Trigonometry Module T09 Solutions of Right Triangles with Practical pplications Copyright This publication The Northern lberta Institute of Technology 2002. ll Rights Reserved. LST REVISED December, 2008
Solution of Right Triangles with Practical pplications Statement of Prerequisite Skills Complete all previous TLM modules before completing this module. Required Supporting Materials ccess to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player. Rationale Why is it important for you to learn this material? Right triangles are often used to measure distances that cannot be measured directly. pplications in surveying, projectiles, navigation, building construction, and many other technologies exist. The skills the student learns in this module will be applied both in the technology of their choice and in the trigonometry modules to follow. Learning Outcome When you complete this module you will be able to Solve right triangles. Learning Objectives 1. Solve a right triangle given an acute angle and a side which is not the hypotenuse. 2. Solve a right triangle given an acute angle and the hypotenuse. 3. Solve a right triangle given the hypotenuse and one other side. 4. Solve a right triangle given two sides neither of which is the hypotenuse. 5. Solve applied right triangle problems. 6. Solve right triangle problems dealing with vectors and navigation. 1
Connection ctivity Can you think of any examples of right triangles occurring in the world around you? The angle a building makes with the ground, the corner of a room, and the intersection of two streets are some examples. There are many instances that right triangles can be imposed on a situation to provide an easy way to measure distances. Consider the diagram below. How could this triangle be used to determine the width of the creek from to C? Creek C 2
OJECTIVE ONE When you complete this objective you will be able to Solve a right triangle given an acute angle and a side which is not the hypotenuse. Exploration ctivity SOLVE RIGHT TRINGLE In general, when you are asked to solve a triangle, this means to find all unknown sides and angles. The following review will be of assistance: 1. The two acute angles in a right triangle are complements of each other. That is, the sum of the two acute angles is 90º. 2. The hypotenuse is greater than either of the other two sides. 3. The greater angle is opposite the greater side. 4. side is opposite an angle if that side is not an arm of the angle. EXMPLE 1 Given ΔC, side b = 30.0, = 25.0º and a right angle at C. Solve ΔC. c a So we are to find all remaining sides and angles, which are: a, c, b C Solve for since: + = 90.0º and = 25º therefore: 25.0º + = 90.0º solving: = 65.0º 3
Solve for side a Here we must find a trig function that involves side "a" and a known side and a known angle. sin = a c doesn't work since we don't know sides a and c. cos = b c doesn't have an a in it. tan = a b should work a tan 25.0º = 30.0 solving: a = (30.0) (tan 25.0º) a = 13.9892 Note: Where possible, always use given values to find missing values. That is, do not use tan = a b because was a calculated angle. Solve for side c y examining the functions, we decide to use the cosine, solving: cos = b c cos 25.0º = 30.0 c c = 30.0 cos25.0 c = 33.1013 lways check your solutions. 1. Use the Pythagorean Theorem to check the sides (13.9892) 2 + (30.0) 2 = (33.1013) 2 2. The sum of the three angles of a triangle is 180º 25º + 65º + 90º = 180º 4
EXMPLE 2 Solve the given triangle ΔPRQ. Given Q = 48.1º and q = 1290 R q = 1290 P We have to find P, r, and p. p 48.1º r Q Solve for P P + Q = 90º P = 90.0º Q P = 90.0º 48.1º P = 41.9º Solve for side r sin Q = q r sin 48.1º = 1290 r 1290 r = sin 48.1 r = 1733.1452 r = 1733 Solve for side p tan Q = q p tan 48.1º = 1290 p 1290 p = tan 48.1 Note: we could have used cosine of Q to solve for p which is cos Q = r p. However, r is a calculated value and we should always use given values where possible. y using the tangent function, we are using only given values. p = 1157.4508 p = 1157 5
