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Solution Section 4.1 – Law of Sines Exercise In triangle ABC, B 110 , C 40 and b 18 in . Find the length of side c. Solution A 180 ( B C ) 180 110 40 30 a b sin A sin B c b sin C sin B a 18 sin 30 sin110 c 18 sin 40 sin110 a 18sin 30 sin110 c 18sin 40 sin110 9.6 in 12.3 in Exercise In triangle ABC, A 110.4 , C 21.8 and c 246 in . Find all the missing parts. Solution B 180 A C 180 110.4 21.8 47.8 a c sin A sin C b c sin B sin C a 246 sin110.4 sin 21.8 b 246 47.8 sin 21.8 a 246sin110.4 sin 21.8 b 246sin 47.8 sin 21.8 621 in 491 in 1 Exercise Find the missing parts of triangle ABC if B 34 , C 82 , and a 5.6 cm . Solution A 180 ( B C ) 180 (34 82) 180 116 64 b sin B a c sin A sin C a sin A c a sin C sin A b a sin B sin A 5.6 sin 82 sin 64 5.6 sin 34 sin 64 6.2 cm 3.5 cm Exercise Solve triangle ABC if B = 55°40′, b = 8.94 m, and a = 25.1 m. Solution sin A sin B a b sin A 25.1 sin A sin 55 40 60 8.94 25.1sin 55.667 2.3184 1 8.94 Since sin A > 1 is impossible, no such triangle exists. 2 Exercise Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft Solution sin B sin A b a sin B 24.9sin 55.3 0.89787 22.8 B sin 1 0.89787 B 63.9 and B 180 63.9 116.1 C 180 A B C 180 55.3 63.9 C 180 55.3 116.1 C 60.8 C 8.6 c a sin C sin A 22.8 sin 60.8 sin 55.3 24.2 ft c a sin C sin A 22.8 sin8.6 sin 55.3 4.15 ft 3 Exercise Solve triangle ABC given A = 43.5°, a = 10.7 in., and c = 7.2 in Solution sin C sin A c a sin C 7.2sin 43.5 0.4632 10.7 C sin 1 0.4632 C 27.6 and C 180 27.6 152.4 B 180 A C B 180 43.5 27.6 B 180 43.5 152.4 B 108.9 B 15.9 b a sin B sin A 10.7 sin108.9 sin 43.5 14.7 in Is not possible Exercise If A 26, s 22, and r 19 find x Solution C s rad 22 180 66 19 r D 180 A C 180 26 66 88 rx r sin D sin A 19 x 19sin 88 sin 26 x 19sin 88 19 24 sin 26 4 Exercise A man flying in a hot-air balloon in a straight line at a constant rate of 5 feet per second, while keeping it at a constant altitude. As he approaches the parking lot of a market, he notices that the angle of depression from his balloon to a friend’s car in the parking lot is 35. A minute and a half later, after flying directly over this friend’s car, he looks back to see his friend getting into the car and observes the angle of depression to be 36. At that time, what is the distance between him and his friend? Solution 5 ft/sec car 180 35 36 109 35° 36° d 450 sin 35 sin109 d d 450sin 35 273 ft sin 109 car Exercise A satellite is circling above the earth. When the satellite is directly above point B, angle A is 75.4. If the distance between points B and D on the circumference of the earth is 910 miles and the radius of the earth is 3,960 miles, how far above the earth is the satellite? Solution s r C arc length BD divides by radius C 910 rad 3960 910 180 3960 13.2 D 180 ( A C ) 180 (75.4 13.2) 5 91.4 CA 3960 sin D sin A x 3960 sin 91.4 3960 sin 75.4 x 3960 3960 sin 91.4 sin 75.4 x 3960 sin 91.4 3960 sin 75.4 x 130 mi Exercise The dimensions of a land are given in the figure. Find the area of the property in square feet. Solution A 1 148.7 93.5 sin 91.5 6949.3 ft 2 1 2 A 1 100.5155.4 sin87.2 7799.5 ft 2 2 2 The total area = A A 6949.3 7799.5 14,748.8 ft 2 1 2 6 Exercise A pilot left Fairbanks in a light plane and flew 100 miles toward Fort in still air on a course with bearing of 18. She then flew due east (bearing 90) for some time drop supplies to a snowbound family. After the drop, her course to return to Fairbanks had bearing of 225. What