Solution Manual for Trigonometry, 11th Edition

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Solution Manual for Trigonometry, 11th Edition

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S OLUTIONS M ANUAL T IM B RITT Jackson State Community College T RIGONOMETRY : A U NIT C IRCLE A PPROACH E LEVENTH E DITION Michael Sullivan Chicago State University

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Table of Contents Preface Chapter 1 Graphs and Functions 1.1 The Distance and Midpoint Formulas ......................................................................................... 1 1.2 Graphs of Equations in Two Variables; Circles........................................................................ 13 1.3 Functions and Their Graphs ...................................................................................................... 37 1.4 Properties of Functions ............................................................................................................. 55 1.5 Library of Functions; Piecewise-defined Functions ................................................................. 70 1.6 Graphing Techniques: Transformations ................................................................................... 82 1.7 One-to-One Functions; Inverse Functions ................................................................................ 98 Chapter Review.............................................................................................................................. 119 Chapter Test ................................................................................................................................... 129 Chapter Projects ............................................................................................................................. 132 Chapter 2 Trigonometric Functions 2.1 Angles, Arc Length, and Circular Motion .............................................................................. 135 2.2 Trigonometric Functions: Unit Circle Approach .................................................................... 144 2.3 Properties of the Trigonometric Functions ............................................................................. 162 2.4 Graphs of the Sine and Cosine Functions ............................................................................... 176 2.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions...................................... 196 2.6 Phase Shift; Sinusoidal Curve Fitting ..................................................................................... 206 Chapter Review.............................................................................................................................. 217 Chapter Test ................................................................................................................................... 225 Cumulative Review........................................................................................................................ 228 Chapter Projects ............................................................................................................................. 231 Chapter 3 Analytic Trigonometry 3.1 The Inverse Sine, Cosine, and Tangent Functions .................................................................. 234 3.2 The Inverse Trigonometric Functions (Continued) ................................................................ 247 3.3 Trigonometric Equations ........................................................................................................ 259 3.4 Trigonometric Identities ......................................................................................................... 280 3.5 Sum and Difference Formulas ................................................................................................ 292 3.6 Double-angle and Half-angle Formulas .................................................................................. 317 3.7 Product-to-Sum and Sum-to-Product Formulas...................................................................... 343 Chapter Review.............................................................................................................................. 356 Chapter Test ................................................................................................................................... 371 Cumulative Review........................................................................................................................ 376 Chapter Projects ............................................................................................................................. 379 Chapter 4 Applications of Trigonometric Functions 4.1 Right Triangle Trigonometry; Applications ........................................................................... 383 4.2 The Law of Sines .................................................................................................................... 397 4.3 The Law of Cosines ................................................................................................................ 411 4.4 Area of a Triangle ................................................................................................................... 424 4.5 Simple Harmonic Motion; Damped Motion; Combining Waves ........................................... 433 Chapter Review.............................................................................................................................. 443 Chapter Test ................................................................................................................................... 449 Cumulative Review........................................................................................................................ 453 Chapter Projects ............................................................................................................................. 456

