( )( ) Section 7.3 The Law of Cosines. 42. If A = 18 and B = 36 we have 5sin18 sin A, B, and c R R. sin (

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1 Section 7.3 The Law of Cosines If A 8 and B 36 we have 5sin8 sin36 A R.57R sin(8 + 36) (b) law of sines 5. A, B, and c 43. (a) 0.4 in. in in. 3 (b) 44. red sin8 sin 36 A in. sin ( ) Section 7.3 The Law of Cosines. a, b, and C (a) ASA (b) law of sines 6. a, c, and A (a) SAS (b) law of cosines. A, C, and c (a) SSA (b) law of sines 7. a, b, and c (a) SAA (b) law of sines 3. a, b, and A (a) SSS (b) law of cosines 8. b, c, and A (a) SSA (b) law of sines 4. a, B, and C (a) ASA (a) SAS (b) law of cosines 9. a ( ) ()( ) cos a a cos a 49 7 Copyright 03 Pearson Education, Inc.

2 354 Chapter 7 Applications of Trigonometry and Vectors.. b + c a cosθ bc ()() θ 0 30 ( ) b + c a + 3 cosθ bc 3 ()( ) θ A, b 5, c 3 Start by finding a with the law of cosines. a b + c bccosa a ( 5)( 3) cos 49.5 a Of the remaining angles B and C, C must be smaller since it is opposite the shorter of the two sides b and c. Therefore, C cannot be obtuse. sin A sin C sin sin C a C sin sin C C Thus, B A 6, b 4, c 6 Start by finding a with the law of cosines. a b + c bccosa a ( 4)( 6) cos a Of the remaining angles B and C, B must be smaller since it is opposite the shorter of the two sides b and c. Therefore, B cannot be obtuse. sin A sin B sin 6 sin B a b sin6 sin B B Thus, C a, b 0, c 0 We can use the law of cosines to solve for any angle of the triangle. Since b and c have the same measure, so do B and C since this would be an isosceles triangle. If we solve for B, we obtain b a + c accosb a + c b cos B ac cos B ( )( 0) 40 5 B 53. Therefore, C B 53. and A If we solve for A directly, however, we obtain a b + c bccosa b + c a cos A bc cos A 0 ( )( 0) 00 5 A 73.7 The angles may not sum to 80 due to rounding. 6. a 4, b 0, c 8 We can use the law of cosines to solve for any angle of the triangle. We solve for B, the largest angle. We will know that B is obtuse if cos B < 0. b a + c accosb a + c b cos B ac cos B 4 ( )( 8) 64 6 B 08. We can now use the law of sines or the law of cosines to solve for either A or C. Using the law of cosines to solve for A, we have a b + c bccosa b + c a cos A bc cos A A Thus, C B 55, a 90, c 00 Start by finding b with the law of cosines. b a + c accosb b ( 90)( 00) cos b 88.8 (will be rounded as 88.) Of the remaining angles A and C, A must be smaller since it is opposite the shorter of the two sides a and c. Therefore, A cannot be obtuse. sin A sinb sin A sin55 a b sin 55 sin A A Thus, C Copyright 03 Pearson Education, Inc.

3 Section 7.3 The Law of Cosines a 5, b 7, c 9 We can use the law of cosines to solve for any angle of the triangle. We solve for C, the largest angle. We will know that C is obtuse if cos C < 0. c a + b abcosc a + b c cos C ab cos C C We can now use the law of sines or the law of cosines to solve for either A or B. Using the law of sines to solve for A, we have sin A sin C sin A sin 95.7 a C 5 9 5sin95.7 sin A A Thus, B A 4.4º, b.78 yd, c 3.9 yd First find a. a b + c bccosa a (.78)( 3.9) cos a.60 yd Find B next, since angle B is smaller than angle C (because b < c), and thus angle B must be acute. sin B sin A sin B sin 4.4 b a sin 4.4 sin B B 45. Finally, C C 8.3, b 5.7 in., a 4. in. First find c. c a + b abcosc c ( 4.)( 5.7) cos c.83 in. Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sinc sin A sin8.3 a C sin 8.3 sin A A 44.9 Finally, B C 45.6, b 8.94 m, a 7.3 m First find c. c a + b abcosc c ( 7.3)( 8.94) cos c 6.46 m Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sin C sin A sin 45.6 a c sin 45.6 sin A A 53. Finally, B A 67.3, b 37.9 km, c 40.8 km First find a. a b + c bccosa a ( 37.9)( 40.8) cos a 43.7 km Find B next, since angle B is smaller than angle C (because b < c), and thus angle B must be acute. sin B sin A sin B sin 67.3 b a sin 67.3 sin B B 53. Finally, C a 9.3 cm, b 5.7 cm, c 8. cm We can use the law of cosines to solve for any of angle of the triangle. We solve for A, the largest angle. We will know that A is obtuse if cos A < 0. a b + c bccosa cos A ( )( 8.) A 8 Find B next, since angle B is smaller than angle C (because b < c), and thus angle B must sin B sin A sin B sin 8 b a sin 8 sin B B Thus, C Copyright 03 Pearson Education, Inc.

