and opposite sides will be will depend on whether it is θ1 or θ2. One of the functions for each angle of the first triangle has been done for you as an example of what I want. Remember also that, as ratios of lengths of sides of a right triangle, the final values will have no units they are simply ratios. For the example I have completed, the sin of angle θ1 is simply a ratio telling us that side VA is.225 times as long as side V. Recognize also that there will only be one angle that can give us that specific ratio. Round all values to the thousandths place. 18

19 Triangle 1 For angleθ 1 sinθ 1 = o h = V A V = =.225 cos θ 1 tan θ 1 For angleθ 2 sinθ 1 = o h = V B V = =.973 cos θ 2 tan θ 2 Triangle 2 19

20 For angleθ 1 sin θ 1 cos θ 1 tan θ 1 For angleθ 2 sin θ 2 cos θ 2 tan θ 2 Triangle 3 20

21 For angleθ 1 sin θ 1 cos θ 1 tan θ 1 For angleθ 2 sin θ 2 cos θ 2 tan θ 2 8. Now lets use our knowledge to find missing pieces of information. If we are given an angle and one side of a right triangle, we can always find the the lengths of the other two sides (if we have a calculator or a trig table). The basic strategy is this (this will seem confusing at first but we will go through several examples step by step): - - It is always a good idea to label a right triangle with h, o, and a. - - Ask the question, related to the given angle, what side of the triangle have I been given, the hypotenuse, the adjacent side or the opposite side? - - Then ask the question, related to the given angle, what side of the triangle do I want, the hypotenuse, the adjacent side or the opposite side? - - Then ask the question, because I can remember which trig functions are related to which sides because I know the mnemonic, SOH CAH TOA, which trig function contains both the side I have been given and the side I want? 21

22 - - Then write down the basic formula for that trig function (as you did in problem 5). - - Then substitute in the labels on the diagram of the triangle for the abbreviations of the sides in the formula. - - Then recognize that to get the length of the side you want, isolate that variable on one side of the equation. (We will develop the habit of not ever substituting values into a formula until the variable you want has been isolated on one side of the equation) - - Then plug in the values and solve. Using this strategy, find the missing pieces of information in the following problems the first one has been done for you. In each successive problem, I have decreased the amount of guidance. (PLEASE NOTE I am not asking you to write down these questions they are questions you will ask yourself in your brain as you develop the ability use this skill). Triangle 1 Given side V = 23.9 m/s and θ2 = 71 o, find the value of side VB. Label the triangle with h, o, and a. Related to angle θ2, what side have you been given? Hypotenuse Related to angle θ2, what side do you want? Opposite What trig function contains both of these sides? Sin What is the formula for this trig function? sinθ 2 = o h Substitute labels from the diagram into this formula: sinθ 2 = o h = V B V Isolate the desired value on one side of the equation: We want the length of side VB this means that before we plug values into the formula, we isolate VB on one side of the formula. 22

23 In this particular circumstance, in order to get VB by itself on one side of the equation, we need to multiply both sides by the denominator, V. (V )(sinθ 2 ) = o h = (V ) V B V (I like the isolated variable on the left) V sinθ 2 = V B V B = V sinθ 2 Plug in given values and solve: V B = (23.9 m / s)(sin71 0 ) = 22.6 m / s Triangle 2 Given side DA = 414 m/s and θ1 = 32 o, find the value of side DB. Label the triangle with h, o, and a. Related to angle θ1, what side have you been given? Related to angle θ1, what side do you want? What trig function contains both of these sides? What is the formula for this trig function? Substitute labels from the diagram into this formula: Isolate desired value on one side of the equation: Plug in given values and solve: 23

24 Triangle 3 VA =? VB = 191 km/h θ2 θ1 V =? Given side VB = 191 km/h and θ2 = 74 o, find the value of side VA. Note even though I have left out some of the guiding questions you should still be thinking them in your mind to get to the answer of the questions I have left in. What trig function contains both the given side and the side you want? Write the formula for this function using the labels from the given triangle: Isolate the desired value on one side of the equation: Plug in given values and solve: Triangle 4 Given side VA =7.45 m/s and θ 1 = 27 0, find the value of side V. You re on your own for this one just follow the strategy and it should be no problem. You need to show me the formula you are using to solve the problem, how you isolate the desired variable, how you plug the correct values into the formula, and the answer. 24

25 9. Now lets reverse the process. Previous problems have shown us (a) that if we know the lengths of the sides of a right triangle, we can find ratio of the length of the opposite side compared to the hypotenuse (sin), the ratio of the length of the adjacent side compared to the hypotenuse (cos), and the ratio of the length of the opposite side compared to the adjacent side (tan) (b) that if we know the length of a side and an angle, we can find the lengths of the other two sides. We have also stated that for any non- right angle of a right triangle, whether it s 1.7 o, 59.6 o, 74.3 o, or whatever the angle, for that angle, there is one and only one sin, cos and tan that corresponds to that angle. Therefore, for a given value of sin, cos, and tan, there is one and only one angle that can correspond with that value. Therefore, if I have a calculator or a trig table, and I know a value for sin, cos or tan, or I can calculate one of these values using the lengths of sides, I can find the angle by finding the inverse of that function, denoted: sin - 1, cos - 1, and tan - 1. The basic strategy is this given two sides of a right triangle, or, given information that will allow me to calculate the length of two sides of a right triangle: - - Label the triangle h, o, and a. - - Ask the question, related to the angle I want to find, what two sides of the right triangle do I have, the opposite and hypotenuse, the adjacent and hypotenuse, or the opposite and adjacent? - - Then ask the question, what trig function contains both of these sides? - - Then write down the standard formula for that trig function, remembering to include the θ symbol after the name of the function. - - Then move the name of the function to the other side of the equation, in front of the ratio for that function, and place a superscript - 1 after the name of the function, so that just the θ symbol is on one side of the equation, and the inverse designation for the trig function, and its ratio are on the other side. - - Then substitute in the labels on the diagram of the triangle for the abbreviations of the sides in the formula. - - Then plug in the values given and solve. In the following problems, find the requested angle the first one has been done for you. The remaining three I have given you no additional help on. 25

26 Triangle 1 Find angleθ 1 In relationship to angle θ1, what two sides have you been given? opposite and hypotenuse What trig function contains both of these sides? sin What is the formula for this trig function? sinθ 1 = o h Write the inverse form of this formula: θ 1 = sin 1 o h o Substitute in labels from the triangle: θ 1 = sin 1 h = V A sin 1 V Plug in values: θ 1 = sin = Triangle 2 Find angleθ 2 (on your own) Triangle 3 26

27 Find angleθ 1 (on your own) Triangle 4 Find angleθ 2 (on your own) VA = 7.45 m/s θ2 V = 16.4 m/s VB =? θ1 27