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Chapter 5 Polar Coordinates; Vectors 5.1 Polar Coordinates.................................................................................................................... 460 5.2 Polar Equations and Graphs .................................................................................................... 469 5.3 The Complex Plane; De Moivre’s Theorem ........................................................................... 498 5.4 Vectors .................................................................................................................................... 511 5.5 The Dot Product ...................................................................................................................... 524 5.6 Vectors in Space ..................................................................................................................... 530 5.7 The Cross Product................................................................................................................... 536 Chapter Review.............................................................................................................................. 547 Chapter Test ................................................................................................................................... 556 Cumulative Review........................................................................................................................ 560 Chapter Projects ............................................................................................................................. 562 Chapter 6 Analytic Geometry 6.2 The Parabola ........................................................................................................................... 566 6.3 The Ellipse .............................................................................................................................. 581 6.4 The Hyperbola ........................................................................................................................ 598 6.5 Rotation of Axes; General Form of a Conic ........................................................................... 618 6.6 Polar Equations of Conics ....................................................................................................... 631 6.7 Plane Curves and Parametric Equations ................................................................................. 640 Chapter Review.............................................................................................................................. 654 Chapter Test ................................................................................................................................... 664 Cumulative Review........................................................................................................................ 668 Chapter Projects ............................................................................................................................. 670 Chapter 7 Exponential and Logarithmic Functions 7.1 Exponential Functions ............................................................................................................ 674 7.2 Logarithmic Functions ............................................................................................................ 695 7.3 Properties of Logarithms ........................................................................................................ 717 7.4 Logarithmic and Exponential Equations ................................................................................. 726 7.5 Financial Models .................................................................................................................... 745 7.6 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models .................................................................................................................... 753 7.7 Building Exponential, Logarithmic, and Logistic Models from Data..................................... 763 Chapter Review.............................................................................................................................. 768 Chapter Test ................................................................................................................................... 777 Cumulative Review........................................................................................................................ 780 Chapter Projects ............................................................................................................................. 782 Appendix A Review A.1 Algebra Essentials .................................................................................................................. 784 A.2 Geometry Essentials ............................................................................................................... 789 A.3 Factoring Polynomials; Completing the Square..................................................................... 795 A.4 Solving Equations .................................................................................................................. 799 A.5 Complex Numbers; Quadratic Equations in the Complex Number System .......................... 813 A.6 Interval Notation; Solving Inequalities .................................................................................. 818 A.7 n th Roots; Rational Exponents ............................................................................................... 830 A.8 Lines....................................................................................................................................... 840

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Appendix B Graphing Utilities B.1 The Viewing Rectangle .......................................................................................................... 857 B.2 Using a Graphing Utility to Graph Equations ........................................................................ 858 B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry .............................. 863 B.5 Square Screens ....................................................................................................................... 865

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1 Chapter 1 Graphs and Functions Section 1.1 1. 0 2.   5 3 8 8     3. 2 2 3 4 25 5    4. 2 2 2 11 60 121 3600 3721 61      Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle. 5. 1 2 bh 6. true 7. x- coordinate or abscissa; y -coordinate or ordinate 8. quadrants 9. midpoint 10. False; the distance between two points is never negative. 11. False; points that lie in quadrant IV will have a positive x -coordinate and a negative y -coordinate. The point   1, 4  lies in quadrant II. 12. True; 1 2 1 2 , 2 2 x x y y M          13. b 14. a 15. (a) Quadrant II (b) x -axis (c) Quadrant III (d) Quadrant I (e) y -axis (f) Quadrant IV 16. (a) Quadrant I (b) Quadrant III (c) Quadrant II (d) Quadrant I (e) y -axis (f) x -axis 17. The points will be on a vertical line that is two units to the right of the y -axis.

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Chapter 1: Graphs and Functions 2 18. The points will be on a horizontal line that is three units above the x -axis. 19. 2 2 1 2 2 2 ( , ) (2 0) (1 0) 2 1 4 1 5 d P P          20. 2 2 1 2 2 2 ( , ) ( 2 0) (1 0) ( 2) 1 4 1 5 d P P            21. 2 2 1 2 2 2 ( , ) ( 2 1) (2 1) ( 3) 1 9 1 10 d P P            22.   2 2 1 2 2 2 ( , ) 2 ( 1) (2 1) 3 1 9 1 10 d P P           23.       2 2 1 2 2 2 ( , ) (5 3) 4 4 2 8 4 64 68 2 17 d P P            24.         2 2 1 2 2 2 ( , ) 2 1 4 0 3 4 9 16 25 5 d P P            25.   2 2 1 2 2 2 ( , ) 4 ( 7) (0 3) 11 ( 3) 121 9 130            d P P 26.     2 2 1 2 2 2 ( , ) 4 2 2 ( 3) 2 5 4 25 29 d P P           27.   2 2 1 2 2 2 ( , ) (6 5) 1 ( 2) 1 3 1 9 10           d P P 28.     2 2 1 2 2 2 ( , ) 6 ( 4) 2 ( 3) 10 5 100 25 125 5 5 d P P             29.     2 2 1 2 2 2 ( , ) 2.3 ( 0.2) 1.1 (0.3) 2.5 0.8 6.25 0.64 6.89 2.62            d P P 30.     2 2 1 2 2 2 ( , ) 0.3 1.2 1.1 2.3 ( 1.5) ( 1.2) 2.25 1.44 3.69 1.92              d P P 31. 2 2 1 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) d P P a b a b a b           32. 2 2 1 2 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) 2 2 d P P a a a a a a a a             33. ( 2,5), (1,3), ( 1, 0) A B C            2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 1 ( 2) (3 5) 3 ( 2) 9 4 13 ( , ) 1 1 (0 3) ( 2) ( 3) 4 9 13 ( , ) 1 ( 2) (0 5) 1 ( 5) 1 25 26 d A B d B C d A C                                   