Check: Sides Since the answers calculated for sides r and p were rounded to 4 significant digits, the check of the solution for sides will reflect this approximation. r 2 = p 2 + q 2 (1733) 2 = (1157) 2 + (1290) 2 3003289 = 1338649 + 1664100 3003289 = 3002749 (approximate ) This approximation can be improved by storing the original calculator values when calculating r and p and applying then Pythagorean theorem. ngles P + Q + R = 180º 41.9º + 48.1º + 90º = 180º 180º = 180º 6
Experiential ctivity One Solve the following right triangles for the unknown elements. Check each solution. Each is labeled as an C triangle with the right angle at C. Round all sides to 3 significant digits. Round all angles to one decimal place. 1. a = 11.5 = 27.1º 2. a = 15.0 = 74.9º 3. b = 424 = 74.9º Show Me. 4. b = 8.4 = 77.6º 5. a = 11.2 = 82.9º c a b C Experiential ctivity One nswers 1. = 62.9º b = 5.88 c = 12.9 2. = 15.1º b = 55.6 c = 57.6 3. = 15.1º a = 1570 c = 1630 4. = 12.4º a = 1.85 c = 8.60 5. = 7.1º b = 1.40 c = 11.3 7
OJECTIVE TWO When you complete this objective you will be able to Solve a right triangle given an acute angle and the hypotenuse. Exploration ctivity EXMPLE 1 In ΔC, c = 45.3, and = 20.3º. Find b, a, and. a C b c = 45.3 20.3º Solve for = 90.0º = 90.0º 20.3º = 69.7º Check: Sides (15.72) 2 + (42.49) 2 = (45.3) 2 2053 = 2053 Solve for side b cos = b c b = c cos b = 45.3 cos20.3 b = 42.49 ngles + + C = 180º 20.3º + 69.7º + 90º = 180º 180º = 180º Solve for side a sin = a c a = c sin a = 45.3 sin 20.3 a = 15.72 8
Experiential ctivity Two Solve the following right angle triangles. Check your solutions. Solve all for all unknowns in right triangle C where angle C is 90º. Round all sides to 3 significant digits. Round all angles to one decimal place. 1. c = 14.4 = 30.3º 2. c = 14.6 = 36.1º 3. c = 3.30 = 38.7º 4. c = 7.30 = 13.8º 5. c = 22.2 = 58.1º Show Me. c a b C Experiential ctivity Two nswers 1. = 59.7º a = 7.27 b = 12.4 2. = 53.9º a = 11.8 b = 8.60 3. = 51.3º a = 2.58 b = 2.06 4. = 76.2º a = 1.74 b = 7.09 5. = 31.9º b = 18.8 a = 11.7 9
OJECTIVE THREE When you complete this objective you will be able to Solve a right triangle given the hypotenuse and one other side. Exploration ctivity EXMPLE Given ΔC where c = 38.3 and b = 23.1. Find a,, and. c b a C Solve for side a a 2 + b 2 = c 2 a 2 = c 2 b 2 2 2 a = c b 2 2 a = (38.3) (23.1) a = 30.55 Solve for cos = b c cos = 23.1 38.3 1 23.1 = cos 38.3 = 52.91º Solve for. sin = b c sin = 23.1 38.3 1 23.1 = sin 38.3 = 37.09º Check: ngle = 90.0º = 90.0º 52.91º = 37.09º Since we solved for another way, and produced the same result, it should be correct. Sides Since we now have checked our angles and know they are correct, we will use them to solve for a again. a cos = 38.30 a = 38.3 cos a = 38.3 cos 37.09º a = 30.55 10
Experiential ctivity Three Solve all for all unknowns in right triangle C where angle C is 90º. Round all sides to 3 significant digits. Round all angles to one decimal place. 1. c = 12.8 b = 7.80 2. c = 11.5 a = 6.50 3. c = 13.9 a = 8.90 4. c = 14.2 a = 9.20 5. c = 10.2 b = 5.20 Show Me c a b C Experiential ctivity Three nswers 1. a = 10.1 = 37.5º = 52.5º 2. b = 9.49 = 34.4º = 55.6º 3. b = 10.7 = 50.2º = 39.8º 4. b = 10.8 = 40.4º = 49.6º 5. a = 8.77 = 30.7º = 59.3º 11
