9/16: Special Right Triangles; Classwork & Homework KEYS
9/17: Intro to Right Triangle Trig – Trigonometric Ratios 9/18: Solving for missing sides in right triangles with applications; KEYS: Doc 1 Doc2 9/19: Solving for missing angles in right triangles with applications & review 9/20: MidUnit Quiz on Special Right Triangles & Right Triangle Trig. 9/23 – 9/27: September Break 9/30: Angles of Elevation & Depression Class Notes/CW Key; HW – Test Review due Wednesday 10/1: Sine & Cosine of Complementary Angles Classwork KEY 10/2: Right Triangle Trig Review (Sum Em Game); Test Review HW KEY 10/3: Right Triangle Trig Unit Test 10/4: Begin Midterm Review – Sub day 10/7: Midterm Review 10/8: Midterm Exam 10/9: 3D Volume Unit Begins 10/10: 3D Volume/Early Release 10/11: 10/13: 10/14: 10/15: PSAT/Unit Review 10/16: 3D Volume Unit Test KEY Oct 08 Download Unit 3 – Right Triangle Trigonometry and more Trigonometry Study Guides, Projects, Research in PDF only on Docsity! 3. Right Angle Trigonometry - 1 - www.mastermathmentor.com - Stu Schwartz Unit 3 – Right Triangle Trigonometry - Classwork We have spent time learning the definitions of trig functions and finding the trig functions of both quadrant and special angles. But what about other angles? To understand how to do this, and more importantly, why we do it, we introduce a concept called the unit circle. A unit circle is a circle whose radius is one. To the left is a unit circle. The angle ! " is drawn in the first quadrant but could be drawn anywhere. Suppose ! " = 40°. If we were to find ! sin40°, we know that it would be defined as ! y 1 = y . So when we take ! sin40°, we are finding the height of the triangle in a unit circle. The same argument holds when we take ! cos40° … we are actually finding the x variable in a unit circle. When we take ! tan40° , we are finding the ratio of y to x in a unit circle. On your calculator, be sure you are in Degree Mode and set your decimal accuracy to FLOAT. Use your calculator to find ! sin40° and ! cos40° . Remember what it is you are finding: the y and x variables in the triangle above. And since ! x = cos40° and y = sin40° , let us show that the Pythagorean theorem holds in this triangle based on the unit circle. Taking trig functions on the calculator is straightforward: type in the trig function (you will get a left parentheses) and the angle. You do not need to complete the parentheses. Press ENTER and out it comes. Although we can get extreme accuracy, we will find that four decimal places is usually enough. So set your calculator to 4 decimal places. Remember that angles are assumed to be in radians unless in degree format. Example 1) Find the following: a) ! sin29° = .4648 b) ! cos131° = ".6561 c) ! tan 7" 8 # $ % & ' ( = ).4142 If angles are input with more accuracy, it is assumed that they are in decimal degrees. Note that parentheses can be used to make the problems clearer in intent. Example 2) Find the following: a) ! tan12.8° = .2272 b) ! sin "32.35°( ) = ".5351 c) ! cos 0.724°( ) = .9999 If trig functions of angles that are in degree-minute-second form, use the Angle menu to input them. Remember that seconds are input with ALPHA +. ! cos38°4 " 0 2 " " 9 would be input to the calculator thusly: Example 3) Find the following: a) ! sin82°1 " 2 = .9907 b) ! cos126°4 " 2 5 " " 3 = #.5978 c) ! tan "8°5 # 7 1 # # 6 ( ) = ".1576 3. Right Angle Trigonometry - 2 - www.mastermathmentor.com - Stu Schwartz Note that there are no keys for the csc, sec, or cot functions on your calculator. To find them we have to use the fact that sin and csc functions are reciprocals of each other, as are the cos and sec functions, and the tan and cot functions. There are three ways to find, for example ! csc37°. Take ! sin37° and