was her maximum distance from Fairbanks? Solution From the triangle ABC: ABC 90 18 108 ACB 360 225 90 45 BAC 90 18 45 27 The length AC is the maximum distance from Fairbanks: b 100 sin108 sin 45 b 100sin108 134.5 miles sin 45 Exercise The angle of elevation of the top of a water tower from point A on the ground is 19.9. From point B, 50.0 feet closer to the tower, the angle of elevation is 21.8. What is the height of the tower? Solution ABC 180 21.8 158.2 ACB 180 19.9 158.2 1.9 Apply the law of sines in triangle ABC: BC 50 BC 50sin19.9 513.3 sin19.9 sin1.9 sin1.9 y Using the right triangle: sin 21.8 BC y 513.3sin 21.8 191 ft 7 Exercise A 40-ft wide house has a roof with a 6-12 pitch (the roof rises 6 ft for a run of 12 ft). The owner plans a 14-ft wide addition that will have a 3-12 pitch to its roof. Find the lengths of AB and BC . Solution 12 tan 3 tan 1 3 14.036 12 12 tan 6 tan 1 6 26.565 12 B 180 180 26.565 153.435 180 14.036 153.435 12.529 AB 14 sin153.435 sin12.529 A AB 14sin153.435 28.9 ft sin12.529 BC 14 sin14.036 sin12.529 BC 14sin14.036 15.7 ft sin12.529 8 14 C 20 Exercise A hill has an angle of inclination of 36. A study completed by a state’s highway commission showed that the placement of a highway requires that 400 ft of the hill, measured horizontally, be removed. The engineers plan to leave a slope alongside the highway with an angle of inclination of 62. Located 750 ft up the hill measured from the base is a tree containing the nest of an endangered hawk. Will this tree be removed in the excavation? Solution ACB 180 62 118 B ABC 180 118 36 26 0f 75 Using the law of sines: x 400 sin118 sin 26 t 36° A x 400sin118 805.7 ft sin 26 Yes, the tree will have to be excavated. 9 26° x 400 ft 118° C 62° Exercise A cruise missile is traveling straight across the desert at 548 mph at an altitude of 1 mile. A gunner spots the missile coming in his direction and fires a projectile at the missile when the angle of elevation of the missile is 35. If the speed of the projectile is 688 mph, then for what angle of elevation of the gun will the projectile hit the missile? Solution ACB 35 BAC 180 35 B After t seconds; 688t mi The cruise missile distance: 548 t miles 3600 The Projectile distance: 688 t miles 3600 Using the law of sines: 548t 688t 3600 3600 sin 145 sin 35 548t mi C 1 mi 35° A 548t sin 35 688t sin 145 3600 3600 548sin 35 688sin 145 sin 145 548 sin 35 688 145 sin 1 548 sin 35 117.8 688 145 sin 1 548 sin 35 688 BAC 180 35 117.8 27.2 The angle of elevation of the projectile must be 35 27.2 62.2 10 D Exercise When the ball is snapped, Smith starts running at a 50 angle to the line of scrimmage. At the moment when Smith is at a 60 angle from Jones, Smith is running at 17 ft/sec and Jones passes the ball at 60 ft/sec to Smith. However, to complete the pass, Jones must lead Smith by the angle . Find (find in his head. Note that can be found without knowing any distances.) Solution ABD 180 60 50 70 A ABC 180 70 110 60° 50° Using the law of sines: 60t ft 17t 60t sin sin110 17 60 sin sin110 sin 17 sin110 60 C sin 1 17 sin110 15.4 60 11 17 t ft B D Exercise A rabbit starts running from point A in a straight line in the direction 30 from the north at 3.5 ft/sec. At the same time a fox starts running in a straight line from a position 30 ft to the west of the rabbit 6.5 ft/sec. The fox chooses his path so that he will catch the rabbit at point C. In how many seconds will the fox catch the rabbit? Solution BAC 90 30 120 C 6.5t 3.5t sin120 sin B 6.5 3.5 sin120 sin B sin B 3.5sin120 6.5 B B sin 1 3.5sin120 28 6.5 C 180 120 28 32 3.5t 30 sin 28 sin 32 t 30sin 28 7.6 sec 3.5sin 32 It will take 7.6 sec. to catch the rabbit. 