OJECTIVE FOUR When you complete this objective you will be able to Solve a right triangle given two sides neither of which is the hypotenuse. Exploration ctivity EXMPLE Given ΔC where a = 37.4 and b = 76.0. Find c,, and c a b C Solve for side c Solve for Solve for a2 + b2 = c2 c = a + b 2 2 2 2 c = (37.4) + (76.0) c = 84.70 tan = a b tan = 37.4 76.0 = tan 1 37.4 76.0 = 26.2º tan = b a 76.0 tan = 37.4 = tan 1 76.0 37.4 = 63.8º Check: ngles = 63.80º = 90.0º = 90.0º 26.20º = 63.80º Sides sin = b c b = c sin b c = sin 76.0 c = sin 63.8 c = 84.70 12
Experiential ctivity Four 1. Solve all for all unknowns in right triangle C where angle C is 90º. Round all sides to 3 significant digits. Round all angles to one decimal place. a) a = 27.8 b = 52.8 b) a = 15.2 b = 31.4 c) b = 35.1 a = 14.0 d) b = 25.7 a = 8.7 e) b = 7.7 a = 35.9 c a b C 2. Solve the following triangles, assuming angle C is 90º. Round all sides to 4 significant digits. Round all angles to two decimal places. Side a Side b Side c ngle ngle a) 3.327 69.08º b) 3.578 7.981 c) 7.833 37.61º d) 8.042 53.93º e) 6.731 7.351 f) 4.647 4.830 g) 8.958 79.75º h) 8.852 1.984 i) 1.653 12.83º j) 9.272 65.98º 13
3. Solve the following triangles. Round all sides to 3 significant digits. Round all angles to one decimal place. a) b) d F E a = 7.515 c e = 2.103 f = 3.194 28.28º C b D c) G h = 6.936 I d) L j K i g = 2.489 H k 70.85º l = 2.881 J e) M o = 6.996 N f) R n = 2.883 m q = 9.854 p O P r = 8.697 Q g) U h) X 50.64º s t = 6.258 v = 3.567 w T 48.82º u S W x V i) z 26.1º Y j) y = 1.411 aa cc = 2.317 dd Z DD 54.94º bb CC 14
Experiential ctivity Four nswers 1. a) = 62.2º = 27.8º c = 59.7 b) = 64.2 = 25.8º c = 34.9 c) = 68.3º = 21.7º c = 37.8 d) = 18.7º = 71.3º c = 27.1 e) = 12.1º = 77.9º c = 36.7 2. a) b = 1.27 c = 3.56 = 20.9º b) a = 7.13 = 63.4º = 26.6º c) a = 6.03 c = 9.89 = 52.4º d) a = 4.73 b = 6.50 = 36.1º e) c = 9.97 = 42.5º = 47.5º f) b = 1.32 = 74.2º = 15.8º g) a = 1.62 c = 9.10 = 10.3º h) c = 9.07 = 77.4º = 12.6º i) b = 7.26 c = 7.44 = 77.2º j) a = 3.77 b = 8.47 = 24.0º 3. a) b = 14.0 c = 15.9 = 61.7º b) d = 2.40 E = 41.2º D = 48.8º c) i = 7.37, G = 19.7º H = 70.3º d) k = 8.78 j = 8.30 L = 19.2º e) m = 7.57 N = 22.4º O = 67.6º f) p = 4.63 P = 28.0º R = 62.0º g) s = 4.71 u = 4.12 U = 41.2º h) x = 2.76 w = 2.26 W = 39.4º i) a = 3.21 z = 2.88 Z = 63.9º j) bb = 1.33 dd = 1.90 = 35.1º 15
OJECTIVE FIVE When you complete this objective you will be able to Solve applied right triangle problems. Exploration ctivity PROLEM SOLVING The following definitions will be useful in this objective: 1. ngle of elevation: The angle between the horizontal and the line of sight looking upward at an object. Observation point ngle of elevation 2. ngle of depression: The angle between the horizontal and the line of sight looking downward at an object Observation point θ ngle of depression Line of sight Rules for solving applied right triangle problems: 1. Construct a diagram and label the three sides and angles given in the problem. 2. Select the most convenient trigonometric function(s) to solve for the unknown sides and/or angles. Where possible, choose the function that will derive the unknown parts directly, without having to find other parts first. 3. Round off side lengths and angle measurements after the final calculation as indicated. 16
Example 1: transmission line rises 2.65 m in a run of 36.15 m. What is the angle the line makes with the horizontal? Solution: Transmission line θ 36.15 m 2.65 m To find θ Use the tangent function opposite tan θ = adjacent tan θ = 2.65 36.15 Therefore: θ = tan 1 2.65 36.15 θ = 4.2º 17
Example 2 22.0 m high tree casts a shadow 15.6 m long. What is the angle of elevation of the sun? 22.0 m Solution: opposite tan θ = adjacent 22.0 tan θ = 15.6 θ = tan 1 22.0 15.6 θ = 54.7º 15.6 m Example 3: From the top of a bridge 30.5 m high, the angle of depression of an adjacent building is 15.5º. Determine the horizontal distance to the building from a point directly beneath this observation point. Observation Point 30.5 m ridge 15.5º uilding (10 m) Distance Solution: tan θ = opposite adjacent x tan 74.5º = 20.5 x = 20.5 tan 74.5º x = 73.9 m 18