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Section 1.1: The Distance and Midpoint Formulas 3 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 13 13 26 13 13 26 26 26 d A B d B C d A C        The area of a triangle is 1 2 A bh   . In this problem,     1 ( , ) ( , ) 2 1 1 13 13 13 2 2 13 square units 2 A d A B d B C          34. ( 2, 5), (12, 3), (10, 11) A B C            2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 12 ( 2) (3 5) 14 ( 2) 196 4 200 10 2 ( , ) 10 12 ( 11 3) ( 2) ( 14) 4 196 200 10 2 ( , ) 10 ( 2) ( 11 5) 12 ( 16) 144 256 400 20 d A B d B C d A C                                       Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 10 2 10 2 20 200 200 400 400 400 d A B d B C d A C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 10 2 10 2 2 1 100 2 100 square units 2 A d A B d B C           35. ( 5,3), (6, 0), (5,5) A B C           2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 6 ( 5) (0 3) 11 ( 3) 121 9 130 ( , ) 5 6 (5 0) ( 1) 5 1 25 26 ( , ) 5 ( 5) (5 3) 10 2 100 4 104 2 26 d A B d B C d A C                                 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:

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Chapter 1: Graphs and Functions 4             2 2 2 2 2 2 ( , ) ( , ) ( , ) 104 26 130 104 26 130 130 130 d A C d B C d A B        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 104 26 2 1 2 26 26 2 1 2 26 2 26 square units A d A C d B C              36. ( 6, 3), (3, 5), ( 1, 5) A B C             2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 3 ( 6) ( 5 3) 9 ( 8) 81 64 145 ( , ) 1 3 (5 ( 5)) ( 4) 10 16 100 116 2 29 ( , ) 1 ( 6) (5 3) 5 2 25 4 29 d A B d B C d A C                                     Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 29 2 29 145 29 4 29 145 29 116 145 145 145 d A C d B C d A B           The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 29 2 29 2 1 2 29 2 29 square units A d A C d B C           37. (4, 3), (0, 3), (4, 2) A B C              2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (0 4) 3 ( 3) ( 4) 0 16 0 16 4 ( , ) 4 0 2 ( 3) 4 5 16 25 41 ( , ) (4 4) 2 ( 3) 0 5 0 25 25 5 d A B d B C d A C                                   Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:

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Section 1.1: The Distance and Midpoint Formulas 5         2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 5 41 16 25 41 41 41 d A B d A C d B C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 4 5 2 10 square units A d A B d A C        38. (4, 3), (4, 1), (2, 1) A B C             2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (4 4) 1 ( 3) 0 4 0 16 16 4 ( , ) 2 4 1 1 ( 2) 0 4 0 4 2 ( , ) (2 4) 1 ( 3) ( 2) 4 4 16 20 2 5 d A B d B C d A C                                   Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:         2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 2 2 5 16 4 20 20 20 d A B d B C d A C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 4 2 2 4 square units A d A B d B C        39. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 4 3 5 , 2 2 8 0 , 2 2 (4, 0) x x y y x y                            40. The coordinates of the midpoint are:   1 2 1 2 ( , ) , 2 2 2 2 0 4 , 2 2 0 4 , 2 2 0, 2 x x y y x y                            41. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 1 8 4 0 , 2 2 7 4 , 2 2 7 , 2 2                                  x x y y x y