then take its reciprocal or simply finding 1/ ! sin37°. The screen on the right shows these two methods. You could also find ! sin37° and then press the reciprocal key ! x "1. Example 4) Find the following: a) ! csc81° =1.0125 b) ! sec122° = "1.8871 c) ! cot 34.2° =1.4715 d) ! sec338.292° =1.0763 e) ! cot14°2 " 9 3 " " 6 = 3.8686 f) ! csc149°5 " " 0 =1.9424 Many times, we want to reverse the process. We know the sine of an angle and we wish to find the angle itself. To accomplish this, we use inverse trig functions or arc trig functions. These are found on your calculator above the sin, cos, and tangent keys. We use the blue (2nd) key to input them. For instance, let us find the first quadrant angle whose sine is .7523. Note the screen on the right. Our answer would be ! 48.79° (expressed in decimal degrees). If we wanted our answer in degree – minute – second format, note how we would accomplish that by using the Angle menu. Example 5) Find the following (decimal degrees): a) ! sin "1 .9099 = 65.4915° b) ! arccos0.4231= 64.9695° c) ! tan "1 1.8089 = 61.0652° Example 6) Find the following (Degrees – minutes – seconds) a) ! arctan4.002 = 75°5 " 8 1 " " 4 b) ! sin "1 .0809 = 4°3 # 8 2 # # 5 c) ! cos "1 .4998 = 60°4 # # 8 Finally, if we wish to find an arccsc, arcsec, or arctan function, again, there is no one keystroke that will give it to you. To find ! csc "1 2.3552 , for instance, we must first take the reciprocal of 2.3552, and then take the arcsin of that. On the right is the way you would accomplish this (with the optional changing into degrees-minutes-seconds): Example 7) Find the following (decimal degrees): a) ! sec "1 1.76 = 55.3765° b) ! arccot 3.4221=16.2893° c) ! csc "1 1.1102 = 64.2553° Example 8) Find the following (Degrees – minutes – seconds) a) ! arccsc 3.8621 =15° " 0 2 " " 3 b) ! cot "1 0.7501= 53° # 7 3 # # 5 c) ! arcsec5.8621= 80°1 " 0 4 " " 1 About errors: Your calculator can take trig functions and arc trig functions to extreme accuracy. However, if you input the problem into the calculator incorrectly, one of two things will happen. One – an error. Assuming you typed it in the correct syntax, the calculator is telling you that it cannot perform the operation. This is actually good for you. For instance, if you take ! cos "1 1.4231 the calculator gives you a domain error because we know that the 3. Right Angle Trigonometry - 5 - www.mastermathmentor.com - Stu Schwartz Multi-Step Problems: Example 14) Consider the picture below. I want to find the length of segment AB. Suppose ! "A = 25°,"B = 40° and BC =12 . Do the necessary work on the right to find ! AB . ! CD =12tan40° =10.07 AC = 10.07 tan25° = 21.59 AB= 21.59 "12 = 9.59 Example 15) Let’s tweak the problem slightly. I want to find the length of segment CD. ! "A = 25°,"B = 40° and AB =12. Note that we do not have any information about sides of either right triangle. And yet, it is possible to solve this problem. How? ! CD = ACtan25° = ABtan25° + BCtan25° CD = BCtan40° 12tan25° + BCtan25° = BCtan40° 12tan25° = BC tan40°" tan25°( ) BC = 12tan25° tan40°" tan25° =15.01 CD =15.01tan40° =12.59 Example 16) Using the same picture, ! "A = 35°,"B = 68° and AB = 76.5, find the length of segment CD. ! CD = ACtan35° = ABtan35° + BCtan35° CD = BCtan68° 76.5tan35° + BCtan35° = BCtan68°" 76.5tan35° = BC tan68°# tan35°( ) BC = 76.5tan35° tan68°# tan35° = 30.18" CD = 30.18tan68° = 74.72 Real-Life Applications Guidelines for solving a triangle problem: a) Draw a sketch of the problem situation. Don’t be afraid to make it large. b) Look for right triangles and sketch them in. c) Mark the known sides and angles and unknown sides and angles using variables. d) Express the desired sides or angles in terms of trig functions with known quantities using the variables in the sketch. e) Solve the trig equation you generated and express the answer using correct units. 