Experiential ctivity Five Solve the following problems. Round all answers to 3 significant digits 1. tower casts a shadow 143 m long, and at the same time the angle of elevation of the sun is 41.7º. What is the height of the tower? 2. n antenna 120 m tall casts a shadow 82 m long. What is the angle of elevation of the sun? 3. t a horizontal distance of 90 m from the foot of a water tower, the angle of elevation of the top to the tower is found to be 31º. How high is the tower? 4. light pole 14.0 m high is to be guy wired from its middle, and the guy wire is to make an angle of 45º with the ground. llowing 2 m extra for splicing, how long must the guy wire be? 5. n extension ladder 17 m long rests against a vertical wall with its base 6 m from the wall. a) What angle does the ladder make with the ground? b) How far up the wall does the ladder reach? 6. From the top of a hill 60 m high, the angle of depression of a boat is 27.5º. How far away is the boat from the viewing point at the top of the hill? 7. In order to find the width C of a creek, a distance was laid off along the bank, the point being directly opposite a steel pole C on the opposite side, as shown in Figure 1. If the angle C was observed to be 59.4º and was measured to be 47 m, find the width of the creek. Creek C 8. n aircraft flying at 50 m/s climbs a vertical distance of 400 m in one minute. t what angle is the airplane climbing? 9. n airplane traveling horizontally at 250 km/h forms an angle of depression of 20º with a ship sailing below. Six minutes later, the airplane is directly above the ship, which is stationary. Determine how high the airplane is flying. Show Me. 10. n electric light pole is braced with a wire 20 m long. One end of the wire is fastened to the pole 1 m from the top and the other end to a stake in the ground some distance from the foot of the pole. If the wire makes an angle of 66 degrees with the ground, find the height of the pole. 19
11. man observes the angle of elevation of a hovering helicopter is 70.0º when he is standing 750 m from its shadow while the sun is directly overhead. How high is the helicopter flying above the level ground? 12. pilot observes an object on the ground to have an angle of depression of 35º and its distance along the line of sight to be 4000 m. Two minutes later the same object has an angle of depression of 55º and is 450 m away. ssuming that the object was in front of the aircraft in both observations, what is the horizontal distance the aircraft traveled between the observation points? Show Me. 13. helicopter rises vertically from level ground and is observed from a point 150 m from a point directly beneath the helicopter to have an angle of elevation of 35º. Two minutes later, the helicopter has an angle of elevation of 84º. How far did the helicopter travel upward between sightings? 14. room is 5 m long, 4 m wide, and 3 m high. What is the diagonal distance across the room from a lower corner to an upper corner? 15. circular base conical storage tank has a slant height of 5350 mm and is 2540 mm in diameter. Determine the vertical height of the tank. 16. hexagonal head bolt measures 20 mm between parallel faces. Find the distance y across the corners. 20 mm 17. Determine the distances L 1, and L 2 in the metal frame shown below. y L 1 L 2 120.0 15.5º 11 000 cm 20
Experiential ctivity Five nswers 1. 127 m 2. 55.7º 3. 54.1 m 4. 11.9 m 5. a) 69.3º b) 15.9 m 6. 130 m 7. 79.5 m 8. 7.66º 9. 9.10 km 10. 19.3 m 11. 2060 m 12. 3020 m 13. 1320 m 14. 7.07 m 15. 5200 mm 16. 23.1 mm 17. L 1 = 28 900 cm L 2 = 29 400 cm 21