4 356 Chapter 7 Applications of Trigonometry and Vectors 4. a 8 ft, b 47 ft, c 58 ft Angle C is the largest, so find it first. c a + b abcosc cos C (8)(47) C 98 Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sin C sin A sin 98 a c sin 98 sin A A 9 58 Thus, B a 4.9 m, b 37.6 m, c 6.7 m Angle C is the largest, so find it first. c a + b abcosc cos C ( 4.9)( 37.6) C Find B next, since angle B is smaller than angle A (because b < a), and thus angle B must be acute. sin B sin C sin B sin0. b c sin0. sin B B Thus, A a 89 yd, b 4 yd, c 35 yd Angle C is the largest, so find it first. c a + b abcosc cos C ( 89)( 4) C Find B next, since angle B is smaller than angle A (because b < a), and thus angle B must be acute. sin B sin C sin B sin07 0 b c sin07.3 sin B B Thus, A a 965 ft, b 876 ft, c 40 ft Angle C is the largest, so find it first. c a + b abcosc cos C ( 965)( 876) C 84.5 or Find B next, since angle B is smaller than angle A (because b < a), and thus angle B must be acute. sinb sinc sinb sin84 30 b c sin sin B B 44.7 or Thus, A a 34 m, b 4 m, c 98 m Angle B is the largest, so find it first. b a + c accosb cos B ( 34)( 98) B Find C next, since angle C is smaller than angle A (because c < a), and thus angle B must sin C sin B sin C sin 85. c b sin 85. sin C C Thus, A A b 43 cm, c 89.6 cm First find a. a b + c bccosa a ( 43)( 89.6) cos ,3.5 a 56 cm Find C next, since angle C is smaller than angle B (because c < a), and thus angle C must sin C sin A sin C sin c a sin sin C C Finally, B Copyright 03 Pearson Education, Inc.

5 Section 7.3 The Law of Cosines C 7 40, a 37 ft, b 5 ft First find c. c a + b abcosc c ( 37)( 5) cos 7 40,03.55 c 348 ft Find B next, since angle B is smaller than angle A (because b < a), and thus angle B must sin B sin C sin B sin 7 40 b C sin 7 40 sin B B Finally, A B 74.8, a 8.9 in., c 6.43 in. First find b. b a + c accosb b ( 8.9)( 6.43) cos b 9.53 in. Find C next, since angle C is smaller than angle A (because c < a), and thus angle C must sin C sin B sin C sin 74.8 c b sin 74.8 sin C C 40.6 Thus, A C 59.7, a 3.73 mi, b 4.70 mi First find c. c a + b abcosc c ( 3.73)( 4.7) cos c 4.8 mi Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sin C sin A sin 59.7 a c sin 59.7 sin A A 48.8 Thus, B A.8, b 6.8 m, c. m First find a. a b + c bccosa a ( 6.8)(.) cos a 5.7 m Find B next, since angle B is smaller than angle C (because b < c), and thus angle B must sinb sin A sinb sin.8 b a sin.8 sin B B.6 Finally, C B 68., a 5. cm, c 9. cm First find b. b a + c accosb b cos b 34. cm Find A next, since angle A is smaller than angle C (because a < c), and thus angle A must be acute. sin A sin B sin A sin68. a b sin68. sin A A 5. Thus, C a 3.0 ft, b 5.0 ft, c 6.0 ft Angle C is the largest, so find it first. c a + b abcosc cos C 3.0 ( )( 5.0) C 94 Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sin C sin A sin 94 a c 3 6 3sin94 sin A A 30 6 Thus, B Copyright 03 Pearson Education, Inc.

