Chapter 7 - Pearson

104 Chapter 7 Applications of Trigonometry and Vectors47.48.49.50.51.52.2722.9 in.2289.9 m265.94 cm284.41 m2100 m2373 m53. a= sin A, b= sin B, and c=sin C54. Answers will vary55.d sinαsinβxsin( β-α )=56. (a) The coordinates of the house are(1131.8, 4390.2)The coordinates of the forest fire are(2277.5, −2596.2)(b) 7079.7 ftSection 7.2 The Ambiguous Case of theLaw of Sines1. A2. D3. (a) 4 < h < 5.(b) h = 4 or h > 5(c) h < 44. (a) none5. 16. 17. 28. 29. 010. 011. 45°12. 30°(b) h 5(c) h £ 513. B 1 = 49.1 ; B 2 = 130.914. A 1 = 72.2 ; A 2 = 107.8C 1 = 59.6 ;C 2 = 24.015. B 1 = 2630 ¢ ; A = 112 10¢.16. A 1 = 37 50 ¢ ; C = 9320¢17. no such triangle exists18. no such triangle exists19. B » 27.19 ; C » 10.6820. A = 45.40 ; C = 20.8821. B 1 = 20.6 ; C = 116.9 ; c » 20.6 ft22. A 1 = 25.5 ; B = 102.2 ; b » 73.9 yd23. no such triangle exists24. no such triangle exists25. B 1 » 4920';C 1 = 9200'c 1 » 15.5 m; c 2 » 2.88 m26. A 1 » 55 20 ¢ ; B 1 = 9450¢b 1 » 10.4 m; b 2 » 4.51 m27. B 1 » 37.77; C = 45.43 ;c » 4.174 ft28. B 1 » 30.39 ;A = 60.91 ; a » 98.25 m29. A 1 = 53.23 ; C 1 = 87.09c 1 » 37.16 m; A 2 = 126.77C 2 = 13.55 ;c 2 » 8.719 m30. C 1 = 59.71 ; B 1 = 69.09b 1 » 8640 cm; B 2 = 8.51 ;b 2 » 1369 cm31. 1; 90 ;right triangle.32. Answers will vary33. Answers will vary34. Answers will vary35. 664 m36. 87.3 ft37. 218 ft38. 7500 ¢ ; 274 miC 1 = 101.2 ;C 2 = 19.4Copyright © 2013 Pearson Education, Inc.

Section 7.3 The Law of Cosines 105a+ b sin A+sin B39. Prove that = .b sin BStart with the law of sines.a sin= b a =b Asin A sin B sin Ba+bSubstitute for a in the expression .bbsinA bsinA+ b + ba+ b sin sin sin B=B =B ⋅b b b sin BbsinA+bsinB b( sin A+sin B)= =bsinB sin Bsin A+sin B=sin Ba-b sin A-sinB40. Prove that =.a+ b sin A+sin BStart with the law of sines.a sin= b a =b Asin A sin B sin BSubstitute for a in the expression a - ba+ b.bsinA bsinA-b-ba- b sin sin sin B=B =B ⋅a+b bsin A bsin A+ b + bsin Bsin B sin BbsinA-bsinB b( sin A-sinB)= =bsin A+ bsin B b( sin A+sin B)sin A-sinB=sin A+sin B41.42.43. (a)é sin AsinB ùA= 5 Rsin ( A+B)êëúû2A=1.12257R(b)28.77 in.25.32 in.44. redSection 7.3 The Law of Cosines1. (a) SAS(b) law of cosines2. (a) SAA(b) law of sines23. (a) SSA(b) law of sines4. (a) ASA(b) law of sines5. (a) ASA(b) law of sines6. (a) SSA(b) law of sines7. (a) SSS(b) law of cosines8. (a) SAS(b) law of cosines9. 510. 711. 12012. 3013. a » 7.0; C » 21.4 ;B = 37.6°14. a » 5.4 B » 40.7 ;C = 78.315. B » 53.1 ;C = 53.1 ; A » 73.7The angles may not sum to 180° due torounding.16. B » 108.2 ;A » 22.3 ;C = 49.5°17. b » 88.18; A » 56.7 ;C = 68.3°18. C » 95.7 ;A » 33.6 ;B = 50.7°19. aa » 2.60 yd; B » 45.1 ;C = 93.520. c » 2.83 in; A » 44.9 ;B = 106.821. c » 6.46 m; A » 53.1 ;B = 81.322. a » 43.7 km; B » 53.2 ;C = 59.523. A » 82 ;B » 37 ;C = 6124. C » 98 ;A » 29 ;B = 5325. C » 10210 ¢ ; B » 3550 ¢ ; A = 4200¢26. C » 10720 ¢ ; B » 3900 ¢ ; A = 3340¢27. C » 8430 ¢ ; B » 4440 ¢ ; A = 5050¢28. B » 8510 ¢ ; C » 4450 ¢ ; A = 5000¢Copyright © 2013 Pearson Education, Inc.