12 A Solution Section 4.2 - Law of Cosines Exercise If a = 13 yd, b = 14 yd, and c = 15 yd, find the largest angle. Solution 2 2 2 C cos1 a b c 2ab 2 2 2 cos1 13 14 15 2(13)(14) 67 Exercise Solve triangle ABC if b = 63.4 km, and c = 75.2 km, A = 124 40 Solution a 2 b2 c2 2bc cos A 63.4 2 75.2 2 2 63.4 75.2 cos 124 40 15098 a 122.9 km sin B b sin A a 63.4sin124.67 122.9 B sin 1 63.4sin124.67 122.9 25.1 C 180 A B 180 124.67 25.1 30.23 13 60 Exercise Solve triangle ABC if a = 832 ft, b = 623 ft, and c = 345 ft Solution 2 2 2 C cos1 a b c 2ab 2 2 2 cos1 832 623 345 2(832)(623) 22 sin B b sin C c 623sin 22 345 B sin 1 623sin 22 345 43 A 180 22 43 115 Exercise Solve triangle ABC if A 42.3, b 12.9m, and c 15.4m Solution a 2 b2 c2 2bc cos A 12.92 15.42 2(12.9)(15.4)cos 42.3 109.7 a 10.47 m sin B b sin A 12.9sin 42.3 10.47 a B sin 1 12.9sin 42.3 56.0 10.47 C 180 42.3 56 81.7 14 Exercise Solve triangle ABC if a 9.47 ft , b 15.9 ft , and c 21.1 ft Solution 2 2 2 C cos1 a b c 2ab 2 2 2 cos1 9.47 15.9 21.1 2(9.47)(15.9) 109.9 sin B b sin C c 15.9sin109.9 21.1 B sin 1 15.9sin109.9 21.1 25.0 A 180 25 109.9 45.1 Exercise The diagonals of a parallelogram are 24.2 cm and 35.4 cm and intersect at an angle of 65.5. Find the length of the shorter side of the parallelogram Solution x 2 17.7 2 12.12 2(17.7)(12.1) cos 65.5 282.07 x 16.8 cm 15 Exercise An engineer wants to position three pipes at the vertices of a triangle. If the pipes A, B, and C have radii 2 in, 3 in, and 4 in, respectively, then what are the measures of the angles of the triangle ABC? Solution AC 6 AB 5 BC 7 2 2 2 A cos1 5 6 7 78.5 2(5)(6) 6 7 sin B sin 78.5 sin B 6sin 78.5 7 B sin 1 6sin 78.5 57.1 7 C 180 78.5 57.1 44.4 Exercise Andrea and Steve left the airport at the same time. Andrea flew at 180 mph on a course with bearing 80, and Steve flew at 240 mph on a course with bearing 210. How far apart were they after 3 hr.? Solution After 3 hrs., Steve flew: 3(240) = 720 mph Andrea flew: 3(180) = 540 mph 80° 720 210° x2 7202 5402 2 720 540 cos 210 x x 7202 5402 2 720 540 cos 210 1144.5 miles 16 540 Exercise A solar panel with a width of 1.2 m is positioned on a flat roof. What is the angle of elevation of the solar panel? Solution 2 2 2 cos1 1.2 1.2 0.4 19.2 2(1.2)(1.2) or s 0.4 r 1.2 Exercise A submarine sights a moving target at a distance of 820 m. A torpedo is fired 9 ahead of the target, and travels 924 m in a straight line to hit the target. How far has the target moved from the time the torpedo is fired to the time of the hit? Solution x 8202 9242 2 820 924 cos 9 171.7 m 17 Exercise A tunnel is planned through a mountain to connect points A and B on two existing roads. If the angle between the roads at point C is 28, what is the distance from point A to B? Find CBA and CAB to the nearest tenth of a degree. Solution By the cosine law, AB 5.32 7.62 2(5.3)(7.6) cos 28 3.8 2 2 2 CBA cos1 3.8 5.3 7.6 112 2(3.8)(5.3) BAC 180 112 28 40 Exercise A 6-ft antenna is installed at the top of a roof. A guy wire is to be attached to the top of the antenna and to a point 10 ft down the roof. If the angle of elevation of the roof is 28, then what length guy wire is needed? Solution 90 28 62 x 180 62 118 By the cosine law, 28° x 62 1002 2(6)(10) cos118 13.9 ft 18 10 6 Exercise On June 30, 1861, Comet