6 358 Chapter 7 Applications of Trigonometry and Vectors 36. a 4.0 ft, b 5.0 ft, c 8.0 ft Angle C is the largest, so find it first. c a + b abcosc cos C 4.0 ( )( 5.0) C 5 Find A next, since angle A is smaller than angle B (because a < b), and thus angle A must sin A sin C sin A sin5 a c 4 8 4sin5 sin A A 4 8 Thus, B There are three ways to apply the law of cosines when a 3, b 4, and c 0. Solving for A: a b + c bccosa cos A Solving for B: b a + c accosb cos B ()( ) Solving for C: c a + b abcosc cos C ()() Since the cosine of any angle of a triangle must be between and, a triangle cannot have sides 3, 4, and Answers will vary. 39. A and B are on opposite sides of False River. We must find AB, or c, in the following triangle. 40. X and Y are on opposite sides of a ravine. We must find XY. XY XZ + YZ XZ YZ cosz XY cos XY 95.3 m 4. Using the law of cosines we can solve for the measure of angle A cos A 5.9 ( )( 3.5) A Find the diagonals, BD and AC, of the following parallelogram. BD AB + AD AB AD cosa BD cos58 BD BD 5. cm AC AB + BC AB BC cosb AC cos AC AC 8.8 cm The lengths of the diagonals are 5. cm and 8.8 cm. c a + b abcosc c cos 46.3 c 65,98.3 c 57 The length of AB is 57 m. Copyright 03 Pearson Education, Inc.

7 Section 7.3 The Law of Cosines Find AC, or b, in the following triangle. m Angles and are alternate interior angles formed when parallel lines (the north lines) are cut by a transversal, line BC, so m m 5 0. m ABC 90 m b a + c accosb b ( 359)( 450) cos ,06 b 8 km C is about 8 km from A. 44. Sketch a triangle showing the situation as follows. x ( 46.75)( 46.75) cos5 9, x The distance between the ends of the two equal sides is feet. 46. Let B be the harbor, AB is the course of one ship, BC is the course of the other ship. Thus, b the distance between the ships. b b cos b 745 mi 47. Sketch a triangle showing the situation as follows. The angle marked 30 is the corresponding angle that measures The angle marked 55 is the supplement of the 5 angle. Finally, the 75 angle is marked as such because We can use the law of cosines to solve for the side of the triangle marked d. d ( 80)( 00) cos 75 33,08 d 8.9 The distance is approximately 80 mi. (rounded to two significant digits) 45. Let x the distance between the ends of the two equal sides. Use the law of cosines to find x. m A m C m B 80 A C Since we have only one side of a triangle, use the law of sines to find BC a. a b a 5. sin A sin B sin sin sin a 0.8 sin The distance between the ship and the rock is about 0.8 miles. 48. Let d the distance between the submarine and the battleship. α β since the angle of depression to the battleship equals the angle of elevation from the battleship (They are alternate interior angles.) (continued on next page) Copyright 03 Pearson Education, Inc.

8 360 Chapter 7 Applications of Trigonometry and Vectors (continued) θ Since we have only one side of a triangle, use the law of sines to find d. d 50 sin 6 40 sin sin 6 40 d 45.9 sin55 50 The distance between the submarine and the battleship is 450 ft. (rounded to three significant digits) 49. Use the law of cosines to find the angle, θ. 5. AB is the horizontal distance between points A and B. Using the laws of cosines, we have AB (0)(0)cos8 AB 33. AB 8 ft 53. Let A home plate; B first base; C second base; D third base; P pitcher s rubber. Draw AC through P, draw PB and PD cosθ 0 ( )( 6) θ cos β 0 ( )( 3) β Let A the angle between the beam and the 45-ft cable. Let B the angle between the beam and the 60-ft cable cos A (45)(90) A cos B 90 ( )( 60) 0, B 6 In triangle ABC, m B 90, and m A m C 45. AC and PC ft In triangle APB, m A 45. PB AP + AB AP AB cosa PB cos 45 PB PB 63.7 ft Since triangles APB and APD are congruent, PB PD 63.7 ft. The distance to second base is 66.8 ft and the distance to both first and third base is 63.7 ft. 54. Let A home plate; B first base; C second base; D third base; P pitcher s rubber. Draw AC through P, draw PB and PD. (continued on next page) Copyright 03 Pearson Education, Inc.