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Chapter 1: Graphs and Functions 6 42. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 2 4 3 2 , 2 2 6 1 , 2 2 1 3, 2 x x y y x y                                    43. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 7 9 5 1 , 2 2 4 16 , 2 2 (8, 2)                              x x y y x y 44. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 2 3 2 , 2 2 2 1 , 2 2 1 1, 2 x x y y x y                                       45. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a b a b                          46. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a a a a                          47. The x coordinate would be 2 3 5   and the y coordinate would be 5 2 3   . Thus the new point would be   5,3 . 48. The new x coordinate would be 1 2 3     and the new y coordinate would be 6 4 10   . Thus the new point would be   3,10  49. a. If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length. One of the legs of the triangle will be 2+3=5. Thus the other leg will be: 2 2 2 2 2 5 13 25 169 144 12 b b b b       Thus the coordinates will have an y value of 1 12 13     and 1 12 11    . So the points are   3,11 and   3, 13  . b. Consider points of the form   3, y that are a distance of 13 units from the point   2, 1   .             2 2 2 1 2 1 2 2 2 2 2 2 3 ( 2) 1 5 1 25 1 2 2 26 d x x y y y y y y y y                           2 2 2 2 2 2 13 2 26 13 2 26 169 2 26 0 2 143 0 11 13 y y y y y y y y y y                11 0 11 y y    or 13 0 13 y y     Thus, the points   3,11 and   3, 13  are a distance of 13 units from the point   2, 1   .

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Section 1.1: The Distance and Midpoint Formulas 7 50. a. If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length. One of the legs of the triangle will be 2+6=8. Thus the other leg will be: 2 2 2 2 2 8 17 64 289 225 15 b b b b       Thus the coordinates will have an x value of 1 15 14    and 1 15 16   . So the points are   14, 6   and   16, 6  . b. Consider points of the form   , 6 x  that are a distance of 17 units from the point   1, 2 .             2 2 2 1 2 1 2 2 2 2 2 2 1 2 6 2 1 8 2 1 64 2 65 d x x y y x x x x x x x                          2 2 2 2 2 2 17 2 65 17 2 65 289 2 65 0 2 224 0 14 16 x x x x x x x x x x                14 0 14 x x     or 16 0 16 x x    Thus, the points   14, 6   and   16, 6  are a distance of 13 units from the point   1, 2 . 51. Points on the x -axis have a y -coordinate of 0. Thus, we consider points of the form   , 0 x that are a distance of 6 units from the point   4, 3  .           2 2 2 1 2 1 2 2 2 2 2 2 4 3 0 16 8 3 16 8 9 8 25 d x x y y x x x x x x x                        2 2 2 2 2 2 2 6 8 25 6 8 25 36 8 25 0 8 11 ( 8) ( 8) 4(1)( 11) 2(1) 8 64 44 8 108 2 2 8 6 3 4 3 3 2 x x x x x x x x x                             4 3 3 x   or 4 3 3 x   Thus, the points   4 3 3, 0  and   4 3 3, 0  are on the x -axis and a distance of 6 units from the point   4, 3  . 52. Points on the y -axis have an x -coordinate of 0. Thus, we consider points of the form   0, y that are a distance of 6 units from the point   4, 3  .         2 2 2 1 2 1 2 2 2 2 2 2 4 0 3 4 9 6 16 9 6 6 25 d x x y y y y y y y y y                    