OJECTIVE SIX When you complete this objective you will be able to Solve right triangle problems dealing with vectors and navigation. Exploration ctivity DEFINITIONS OF DIRECTION MESUREMENT The following definitions will be useful in this objective:\ 1. The bearing of a point from a point is defined as the acute angle made by the line drawn from through with the north-south line through. The bearing is then read from the north or south line toward the east or west. N N N N S S S S earing: N 40º E S 40º E S 40º W N 40º W This is a surveyor's way of reading bearings, the first letter is always N or S, and the last letter is always E or W, with an acute angle between them. 22
2. The bearing of from in aeronautics is given as the angle made by the line with the north line through, measured clockwise from north. N N N N S S S S earing: 40º 140º 220º 320º Example 1: motorboat moves in the direction N40ºE from for 4 hr at 20 km/h. How far north and how far east does it travel? Solution: N Solve distance traveled (): Since the problem gave us the velocity of the boat and the time traveled, we need to determine the total distance traveled () first. C d = rt d = (20 km/hr)(4 hr) d = 80 km = 80 km. 40º E Solve for how far East (C): Using sin we have: sin = C C sin 40º = 80 C = 80 sin 40º C = 51.4 km. Solve for how far North (C): Using cos we have cos = C C cos 40º = 80 C = 80 cos 40º C = 61.3 km. The boat travels 61.3 km north and 51.4 km east. 23
Example 2: n airplane is moving at 400 km/h when a bullet is shot with a speed of 2750 km/h at right angles to the path of the airplane. Find the resultant speed and direction of the bullet. In the diagram, vector represents the velocity of the airplane, vector C represents the velocity of the bullet, and vector D represents the resultant velocity of the bullet. D 400 km/hr θ 2750 km/hr C Solution: In the right ΔDC: Solve for the resultant velocity of the bullet: D 2 = C 2 + CD 2 D = 400 + 2750 D = 2779 km/h 2 2 Solve for the direction of the bullet: tan θ = 400 2750 θ = tan 1 400 2750 θ = 8.3º The angle between the path of the bullet and the plane is C. To get this angle we subtract θ from 90 to get 81.7. Thus, the bullet travels at 2779 km/h along a path making an angle of 81.7 with the path of the plane. 24
Experiential ctivity Six Solve the given vector problems. Round all angles to one decimal place. Round all other values to 3 significant digits. In questions 1 to 4, find the x and y components of the indicated vectors by the use of trigonometric ratios. 1. Vector, magnitude 5.50, directed 28º above positive direction of x axis. 2. Vector, magnitude 18.3, directed 16º to right of positive direction of y axis. 3. Vector C, magnitude 800, directed 35º above negative direction of x axis. 4. Vector D, magnitude 0.805, directed 32º above negative direction of x axis. 5. motorboat travels 75 km due east from a port and then turns south for 26 km. What is the displacement of the boat from the port? Show Me. 6. ship which travels 8.0 km/h in still water heads directly across a stream that flows at 4.0 km/h. What is the resultant velocity of the ship? 7. rocket is fired at an angle of 80º with the horizontal, with a speed of 2000 km/h; find the horizontal and vertical components of the velocity. 8. n airplane flies 150 km in the direction 143º. How far south and how far east of the starting point is it? Experiential ctivity Six nswers 1. x = 4.86, y = 2.58 2. x = 5.04, y = 17.6 3. x = 655, y = 459 4. x = 0.683, y = 0.427 5. 79.4 km, S 70.9º E 6. 8.94, 26.6º 7. 347 km/h, 1970 km/h 8. 90.3 km E, 120 km S Practical pplication ctivity Complete the Right Triangles assignment in TLM. Summary This module presented the student with some applications of solving right triangles. 25