Tebutt, one of the greatest comets, was visible even before sunset. One of the factors that causes a comet to be extra bright is a small scattering angle . When Comet Tebutt was at its brightest, it was 0.133 a.u. from the earth, 0.894 a.u. from the sun, and the earth was 1.017 a.u. from the sun. Find the phase angle and the scattering angle for Comet Tebutt on June 30, 1861. (One astronomical unit (a.u) is the average distance between the earth and the sub.) Solution By the cosine law: 2 2 cos1 0.133 0.894 1.017 2(0.133)(0.891) 2 156 180 180 156 24 19 Exercise A human arm consists of an upper arm of 30 cm and a lower arm of 30 cm. To move the hand to the point (36, 8), the human brain chooses angle and to the nearest tenth of a degree. 1 2 Solution AC 30 8 22 BC 36 AB AC 2 CB2 A 222 362 42.19 By the cosine law: 2 2 2 ADB cos1 AD DB AB 2 AD DB B C 2 2 2 ADB cos1 30 30 42.19 ADB 89.4 2 30 30 180 89.4 90.6 2 D a tan CAB BC 36 AC 22 CAB tan 1 36 58.57 22 sin DAB sin 89.4 sin DAB 30sin 89.4 30 42.19 42.19 DAB sin 1 30sin 89.4 45.32 42.19 CAB DAB 1 20 58.57 45.32 13.25 Exercise A forest ranger is 150 ft above the ground in a fire tower when she spots an angry grizzly bear east of the tower with an angle of depression of 10. Southeast of the tower she spots a hiker with an angle of depression of 15. Find the distance between the hiker and the angry bear. Solution BEC ECD 10 From triangle EBC: tan10 150 BE BE 150 850.692 tan10 C B D 45° F x BAC ACF 15 From triangle ABC: tan15 150 AB AB 150 559.808 tan15 A x AB2 BE 2 2 AB BE cos 45 559.8082 850.6922 2 559.808850.692 cos 45 603 ft 21 Solution Section 4.3 – Vectors and Dot Product Exercise An arrow is shot into the air so that its horizontal velocity is 25 feet per second and its vertical is 15 feet per second. Find the velocity of the arrow. Solution The magnitude of the velocity is: V 25 2 15 2 29 ft / sec tan Vy Vx 15 0.6 25 tan 1 0.6 31 Exercise A boat travels 72 miles on a course of bearing N 27 E and then changes its course to travel 37 miles at N 55 E. How far north and how far east has the boat traveled on this 109-mile trip? Solution Total distance traveled east = U x V x N N = 72 cos 63 37 cos 35 Vy 55 = 63 mi 37 mi V 72 mi Total distance traveled North = U y V y = 72 sin 63 37 sin 35 Uy 27 U 63 = 85 mi Ux 22 E Vx Exercise Let u = –2, 1 and v = 4, 3. Find the following. a) u + v b) –2u c) 4u – 3v Solution a) u + v = –2, 1 + 4, 3 = 2, 4 b) –2u = -2–2, 1 = 4, -2 c) 4u – 3v = 4–2, 1 - 34, 3 = -8, 4 - 12, 9 = -20, -5 Exercise Given: V 13.8, 24.2 , find the magnitudes of the horizontal and vertical vector components of V, V and V , respectively x y Solution V V cos 13.8cos 24.2 12.6 x V V sin 13.8sin 24.2 5.7 y Exercise Find the angle θ between the two vectors u = 3, 4 and v = 2, 1. Solution cos1 u.v u .v (3)(2) (4)(1) cos1 2 2 2 2 3 4 2 1 26.57 23 Exercise A bullet is fired into the air with an initial velocity of 1,800 feet per second at an angle of 60 from the horizontal. Find the magnitude of the horizontal and vertical vector component as of the velocity vector. Solution Vx 1800cos 60 900 ft / sec Vy 1800sin 60 1600 ft / sec Exercise A bullet is fired into the air with an initial velocity of 1,200 feet per second at an angle of 45 from the horizontal. a) Find the magnitude of the horizontal and vertical vector component as of the velocity vector. b) Find the horizontal distance traveled by the bullet in 3 seconds. (Neglect the resistance of air on the bullet). Solution Vx 1200cos 