3. Right Angle Trigonometry - 6 - www.mastermathmentor.com - Stu Schwartz Surveying Problems: Problems involving finding quantities that would be too difficult to measure using rulers and other instruments. 17) I am standing at point A on one side of a river and wish to measure the distance across a river to a house H on the other side. I walk 50 feet along the riverbank to point B and measure the angle ABH to be ! 18°23. Find the distance across the river (assume a right triangle). ! tan18°23 = AH 50 AH =16.62 ft 18) A guy-wire must be attached to a 50 foot pole. The angle that the guy wire must make with the ground has to be ! 75.5°. Find the length of the wire required to do the job. ! sin75.5° = 50 w w = 51.65 ft Angle of Elevation and Depression. As a person at point A looks up at point B, an angle of elevation with the ground if created. When B looks down at point A, an angle of depression is created (with the horizontal). These angles are congruent. Why? Alternate interior angles. Angles of elevation and depression are always created with the horizontal. 19) A device for measuring cloud height at night consists of a vertical beam of light that makes a spot on the clouds. That spot is viewed from a point 140 feet along the ground and the angle of elevation is ! 64°4 " 0 . Find the height of the cloud. 20) While standing on a cliff 163 feet above the ocean, I see a sailboat in the distance at an angle of depression of ! 21.2°. What is the horizontal distance to the sailboat? ! tan21.2° = 163 h h = 420.24 ft ! tan 64°4 " 0 = h 140 h = 295.73 ft 3. Right Angle Trigonometry - 7 - www.mastermathmentor.com - Stu Schwartz 21) A wire holding up a 40 foot telephone pole is 38 feet long. The wire attaches to the telephone pole 10.5 feet below the top. What is the angle of elevation of the wire? ! sin" = 29.5 38 " = 50.92° 22) From a window in building A, I observe the top of building B across the 50 foot wide street at an angle of elevation of ! 74°2 " 5 . I observe the base of building B at an angle of depression of ! 52°1 " 8 . Find the height of building B. ! tan52°1 " 8 = a 50 tan74°2 " 5 = b 50 a = 64.64 ft b =174.28 ft a + b = 243.97 ft Bearing and Course: When ships or planes navigate, they need to have a simple way of explain what direction they are traveling. One way is called bearing. The bearing will be in the form (N or S angle E or W). A bearing is always drawn from the nearest north or south line. A heading (or course) is always drawn from the north line in a clockwise direction. Following are ship directions, the bearing and course. ! Bearing : N60°E Course : 60° ! Bearing : S70°E Course :110° ! Bearing : S35°W Course : 215° ! Bearing : N72°1 " 8 W Course : 287°4 " 2 Tip: in word problems, whenever you see or are asked for a bearing or heading, always look for the word “from” and draw an x-y axis at that point. Tip: if the bearing from A to B is ! N65°E , the bearing from B to A will be the same angle but the opposite direction: ! S65°W . 23) A jeep leaves its present location and travels along bearing ! N62°W for 29 miles. How far north and west of its original position is it? ! sin28° = n 29 cos28° = w 29 n =13.61 miles w = 25.61 miles 3. Right Angle Trigonometry - 10 - www.mastermathmentor.com - Stu Schwartz d) ! A = 21.75° a = 0.977 in B = 68.25° b = 2.45 inches C = 90° c = 2.638 in ! tan21.75° = a 2.45 sin68.25° = 2.45 c _________________________________________________________________________________ e) ! A = 28.072° a = 8 B = 61.928° b = 15 C = 90° c = 17 ! 64 + b 2 = 289 sin A = 8 17 _________________________________________________________________________________ f) ! A = 46°3 " 5 2 " " 8 a = 9.25 B = 43°2 " 4 3 " " 2 b = 8.75 C = 90° c = 12.733 ! 