106 Chapter 7 Applications of Trigonometry and Vectors29. a » 156 cm; C = 3430 ¢ ; B = 6450¢30. c » 348 ft; B = 4330 ¢ ; A = 6350¢31. b » 9.53 in; C » 40.6 ;A = 64.6°32. c » 4.28 mi; A » 48.8 ;B =71.5°33. a » 15.7 m; B » 21.6 ;C = 45.6°34. b » 34.1 cm; A » 5.2 ;C = 6.6°35. C » 94 ;A » 30 ;B = 56°36. C » 125 ;A » 24 ;B = 31°37. Since the cosine of any angle of a trianglemust be between –1 and 1, a triangle cannothave sides 3, 4, and 10.38. Answers will vary.39. 257 m40. 95.3 m41. 163.542. 5.2 cm and 8.8 cm43. 281 km44. approximately 180 mi45. 438.14 feet46. 745 mi47. 10.8 miles48. 1450 ft49. θ » 4050. β » 5351. A » 36 ;B » 2652. AB » 18 ft53. second base is 66.8 ft and the distance to bothfirst and third base is 63.7 ft54. second base is 38.9 ft and the distance to bothfirst and third base is 42.6 ft55. v » 39.2 km56. 1473 m apart57. 47.5 feet58. 5.9 mi59. 5500 meters long60. 5.99 km61. θ » 16.2662. θ » 14.2563. » 41.57; Both formulas give the same area.64. » 25.98; Both formulas give the same result.65.66.67.68.69.70.2 » 78 m2 » 310 in.2 »12,600 cm2 » 228 yd2 » 3650 ft2 » 83.01 m71. the perimeter and area both equal 36, thetriangle is a perfect triangle.72. (a) = 66, which is an integer(b) = 84, which is an integer(c) = 42, which is an integer.(d) = 36, which is an integer.73. 390,000 2 mi .74. 33 cans75. (a) C 1 = 87.8 ;B 2 = 92.2Both appear to be possible(b) C » 92.2(c) With the law of cosines, we are requiredto find the inverse cosine of a negativenumber; therefore; we know angle C isgreater than 90°.Copyright © 2013 Pearson Education, Inc.