45 848.53 850 ft / sec a) Vy 1200sin 45 850 ft / sec b) x 850 * 3 2,550 ft x V x .t 850*3 2,550 ft Exercise A ship travels 130 km on a bearing of S 42 E. How far east and how far south has it traveled? Solution V east V 130cos 42 96.6 km south 130sin 42 87 km 24 Exercise An arrow is shot into the air with so that its horizontal velocity is 15.0 ft./sec and its vertical velocity is 25.0 ft./sec. Find the velocity of the arrow? Solution V V x2 V y2 V 15 2 25 2 V 29.2 ft / sec tan 25 15 15 tan 1 25 60 Exercise A plane travels 170 miles on a bearing of N 18 E and then changes its course to N 49 E and travels another 120 miles. Find the total distance traveled north and the total distance traveled east. Solution Total distance traveled east = U x V x N N = 170 cos 72 120 cos 41 Vy = 143.1 mi 49 120 V = 140 mi - East 170 Total distance traveled North = U y V y Uy = 170 sin 72 120 sin 41 18 U 72 E = 240 mi - South Ux 25 Vx Exercise A boat is crossing a river that run due north. The boat is pointed due east and is moving through the water at 12 miles per hour. If the current of the river is a constant 5.1 miles per hour, find the actual course of the boat through the water to two significant digits. Solution tan 12 5.1 2.3529 N tan 1 2.3529 67 V 5.1 mi/hr E 12 mi/hr V 12 2 5.12 (using Pythagorean Theorem) 13 Exercise Two forces of 15 and 22 Newtons act on a point in the plane. (A Newton is a unit of force that equals .225 lb.) If the angle between the forces is 100°, find the magnitude of the resultant vector. Solution P Q R S 360 2P 2Q 360 2Q 360 2(100) 2Q 160 Q 80 V 2 152 222 2(15)(22) cos80 594.39 V 24.4 N 26 Exercise Find the magnitude of the equilibrant of forces of 48 N and 60 N acting on a point A, if the angle between the forces is 50°. Then find the angle between the equilibrant and the 48-newton force. Solution The equilibrant is –v. 2 A 2B 360 2B 360 2(50) 2B 260 B 130 V 2 482 602 2(48)(60) cos130 V 98 N sin BAC sin130 sin BAC 60sin130 60 98 98 BAC sin 1 60sin130 28 98 180 28 152 Exercise Find the force required to keep a 50-lb wagon from sliding down a ramp inclined at 20° to the horizontal. (Assume there is no friction.) Solution Vector BC represents the force with which the weight pushes against the ramp. Vector BF represents the force that would pull the weight up the ramp. The vertical force BA represents the force of gravity. sin 20 AC 50 AC 50sin 20 17 A force of approximately 17 lbs. will keep the wagon from sliding down the ramp. 27 Exercise A force of 16.0 lbs. is required to hold a 40.0 lbs. lawn mower on an incline. What angle does the incline make with the horizontal? Solution Vector BE represents the force required to hold the mower on the incline. sin B 16 40 B sin 1 16 23.6 40 The hill makes an angle of about 23.6° with the horizontal. Exercise A ship leaves port on a bearing of 28.0° and travels 8.20 mi. The ship then turns due east and travels 4.30 mi. How far is the ship from port? What is its bearing from port? Solution NAP 90 28 62 PAE 180 62 118 PE 8.202 4.302 2 8.2 4.3 cos118 2 120.6 PE 10.9 The ship is about 10.9 miles from port. The bearing is given by: sin APE sin118 4.3 10.9 sin APE 4.3sin118 10.9 APE sin 1 4.3sin118 20.4 10.9 The bearing: 28 20.4 48.4 28 Exercise Two prospectors are pulling on ropes attached around the neck of a donkey that does not want to move. One prospector pulls with a force of 55 lb, and the other pulls with a force of 75 lb. If the angle between the ropes is 25, then how much