9.25 2 + 8.75 2 = c 2 tan A = 9.25 8.75 _________________________________________________________________________________ g) ! A = 56.044° a = 2.25 miles = 11,880 feet B = 35.956° b = 8,000 feet C = 90° c = 14,322.514 feet or 2.713 miles ! 11880 2 + 8000 2 = c 2 tan A = 11880 8000 _________________________________________________________________________________ 3. In the figure above find segment CD if ! "A = 27°,"B = 51° and AB = 8.2 miles ! CD = ACtan 27° = ABtan 27°+ BCtan 27° CD = BCtan 51° 8.2tan 27°+ BCtan 27° = BCtan 51° 8.2tan 27° = BC tan 51°" tan 27°( ) BC = 8.2tan 27° tan 51°" tan 27° = 5.760 CD = 5.760tan 51° = 7.113 miles 3. Right Angle Trigonometry - 11 - www.mastermathmentor.com - Stu Schwartz For each word problem, draw a picture, fill in given sides and use variables for sides and angles you don’t know, then show equations (and/or trig functions) to find out the desired information and circle your answer(s) being sure to use proper unit. 4. A surveyor wants to determine the horizontal distance that the top of a slope is from a tall house at the bottom of the slope. He measures the distance along the slope to be 125 feet. The angle of depression created from the top of the slope to the top and bottom of the house is ! 17°. Find the horizontal distance from the top of the slope to the house. ! cos17° = a 125 a =119.538 ft 5. What is the angle of elevation of the sun when a 6’3” man casts a 10.5-ft shadow? ! tan" = 75 126 " = 30.763° 6. A street in San Francisco has a 20% grade – that is it rises 20 feet for every 100 feet horizontally. Find the angle of elevation of the street? 7. A kite is 120 feet high when 650 feet of string is out. What angle of elevation does the kite make with the ground? 8. From a balloon 2500 ft high, a command post is seen with an angle of depression of ! 8°4 " 5 . How far is it from a point on the ground below the balloon to the command post? ! tan" = 20 100 " = 11.31° ! sin" = 120 650 " = 10.639° ! tan 8°4 " 5 = 2500 a a = 16242.761 ft 3. Right Angle Trigonometry - 12 - www.mastermathmentor.com - Stu Schwartz 9. An observer on a ladder looks at a building 100 ft away, noting that the angle of elevation of the top of the building is ! 28°4 " 0 while the angle of depression of the bottom of the building is ! 7°3 " 0 . How tall is the building? 10. From a plane 2 miles high, the angles of depression to two towns in line with each other and on the same side of the plane are ! 82°5 " 5 and ! 12° " 8 . How far apart are the two towns? 11. Jay and Sam who are both on one side of a hill are staring at the top of that 180 foot tall hill. Jay, who is nearer the hill sees the top of the hill at an angle of elevation of ! 72°5 " 8 while Sam, who is further from the hill sees the top of the hill at an angle of elevation of ! 41°2 " 3 . How far apart are Jay and Sam? 12. Jay and Sam are still staring at the top of the hill. Jay and Sam are 75 feet apart. The angle of elevation of the top of the hill for Jay is ! 22° while the angle of elevation of the top of the hill for Sam is ! 57°. Find the elevation of the hill. ! tan22° = h a + 75 tan57° = h a h = tan22° a + 75( ) h = atan57°" tan22° a + 75( ) = atan57° a tan57°# tan22°( ) = 75tan22° a = 75tan22° tan57°# tan22° = 26.366 h = 26.366tan57° = 40.6 ft. ! tan 28°4 " 0 = a 100 # a = 54.673 tan 7°3 " 0 = b 100 # b = 13.165 building = 67.838 ft ! tan 82°5 " 5 = 2 a # a = 0.249 tan12° " 8 = 2 a+ b # a+ b = 9.303 distance = 9.054 miles ! tan 72°5 " 8 = 180 a # a = 55.146 tan 41°2 " 3 = 180 a + b # a + b = 204.290 distance = 149.144 ft |
Unit 3 right triangle trigonometry answer key
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