Section 7.4 Vectors, Operations, and the Dot Product 10776. Using the law of cosines, we have2 2 2a + c -bcos B = 2ac2 2 26 + 5 - 4 36+ 25-16 3cos B = = =26 ( )( 5)60 42 2 2b + c -acos A = 2bc2 2 24 + 5 - 6 16+ 25-36 1cos A = = =24 ( )( 5)40 8Since2 æ3ö æ 9 ö 162cos B - 1= 2 - 1= 2-çè4÷ ø èç16ø÷162 1= = = cos A,A is twice the size of16 8B.77.210. 3921 mSection 7.4 Vectors, Operations, and theDot Product1. m and p; n and r.2. m and q, p and q, n and s, r and s.3. m= 1; p m= 2; t n= 1; r p=2t or1 1p= 1 m; t= m; r= 1 n;t=p2 24. m = –1q; p = –1q; r = –1s; q = –2t; n = –1s5.6.7.8.9.a = 34; b = 29; c = 1378. = 9.5 sq units79. A = 9.5 sq units (found using a calculator)80. A = 9.5 sq unitsChapter 7 Quiz(Sections 7.1−7.3)1. A » 1312. a » 201 m3. C » 48.0 4. 15.75 sq units10.11.5.2 = 189 km6. A» 41.6 or A» 138.47. C = 28 ;a» 648; b»4568. 3.6 mi9. 25.24983 mi.Copyright © 2013 Pearson Education, Inc.

108 Chapter 7 Applications of Trigonometry and Vectors12.21. (a) 8,0(b) 0,16(c) -4,-822. (a) 4,0(b) -12,-813.(c) 4, 423. (a) 0,12(b) -16,-414.(c) 8, - 424. (a) 4, 4(b) 12, - 12(c) - 8, 415.25. (a) 4i(b) 7i+3j(c) - 5i+j26. (a) - 2i+4j16.(b) i+j(c) 4i-7j27. (a) - 2, 417. Yes, vector addition is associative.18. Yes, vector addition is commutative.19. (a) - 4,16(b) - 12,0(c) 8, - 820. (a) -4,-8(b) 12,0(b) 7,4(c) 6, - 628. (a) -4,-2(b) - 13,4(c) 3,529. u = 12, v = 20, θ = 27(c) - 4, 4Copyright © 2013 Pearson Education, Inc.

Section 7.4 Vectors, Operations, and the Dot Product 10930. u = 8, v = 12, θ = 2047. v » 4.0958, -2.867948. v » -1.5321, -1.285631. u = 20, v = 30, θ = 3032. u = 50, v = 70, θ = 4049. 530 newtons. (rounded to two significantdigits)50. 29 newtons. (rounded to two significant digits)51. 88.2 lb. (rounded to three significant digits)52. 76.2 lb. (rounded to three significant digits)53. v » 94.2 lb54. v » 158.0 lb55. v » 24.4 lb33. Magnitude: 17; Angle: 331.9(θ lies in quadrant IV)34. Magnitude: 25; Angle: 106.3(θ lies in quadrant II)35. Magnitude: 8; Angle: 120(θ lies in quadrant II)36. Magnitude: 16; Angle: 315(θ lies in quadrant IV)37. x » 47y » 1738. x » 17y » 2039. x » 38.8y » 28.040. x » 13.7y » 7.1141. x » 123y » 15542. x » 198y » 13243.u =5 3 5 ,2 244. u = 4, 4 345. v » -3.0642, 2.571246. v » -1.9284,2.298156. v » 1286.0 lb57. u+ v= a+ c,b+d58. z + z = ( a+ c) + ( b+d)i1 2Additional answers will vary59. u- v = -6, 260. v- u = 6, -261. - 4u= 8, -2062. - 5v= -20,-1563. 3u- 6v= -30,-364. - 2u+ 4v= 20,265. u+ v- 3u= 8, -766. 2u+ v- 6v= -24,-567. - 5, 8 =- 5i+8j68. 6, - 3 = 6i-3j69. 2, 0 = 2i+ 0j=2i70. 0, - 4 = 0i- 4j=-4j71. 772. - 6173. 074. 075. 2076. - 4Copyright © 2013 Pearson Education, Inc.