force must the donkey use in order to stay put? (The donkey knows the proper direction in which to apply his force.) Solution OCB OAB 1 360 2(25) 155 2 OB OC 2 CB2 2(OC )(OB) cos OCB 75 O 752 552 2(75)(55) cos 155 55 B ° 25 C By the cosine law, the magnitude of the resultant force is: 55 A 127 lb So the donkey must pull a force of 127 pounds in the direction opposite that of the resultant’s. Exercise A solid steel ball is placed on a 10 incline. If a force of 3.2 lb in the direction of the incline is required to keep the ball in place, then what is the weight of the ball? w 10 ° ° 10 Solution 3.2 sin10 f w 3.2 18.4 lb w sin10 29 Exercise Find the amount of force required for a winch to pull a 3000-lb car up a ramp that is inclined at 20. sin 20 3000 20 ° ° 20 Solution f f f 3000sin 20 1026.1 lb 3000 Exercise If the amount of force required to push a block of ice up an ice-covered driveway that is inclined at 25 is 100lb, then what is the weight of the block? ° 25 w 25 ° Solution 100 sin 25 100 w 100 236.6 lb w sin 25 Exercise If superman exerts 1000 lb of force to prevent a 5000-lb boulder from rolling down a hill and crushing a bus full of children, then what is the angle of inclination of the hill? w Solution 1000 sin 1000 sin 1 1000 11.5 5000 5000 30 Exercise If Sisyphus exerts a 500-lb force in rolling his 4000-lb spherical boulder uphill, then what is the angle of inclination of the hill? Solution sin 500 sin 1 500 7.2 4000 4000 Exercise A plane is headed due east with an air speed of 240 mph. The wind is from the north at 30 mph. Find the bearing for the course and the ground speed of the plane. Solution 240 30 x The bearing is: tan 1 30 7.1 240 The ground speed can be found by using Pythagorean Theorem: x 2402 302 241.9 mph Exercise A plane is headed due west with an air speed of 300 mph. The wind is from the north at 80 mph. Find the bearing for the course and the ground speed of the plane. Solution 300 80 x tan 1 80 14.9 300 The bearing is: 270 14.9 255.1 The ground speed can be found by using Pythagorean Theorem: x 3002 802 310.5 mph 31 Exercise An ultralight is flying northeast at 50 mph. The wind is from the north at 20 mph. Find the bearing for the course and the ground speed of the ultralight. Solution 45° 20 50 x Using the cosine law, the ground speed is: x 502 202 2(50)(20) cos 45 38.5 mph sin sin 45 sin 20sin 45 20 38.5 38.5 sin 1 20sin 45 21.5 38.5 The bearing is: 45 21.5 66.5 Exercise A superlight is flying northwest at 75 mph. The wind is from the south at 40 mph. Find the bearing for the course and the ground speed of the superlight. Solution x 5° 45° 13 40 75 Using the cosine law, the ground speed is: x 752 402 2(75)(40) cos135 107.1 mph sin sin135 sin 40sin135 40 107.1 107.1 sin 1 40sin135 15.3 107.1 The bearing is: 315 15.3 330.3 32 Exercise An airplane is heading on a bearing of 102 with an air speed of 480 mph. If the wind is out of the northeast (bearing 225) at 58 mph, then what are the bearing of the course and the ground speed of the airplane? Solution x: the airplane’s vector course and ground speed 225 180 45 2° 10 From the parallelogram (from the vectors): 12° 1 90 12 45 57 By the cosine law: x 4802 582 2(480)(58) cos 57 58 2 x 480 451 mph sin sin 57 58 451 sin 58sin 57 451 sin 1 58sin 57 6.2 451 The bearing of the plane: 102 6.2 108.2 Exercise In Roman mythology, Sisyphus revealed a secret of Zeus and thus incurred the god’s wrath. As punishment, Zeus banished him to Hades, where he was doomed for eternity to roll a rock uphill, only to have it roll back on him. If Sisyphus stands in front of a 4000-lb spherical rock on a 20 incline, then what force applied in the direction of the incline would keep the rock from rolling down the incline? 