9 Section 7.3 The Law of Cosines 36 (continued) In triangle ABC, m B 90, and m A m C 45. AC and PC ft In triangle APB, m A 45. PB AP + AB AP AB cosa PB cos 45 PB 8.77 PB 4.6 ft Since triangles APB and APD are congruent, PB PD 4.6 ft. The distance to second base is 38.9 ft and the distance to both first and third base is 4.6 ft. 55. Find the distance of the ship from point A. 57. Let c the length of the property line that cannot be directly measured. Using the law of cosines, we have c ( 4.0)( 3.0) cos c 5.5 ft (rounded to three significant digits) The length of the property line is approximately feet 58. Let A the point where the ship changes to a bearing of 6 ; C the point where it changes to a bearing of 5. m m m + m Use the law of cosines to find v. v cos v 39. km 56. Let A the man s location; B the factory whistle heard at 3 sec after 5:00; C the factory whistle heard at 6 sec after 5:00. Since sound travels at 344 m per sec and the man hears the whistles in 3 sec and 6 sec, the factories are c m and b m from the man. Using the law of cosines we have a ( 03)( 064) cos 4.,69,.3 a 47.8 The factories are about 473 m apart. (rounded to four significant digits) m CAB m FCA 80 m DAC m ACB 360 m FCB m FCA m CBA 80 m ACB m CAB Since we have only one side of a triangle, use the law of sines to find CB. CB 50 50sin 8 CB 9.4 sin 8 sin7 sin7 and AC 50 50sin 5 AC 6.5 sin 5 sin7 sin7 The ship traveled mi. To avoid the iceberg, the ship had to travel mi farther. Copyright 03 Pearson Education, Inc.

10 36 Chapter 7 Applications of Trigonometry and Vectors 59. Let c the length of the tunnel. Use the law of cosines to find c. c ( 3800)( 900) cos0 30, 388,4 c 55.5 The tunnel is 5500 meters long. (rounded to two significant digits) 60. Let x the distance from the plane to the mountain when the second bearing is taken. θ Since we have only one side of a triangle, use the law of sines to find x. x 7.9 sin 4. sin sin 4. x 5.99 sin47.3 The plane is about 5.99 km from the mountain. (rounded to three significant digits) 6. Let a be the length of the segment from (0, 0) to (6, 8). Use the distance formula. a Let b be the length of the segment from (0, 0) to (4, 3). b Let c be the length of the segment from (4, 3) to (6, 8). c a + b c cosθ ab ( ) cosθ ( 0)( 5) θ Let a be the length of the segment from (0, 0) to (8, 6). Use the distance formula. a Let b be the length of the segment from (0, 0) to (, 5). b Let c be the length of the segment from (8, 6) to (, 5). ( ) c a + b c cosθ ab ( ) cosθ ( )( 3) θ Using A bh A ( 6 )( 3 3 ) To use Heron s Formula, first find the semiperimeter, s ( a+ b+ c) ( ) Now find the area of the triangle Both formulas give the same area. 64. Using A bh A ( 0 )( 3 3 ) To use Heron s Formula, first find the semiperimeter, s ( a+ b+ c) ( ) Now find the area of the triangle Both formulas give the same result. Copyright 03 Pearson Education, Inc.

11 Section 7.3 The Law of Cosines a m, b 6 m, c 5 m s a+ b+ c m (rounded to two significant digits) 66. a in., b 45 in., c 3 in. s a+ b+ c in. (rounded to two significant digits) 67. a 54 cm, b 79 cm, c 83 cm s a+ b+ c , 600 cm (rounded to three significant digits) 68. a 5.4 yd, b 38. yd, c 9.8 yd s a+ b+ c ( )( ) ( ) yd (rounded to three significant digits) 69. a 76.3 ft, b 09 ft, c 98.8 ft s a+ b+ c ( )( ) ( ) ft (rounded to three significant digits) 70. a 5.89 m, b.74 m, c 0.9 m s a+ b+ c ( ) ( )( ) s s a s b s c m (rounded to four significant digits) 7. Perimeter: feet, so the semiperimeter is 36 8 feet. Use Heron s Formula to find the area ft Since the perimeter and area both equal 36, the triangle is a perfect triangle (a) s ( a+ b+ c) ( + + ) , which is an integer (b) s ( a+ b+ c) ( + + ) ()()() , which is an integer Copyright 03 Pearson Education, Inc.