110 Chapter 7 Applications of Trigonometry and Vectors77. θ = 135 78. θ » 36.87 79. θ = 90 80. θ = 45 81. θ » 36.8782. θ » 78.93 83. - 684. - 685. - 2486. - 2487. the vectors are orthogonal.88. the vectors are orthogonal.89. the vectors are not orthogonal90. the vectors are not orthogonal.91. the vectors are not orthogonal.92. the vectors are not orthogonal93. Draw a line parallel to the x-axis and thevector u + v (shown as a dashed line)Since θ 1 = 110, its supplementary angle is70°. Further, since θ 2 = 260, the angle α is260-180= 80 . Then the angle CBAbecomes 180 – (80 + 70) = 180 – 150 = 30°.Magnitude: » 9.5208The direction angle is 119.064794. ab , » - 4.1042, 11.276395. cd , » -0.5209, - 2.954496. » - 4.6252,8.321997. Magnitude: » 9.5208; Angle: 119.0647(θ lies in quadrant II)98. They are the same. Preference of method is anindividual choice.Section 7.5Applications of Vectors1. 2640 lb at an angle of 167.2 with the 1480-lbforce2. 1800 lb at an angle of 167 (rounded to threesignificant digits) with the 840-lb force3. 93.94. 70.15. 190 lb and the magnitude of the resultant isabout 283 lb6. The magnitude of the resultant is 117 lb; thesecond force is 93.9 lb.7. 188. 800 lb.9. 2.4 tons10. 2.8 tons.11. 17.512. 22.013. 226 lb.14. The weight of the crate is 64.8 lb; the tensionis 61.9 lb.15. 13.5 mi; 50.416. 6.6 mi; 117.117. 39.2 km18. 14.5 km19. The speed of the current is 3.5 mph and theactual speed of the motorboat is 19.7 mph.20. (a) 6.3(b) 14 seconds(c) 2521. the bearing and ground speed of the planev » 470.1the bearing is 23722. the pilot turned at 1 hr 21 min after 2 P.M., orat 3:21 P.M.23. Let x = the airspeed and d = the groundspeed.x » 156 mph; d » 161 mph24. 173.125. The bearing is 7400¢ ; the ground speed is202 mph.26. The bearing is 6530¢181 mph.; the ground speed is27. The airspeed must be 170 mph.The bearing must be approximately 358°.Copyright © 2013 Pearson Education, Inc.

Chapter 7 Review Exercises 11128. The ground speed is 198 mph. The bearing is186.5°.29. The ground speed is 230 km per hr. Thebearing is 167°.30. (a) v t » 56 mi sec.(b) 87 mi/sec.31. (a) R = 5 » 2.2 andA = 0.25+ 1 » 1.1About 2.2 in. of rain fell. The area of theopening of the rain gauge is about 1.12in. .3(b) V = 1.5; The volume of rain was 1.5 in. .32. R and A should be parallel and point inopposite directions.Summary Exercises on Applications ofTrigonometry and Vectors1. The lengths of the two wires are about 29 ftand 38 ft.2. 38.3 cm3. 5856 m apart.4. x » 15.8 ft per sec; 71.65. 42 lb6. 7200 ft above the ground7. (a) The speed of the wind is 10 mph(b) This represents a 30 mph wind in thedirection of v.(c) u represents a southeast wind of11.3 mph8. the groundspeed is about 380 mph; θ » 2; the bearing of the plane is 64°9. - 1£ sin A £ 1, the triangle cannot exist10. other angles can be 36 10 ¢ ; third side40.5 yd, or other angles can beChapter 714350¢;8º00¢ and third side 6.25 yds1. 63.7 m2. B » 25.03. B = 41.7°4. 70.9 mReview Exercises5. 54 20 ¢ or 12540¢6. A = 4930 ¢ .7. No; If you are given two angles of a triangle,then the third angle is known since the sum ofthe measures of the three angles is 180°. Sinceyou are also given one side, there will only beone triangle that will satisfy the conditions.8. No; the sum of a and b do not exceed c.9. a = 10, B = 30°(a) b = 10 sin 30° = 5. Also, any value of bgreater than or equal to 10 would yield aunique value for A.(b) Any value of b between 5 and 10, wouldyield two possible values for A.(c) If b is less than 5, then no value for A ispossible.10. A = 140º, a = 5, and b =7With these conditions, we can try to solve thetriangle with the law of sines.sin B sin A sin B sin140= = b a 7 57sin140sin B= » 0.89990265 B» 645Since A+ B= 140+ 64= 204> 180 , nosuch triangle exists.11. A » 19.87 or 19 52 ¢ .12. 173 ft13. 55.5 m14. B » 26.5 or 26 30 ¢ .15. 19cm16. 4717. B = 17.3 ; C = 137.5 ;c » 11.0 yd18. B 1 = 74.6 ; C 1 = 43.7 ; c 1 » 61.9 mC 2 = 12.9 ;c 2 » 20.0 m19. c » 18.65cm; B » 4550 ¢ ; A = 91 40¢20. B » 73.9 ;A » 47.7 ;C = 58.421.22.23.2153,600 m220.3 ft20.234 kmCopyright © 2013 Pearson Education, Inc.