33 Solution f f cos 70 4000 70 4000 20 f 4000 cos 70 1368 lb Exercise A trigonometry student wants to cross a river that is 0.2 mi wide and has a current of 1 mph. The boat goes 3 mph in still water. a) Write the distance the boats travels as a function of the angle . b) Write the actual speed of the boat as a function of and . c) Write the time for the trip as a function of . Find the angle for which the student will cross the river in the shortest amount of time. Solution 1 mph Vy 3 mph Vx a) The vector in still water is given by: Vx Vs cos 90 3sin Vy 3cos V 3sin i 3cos j The vector of the current: V c 1i Therefore, the actual direction and speed is determined by the vector: v 3sin 1 i 3cos j 34 The number t of hours it takes the boat to cross the river: t d 0.2 v 3cos y The distance of the boat travels: d t v 0.2 3cos 3sin 12 3cos 2 0.2 3sin 12 3cos 2 3cos 2 3cos 2 0.2 1 3sin 3cos 0.2 tan 2 1 2 1 0.2 tan 2 1 v tan x 3sin 1 vy 3cos sec2 tan 2 1 0.2 sec2 0.2 sec b) The actual speed d t 0.2 sec 0.2 3cos 3cos sec c) The shortest time for the boat to cross river is straight ahead which = 0 t 0.2 0.2 2 1 sec 3cos 0 3 30 15 35 Exercise A male uses three elephants to pull a very large log out of the jungle. The papa elephant pulls with 800 lb. of force, the mama elephant pulls with 500 lb. of force, and the baby elephant pulls with 200 lb. force. The angles between the forces are shown in the figure. What is the magnitude of the resultant of all three forces? If mama is pulling due east, then in what direction will the log move? Solution v p 800 cos 30i 800sin 30 j vm 30° vm 500i 20° vb 200cos 20i 200sin 20 j vb The total force is determined by: F v p vm v b 800 cos 30i 800sin 30 j 500i 200 cos 20i 200sin 20 j 800 cos 30 500 200 cos 20 i 800sin 30 200sin 20 j 1380.76i 331.60 j The magnitude of the resultant: F 1380.762 331.602 1420 lbs The direction: tan 1 331.6 13.5 1380.76 800 lb vp E13.5 N 36 500 lb 200 lb Section 4.4 – Trigonometric Form of Complex Numbers Solution Exercise Write 3 i in trigonometric form. (Use radian measure) Solution x 3 3 i y 1 r 3 tan 2 12 2 y 1 3 x 3 3 The reference angle for is and the angle is in quadrant II. 6 Therefore, 5 6 6 3 i 2 cis 5 6 Exercise Write 3 4i in trigonometric form. Solution 2 2 r 3 (4) 5 3 4i 1 4 53 tan 3 The angle is in quadrant II; therefore, 180 53 127 3 4i 5 cis127 37 Exercise Write 21 20i in trigonometric form. Solution 2 2 r 21 20 29 21 20i tan 1 20 43.6 21 The angle is in quadrant III; therefore, 180 43.6 223.6 21 20i 29 cis223.6 Exercise Write 11 2i in trigonometric form. Solution 2 2 r 11 2 125 5 5 11 2i 1 2 10.3 tan 11 The angle is in quadrant I; therefore, 10.3 11 2i 5 5 cis10.3 Exercise Write 4 cos30 i sin 30 in standard form. Solution 4 cos30 i sin 30 4 3 i 1 2 2 2 3 2i 38 Exercise Write 2 cis 7 in standard form. 4 Solution 2 cis 7 2 cos 7 i sin 7 4 4 4 2 1 i 1 2 2 1 i Exercise Find the quotient 20cis 75 . Write the result in rectangular form. 4cis 40 Solution 20cis 75 20 cis 75 40 4 4cis 40 5cis 35 5 cos35 i sin 35 4.1 2.87i 39 Exercise Divide z 1 i 3 by z 3 i . Write the result in rectangular form. 1 2 Solution z1 1 i 3 z2 3 i or 2 2 r 1 3 1 i 3 : tan 1 3 1 3 2 2 r 3 1 3 i: tan 1 1 3 6 2cis 3 1 z 2 2cis 6 z 1 i 3 3 i 3 i 3 i 