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Chapter 1: Graphs and Functions 8   2 2 2 2 2 2 2 6 6 25 6 6 25 36 6 25 0 6 11 ( 6) (6) 4(1)( 11) 2(1) 6 36 44 6 80 2 2 6 4 5 3 2 5 2 y y y y y y y y y                               3 2 5 y    or 3 2 5 y    Thus, the points   0, 3 2 5   and   0, 3 2 5   are on the y- axis and a distance of 6 units from the point   4, 3  . 53. a. To shift 3 units left and 4 units down, we subtract 3 from the x -coordinate and subtract 4 from the y -coordinate.     2 3,5 4 1,1     b. To shift left 2 units and up 8 units, we subtract 2 from the x -coordinate and add 8 to the y -coordinate.     2 2,5 8 0,13    54. Let the coordinates of point B be   , x y . Using the midpoint formula, we can write   1 8 2,3 , 2 2 x y           . This leads to two equations we can solve. 1 2 2 1 4 5 x x x        8 3 2 8 6 2 y y y       Point B has coordinates   5, 2  . 55.   1 2 1 2 , , 2 2 x x y y M x y           .   1 1 1 , ( 3, 6) P x y    and ( , ) ( 1, 4) x y   , so 1 2 2 2 2 2 3 1 2 2 3 1 x x x x x x            and 1 2 2 2 2 2 6 4 2 8 6 2 y y y y y y        Thus, 2 (1, 2) P  . 56.   1 2 1 2 , , 2 2 x x y y M x y           .   2 2 2 , (7, 2) P x y    and ( , ) (5, 4) x y   , so 1 2 1 1 1 2 7 5 2 10 7 3 x x x x x x        and 1 2 1 1 1 2 ( 2) 4 2 8 ( 2) 6 y y y y y y             Thus, 1 (3, 6) P   . 57. The midpoint of AB is:   0 6 0 0 , 2 2 3, 0 D           The midpoint of AC is:   0 4 0 4 , 2 2 2, 2 E           The midpoint of BC is:   6 4 0 4 , 2 2 5, 2 F             2 2 2 2 ( , ) 0 4 (3 4) ( 4) ( 1) 16 1 17 d C D              2 2 2 2 ( , ) 2 6 (2 0) ( 4) 2 16 4 20 2 5 d B E            2 2 2 2 ( , ) (2 0) (5 0) 2 5 4 25 29 d A F         

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Section 1.1: The Distance and Midpoint Formulas 9 58. Let 1 2 (0, 0), (0, 4), ( , ) P P P x y          2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 , (0 0) (4 0) 16 4 , ( 0) ( 0) 4 16 , ( 0) ( 4) ( 4) 4 ( 4) 16 d P P d P P x y x y x y d P P x y x y x y                             Therefore,   2 2 2 2 4 8 16 8 16 2 y y y y y y y        which gives 2 2 2 2 16 12 2 3 x x x      Two triangles are possible. The third vertex is     2 3, 2 or 2 3, 2  . 59. 2 2 1 2 2 2 ( , ) ( 4 2) (1 1) ( 6) 0 36 6 d P P             2 2 2 3 2 2 ( , ) 4 ( 4) ( 3 1) 0 ( 4) 16 4 d P P             2 2 1 3 2 2 ( , ) ( 4 2) ( 3 1) ( 6) ( 4) 36 16 52 2 13 d P P               Since       2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. 60.   2 2 1 2 2 2 ( , ) 6 ( 1) (2 4) 7 ( 2) 49 4 53 d P P              2 2 2 3 2 2 ( , ) 4 6 ( 5 2) ( 2) ( 7) 4 49 53 d P P               2 2 1 3 2 2 ( , ) 4 ( 1) ( 5 4) 5 ( 9) 25 81 106 d P P             Since       2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. Since     1 2 2 3 , , d P P d P P  , the triangle is isosceles. Therefore, the triangle is an isosceles right triangle. 61.     2 2 1 2 2 2 ( , ) 0 ( 2) 7 ( 1) 2 8 4 64 68 2 17 d P P               2 2 2 3 2 2 ( , ) 3 0 (2 7) 3 ( 5) 9 25 34 d P P               2 2 1 3 2 2 ( , ) 3 ( 2) 2 ( 1) 5 3 25 9 34 d P P            Since 2 3 1 3 ( , ) ( , ) d P P d P P  , the triangle is isosceles. Since       2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle.