112 Chapter 7 Applications of Trigonometry and Vectors24.2680 m25. 58.6 feet26. 11 feet long27. 13 meters tall28. 15.8 ft apart29. 53.2 ft30. C » 77.131. 115 km32. 2.4 miles33. 25 sq units34. The sides of the triangle measure 5, 10, and5 5 , which satisfy the converse of thePythagorean theorem: the area is 25 sq units35. a – b36. a + 3c37. 207 lb38. 209 newtons39. horizontal: 869; vertical: 41840. horizontal: 25 2; vertical: 25 241. magnitude: 15Angle: 126.9(θ lies in quadrant II)42. magnitude: 29Angle: 316.4(θ lies in quadrant IV)43. (a) i(b) 4i-2j(c) 11i-7j44. -9;142.145. The vectors are orthogonal.46. θ = 4547. 29 lb48. The force is 280 newtons (rounded) at an angleof 30.4° with the first boat.49. The pilot should fly on a bearing of 306°. Heractual speed is 524 mph.50. 4º51. 34 lb52. The resulting speed is 21 km per hr (rounded)and bearing is 118°53. Newton’s formula:Therefore,54. Mollweide’s formula:Therefore,a+ b cos ( A-B)=c sin C2+ 3 2+3=2 22- 3 2-3=2 21 2 1 2a- b cos ( A-B)=c cos C12 1 2Copyright © 2013 Pearson Education, Inc.

Chapter 7 Test 11355. Let a = 2, b = 2 3 , A = 30°, B = 60°.Verifytan 12( A- B)a-b= .tan 1 ( A B)a + b2 +tan 1( A-B) tan 1(30-60 )2 2=tan 1( A+ B) tan 1(30+ 60 )2 2tan( -15 )= »-0.26794919tan 45a-b2-2 3 2-2 3 4- 8 3+12= ⋅ =a+ b 2+ 2 3 2-2 3 4-1216-8 3= =- 2 + 3 »-0.26794919-8Thus,tan 12( A- B)a-b=tan 1 ( A B)a + b2values of a, b, A, and B.Chapter 7 Test1. 137.5°2. 179 km.3. 49.0°4. 168 sq units+using the given11. (a) 1, – 3(b) - 6, 18(c) - 20(d) 1012. 41.813. Since the dot product is 0, the vectors areorthogonal.14. The balloon is 2.7 miles off the ground.15. - 346, 45116. 1.91 miles17. 14 m18. 30 lb19. The plane’s bearing is 357°20. 18.75. 18 sq units6. (a) b > 10(b) none(c) b ≤ 107. a » 40 m; B » 41 ;C = 798. B1 » 58 30 ¢ or B2= 12130¢Solving separately for triangles A 1 = 8300¢a 11250 in. ; A 2 = 2000 ¢ ; a 2 » 431 in.9. magnitude: v = 10;angle: 126.9(θ lies in quadrant II)10.Copyright © 2013 Pearson Education, Inc.