3 6 2 cis 2 6 cis 6 cos i sin 6 6 2 3 i 3i 3 i 3 1 2 cis 2 2 3 2i 4 2 3 2i 4 4 3i 2 2 40 Section 4.5 – Polar Coordinates Solution Exercise Convert to rectangular coordinates 2, 60 Solution 2, 60 2 cos 60 i sin 60 2 1 i 3 2 2 1 i 3 Exercise Convert to rectangular coordinates 2, 225 Solution 2, 225 2 cos(225) i sin(225) 2 1 i 1 2 2 1 i Exercise Convert to rectangular coordinates 4 3, 6 Solution 4 3, 6 4 3 cos 6 i sin 6 4 3 3 i 1 2 2 62 3 41 Exercise 3, Convert to polar coordinates 3 r 0 0 360 Solution 2 2 r 3 3 3 2 3, 3 tan 1 3 tan 1 1 45 3 The angle is in quadrant III; therefore, 180 45 225 3, 3 3 2, 225 Exercise Convert to polar coordinates 2, 2 3 r 0 0 360 Solution 2 r 22 2 3 4 2, 2 3 tan 1 2 3 tan 1 3 60 2 The angle is in quadrant IV; therefore, 360 60 300 2, 2 3 4, 300 Exercise Convert to polar coordinates 2, 0 r 0 0 2 Solution 2 2 r 2 0 2 2, 0 tan 1 0 0 2 2, 0 2, 42 Exercise 1, Convert to polar coordinates 3 r 0 0 2 Solution 2 2 r 1 3 2 1, 3 tan 1 3 1 3 The angle is in quadrant III; therefore, 4 3 3 1, 3 2, 43 Exercise Write the equation in rectangular coordinates r2 4 Solution r2 4 x2 y 2 4 Exercise Write the equation in rectangular coordinates r 6cos Solution r 6cos r 6x r r 2 6x x2 y 2 6 x 43 Exercise Write the equation in rectangular coordinates r 2 4cos 2 Solution r 2 4 cos2 sin 2 cos x r sin y r 2 y 2 4 x r r 2 y2 4 x r2 r2 x2 y 2 4 r2 r 4 4 x2 y 2 x2 y 2 4 r 2 x2 y 2 4 x2 4 y 2 Exercise Write the equation in rectangular coordinates r cos sin 2 Solution r cos sin 2 cos x r y r x 2 r r x y r 2 r x y 2 44 sin y r Exercise Write the equation in polar coordinates x y 5 Solution r cos r sin 5 x r cos y r sin r cos sin 5 r 5 cos sin Exercise Write the equation in polar coordinates x2 y 2 9 Solution x2 y 2 9 r 2 x2 y 2 r2 9 Exercise Write the equation in polar coordinates x2 y 2 4 x Solution r 2 4r cos r 2 4r cos r r r 4cos Exercise Write the equation in polar coordinates y x Solution y x x r cos r sin r cos sin cos 45 y r sin Section 4.6 - De Moivre’s Theorem Solution Exercise Find 1 i and express the result in rectangular form. 8 Solution 2 2 r 1 1 2 1 i 1 tan 1 4 1 i 2cis 4 1 i 8 2cis 4 2 8 8 cis 8 4 16cis2 16 cos 2 i sin 2 16 1 i0 16 Exercise Find 1 i 10 and express the result in rectangular form. Solution 1 i 10 2cis 4 2 10 cis 10 4 10 32cis 5 2 32cis 2 32 cos i sin 2 2 32 0 i 32i 46 Exercise Find fifth roots of z 1 i 3 and express the result in rectangular form. Solution 2 2 r 1 3 2 1 i 3 tan 1 3 60 1 1 i 3 1/5 2cis60 1/5 5 2 cis 60 360k 5 5 5 2 cis 12 72k If k 0 5 2 cis 12 720 5 2 cis12 If k 1 5 2 cis 12 72. 1 5 2 cis84 If k 2 5 2 cis 12 72. 2 5 2 cis156 If k 3 5 2 cis 12 72. 3 5 2 cis228 If k 4 5 2 cis 12 72. 4 5 2 cis300 Exercise Find the fourth roots of z 16cis60 Solution 4 z 4 16 cis 60 360 k 4 4 2cis 15 90k If k 0 2 cis 15 90 0 2cis15 If k 1 2 cis 15 90 1 2cis105 If k 2 2 cis 15 90 2 2cis195 If k 3 2 cis 15 90 3 2cis 285 47 Exercise Find the cube roots of 27. Solution 3 27 27cis0 1/3 3 3 27 cis 0 360 k 3 3 cis 0 120k If k 0 z 3 cis 0 120 0 2cis0 If k 1 z 3 cis 0 120 1 2cis120 If k 2 z 3 cis 0 120 2 2cis 240 Exercise Find all complex number solutions of x3 1 0 . Solution x3 1 0 x3 1 2 2 r (1) 0 1 1 1 0 tan 1 x3 1 1 cis x 1 cis 1/3 1 1/3 3 3 cis 2 k cis 2 k 3 3 3 3 3 If k 1 x cis 2 1 cis 3 cis 3 3 3 If k 0 x cis 2 0 cis 3 3 3 2 2 x cos i sin 1 x cos 5 i sin 5 1 i 3 If k 2 x cis 2 2 cis 5 3 x cos i sin 1 i 3 3 3 48 3 2 2 Exercise Find 2cis30 5 Solution 2cis305 25 cis 5(30) 32 cos150 i sin150 32 3 1 i 2 2 16 3 16i 49