12 364 Chapter 7 Applications of Trigonometry and Vectors (c) s ( a b c) ( ) , which is an integer (d) s ( a b c) ( ) , which is an integer. 73. Find the area of the Bermuda Triangle using Heron s Formula. s ( a+ b+ c) ( ) ( ) ( )( ) ,8.8 The area of the Bermuda Triangle is about 390,000 mi. 74. Find the area of the region using Heron s Formula. s ( a+ b+ c) ( ) 8 4 s( s a)( s b)( s c) 4( 4 75)( 4 68)( 4 85) ( 4)( 39)( 46)( 9) m A Number of cans needed (area in m ) cans (m per can) 75 She will need to open 33 cans. 75. (a) Using the law of sines, we have sin C sin A sin C sin 60 c a 5 3 5sin sin C There are two angles C between 0 and 80 that satisfy the condition. Since sin C , to the nearest tenth value of C is C Supplementary angles have the same sine value, so another possible value of C is B (b) By the law of cosines, we have a + b c cos C ab cos C 3 ( )( 7) C 9. (c) With the law of cosines, we are required to find the inverse cosine of a negative number; therefore; we know angle C is greater than Using the law of cosines, we have 77. a + c b cos B ac cos B b + c a cos A bc cos A Since cos B cos A, A is twice the size of B. 6 8 a b c Copyright 03 Pearson Education, Inc.

13 Chapter 7 Quiz Use the law of cosines to find the measure of A. a b + c bccosa ( ) cos A cos A cos A cos A A absin C 9 3 sin C sin cos 9.5 sq units First find the semiperimeter. s ( a+ b+ c) ( ) Using Heron s formula, we have sq units (found using a calculator) 80. ( x, y ) (,5 ); ( x, y ) (,3 ); ( x, y ) ( 4,0) 3 3 A xy yx + xy3 yx 3 + xy 3 yx 3 3 ( ) 5 ( ) + ( )( 0 ) 34 ( ) + 45 ( ) 0 ( ) 9.5 sq units Chapter 7 (Sections ) Quiz. Using the law of sines, we have sin B sin C sin 30.6 sin C b c sin 30.6 sin C C 8. A 80 B C (rounded to three significant digits). Using the law of cosines, we have a b + c bccosa a cos44 40, 3.59 a 0 m (rounded to three significant digits) 3. Using the law of cosines, we have c a + b abcosc cos C cos C cos C C 48.0 (rounded to three significant digits) sin 7 9 sin sq units 4 4. A ab C 5. First find the semiperimeter: s ( ) 3.5 Using Heron s formula, we have ( )( )( ) ,7 89 km 6. Using the law of sines,we have sin A sin C sin A sin 5.4 a c sin 5.4 sin A A 4.6 or A Copyright 03 Pearson Education, Inc.

14 366 Chapter 7 Applications of Trigonometry and Vectors 7. C Using the law of sines, we have a c a 36 sin A sin C sin sin 8 36sin a 648 sin 8 b c b 36 sin B sin C sin 4 sin 8 36sin 4 b 456 sin 8 Note that both a and b have been rounded to three significant digits. 8. The height of the balloon is the length of the altitude of XYZ, AZ. AZ We will use the fact that sin 4 0 to XZ find the length of AZ. First, we must find the length of XZ using the law of sines. Z XZ XY XZ. siny sin Z sin 3 30 sin4 0.sin 3 30 XZ sin4 0 AZ sin AZ sin mi 9. AB.4798 mi, AC mi, A To find the distance between the towns, d, use the law of cosines. d ( 348)( 563) cos ,376,78 d 39.3 The distance between the two towns is about 39 m (rounded to four significant digits.) Section 7.4 Vectors, Operations, and the Dot Product. Equal vectors have the same magnitude and direction. Equal vectors are m and p; n and r.. Opposite vectors have the same magnitude but opposite direction. Opposite vectors are m and q, p and q, n and s, r and s. 3. One vector is a positive scalar multiple of another if the two vectors point in the same direction; they may have different magnitudes. m ; p m ; t n ; r p t or p m; t m; r n; t p 4. One vector is a negative scalar multiple of another if the two vectors point in the opposite direction; they may have different magnitudes. m q; p q; r s; q t; n s This is SAS, so use the law of cosines. BC AC + AB AC AB cosa BC cos BC BC BC is approximately mi. (rounded to seven significant digits) 9. Copyright 03 Pearson Education, Inc.