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Chapter 1: Graphs and Functions 10 62.     2 2 1 2 2 2 ( , ) 4 7 0 2 ( 11) ( 2) 121 4 125 5 5 d P P                2 2 2 3 2 2 ( , ) 4 ( 4) (6 0) 8 6 64 36 100 10 d P P                2 2 1 3 2 2 ( , ) 4 7 6 2 ( 3) 4 9 16 25 5 d P P            Since       2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. 63. Using the Pythagorean Theorem: 2 2 2 2 2 90 90 8100 8100 16200 16200 90 2 127.28 feet d d d d         90 90 90 90 d 64. Using the Pythagorean Theorem: 2 2 2 2 2 60 60 3600 3600 7200 7200 60 2 84.85 feet d d d d          60 60 60 60 d 65. a. First: (90, 0), Second: (90, 90), Third: (0, 90) (0,0) (0,90) (90,0) (90,90) X Y b. Using the distance formula: 2 2 2 2 (310 90) (15 90) 220 ( 75) 54025 5 2161 232.43 feet d           c. Using the distance formula: 2 2 2 2 (300 0) (300 90) 300 210 134100 30 149 366.20 feet d          66. a. First: (60, 0), Second: (60, 60) Third: (0, 60) (0,0) (0,60) (60,0) (60,60) x y b. Using the distance formula: 2 2 2 2 (180 60) (20 60) 120 ( 40) 16000 40 10 126.49 feet d           c. Using the distance formula: 2 2 2 2 (220 0) (220 60) 220 160 74000 20 185 272.03 feet d         

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Section 1.1: The Distance and Midpoint Formulas 11 67. The Focus heading east moves a distance 60 t after t hours. The truck heading south moves a distance 40 t after t hours. Their distance apart after t hours is: 2 2 2 2 2 (60 ) (45 ) 3600 2025 5625 75 miles       d t t t t t t 68. 15 miles 5280 ft 1 hr 22 ft/sec 1 hr 1 mile 3600 sec      2 2 2 100 22 10000 484 feet d t t     100 22 t d 69. a. The shortest side is between 1 (2.6, 1.5) P  and 2 (2.7, 1.7) P  . The estimate for the desired intersection point is:   1 2 1 2 2.6 2.7 1.5 1.7 , , 2 2 2 2 5.3 3.2 , 2 2 2.65, 1.6 x x y y                          b. Using the distance formula: 2 2 2 2 (2.65 1.4) (1.6 1.3) (1.25) (0.3) 1.5625 0.09 1.6525 1.285 units d           70. Let 1 (2013, 102.87)  P and 2 (2017, 126.17)  P . The midpoint is:     1 2 1 2 , , 2 2 2013 2017 102.87 126.17 , 2 2 4030 229.04 , 2 2 2015, 114.52                           x x y y x y The estimate for 2010 is $114.52 billion. The estimate net sales of Costco Wholesale Corporation in 2015 is $0.85 billion off from the reported value of $113.67 billion. 71. For 2009 we have the ordered pair   2009, 21756 and for 2017 we have the ordered pair   2017, 24858 . The midpoint is     2009 2017 21756 24858 year, $ , 2 2 4026 46614 , 2 2 2013, 23307                  Using the midpoint, we estimate the poverty level in 2013 to be $23,307. This is lower than the actual value. 72. Let   1 0, 0 P  ,   2 , 0 P a  , and 3 3 , 2 2 a a P          . Then           2 2 1 2 2 1 2 1 2 2 2 , 0 0 0 d P P x x y y a a a                 2 2 2 3 2 1 2 1 2 2 2 2 2 2 , 3 0 2 2 3 4 4 4 4 d P P x x y y a a a a a a a a                            d 45t 60t

S OLUTIONS M ANUAL T IM B RITT Jackson State Community College T RIGONOMETRY : A U NIT C IRCLE A PPROACH E LEVENTH E DITION Michael Sullivan Chicago State University Table of Contents Preface Chapter 1 Graphs and Functions 1.1 The Distance and Midpoint Formulas ......................................................................................... 1 1.2 Graphs of Equations in Two Variables; Circles........................................................................ 13 1.3 Functions and Their Graphs ...................................................................................................... 37 1.4 Properties of Functions ............................................................................................................. 55 1.5 Library of Functions; Piecewise-defined Functions ................................................................. 70 1.6 Graphing Techniques: Transformations ................................................................................... 82 1.7 One-to-One Functions; Inverse Functions ................................................................................ 98 Chapter Review.............................................................................................................................. 119 Chapter Test ................................................................................................................................... 129 Chapter Projects ............................................................................................................................. 132 Chapter 2 Trigonometric Functions 2.1 Angles, Arc Length, and Circular Motion .............................................................................. 135 2.2 Trigonometric Functions: Unit Circle Approach .................................................................... 144 2.3 Properties of the Trigonometric Functions ............................................................................. 162 2.4 Graphs of the Sine and Cosine Functions ............................................................................... 176 2.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions...................................... 196 2.6 Phase Shift; Sinusoidal Curve Fitting ..................................................................................... 206 Chapter Review.............................................................................................................................. 217 Chapter Test ................................................................................................................................... 225 Cumulative Review........................................................................................................................ 228 Chapter Projects ............................................................................................................................. 231 Chapter 3 Analytic Trigonometry 3.1 The Inverse Sine, Cosine, and Tangent Functions .................................................................. 234 3.2 The Inverse Trigonometric Functions (Continued) ................................................................ 247 3.3 Trigonometric Equations ........................................................................................................ 259 3.4 Trigonometric Identities ......................................................................................................... 280 3.5 Sum and Difference Formulas ................................................................................................ 292 3.6 Double-angle and Half-angle Formulas .................................................................................. 317 3.7 Product-to-Sum and Sum-to-Product Formulas...................................................................... 343 Chapter Review.............................................................................................................................. 356 Chapter Test ................................................................................................................................... 371 Cumulative Review........................................................................................................................ 376 Chapter Projects ............................................................................................................................. 379 Chapter 4 Applications of Trigonometric Functions 4.1 Right Triangle Trigonometry; Applications ........................................................................... 383 4.2 The Law of Sines .................................................................................................................... 397 4.3 The Law of Cosines ................................................................................................................ 411 4.4 Area of a Triangle ................................................................................................................... 424 4.5 Simple Harmonic Motion; Damped Motion; Combining Waves ........................................... 433 Chapter Review.............................................................................................................................. 443 Chapter Test ................................................................................................................................... 449 Cumulative Review........................................................................................................................ 453 Chapter Projects ............................................................................................................................. 456 Chapter 5 Polar Coordinates; Vectors 5.1 Polar Coordinates.................................................................................................................... 460 5.2 Polar Equations and Graphs .................................................................................................... 469 5.3 The Complex Plane; De Moivre’s Theorem ........................................................................... 498 5.4 Vectors .................................................................................................................................... 511 5.5 The Dot Product ...................................................................................................................... 524 5.6 Vectors in Space ..................................................................................................................... 530 5.7 The Cross Product................................................................................................................... 536 Chapter Review.............................................................................................................................. 547 Chapter Test ................................................................................................................................... 556 Cumulative Review........................................................................................................................ 560 Chapter Projects ............................................................................................................................. 562 Chapter 6 Analytic Geometry 6.2 The Parabola ........................................................................................................................... 566 6.3 The Ellipse .............................................................................................................................. 581 6.4 The Hyperbola ........................................................................................................................ 598 6.5 Rotation of Axes; General Form of a Conic ........................................................................... 618 6.6 Polar Equations of Conics ....................................................................................................... 631 6.7 Plane Curves and Parametric Equations ................................................................................. 640 Chapter Review.............................................................................................................................. 654 Chapter Test ................................................................................................................................... 664 Cumulative Review........................................................................................................................ 668 Chapter Projects ............................................................................................................................. 670 Chapter 7 Exponential and Logarithmic Functions 7.1 Exponential Functions ............................................................................................................ 674 7.2 Logarithmic Functions ............................................................................................................ 695 7.3 Properties of Logarithms ........................................................................................................ 717 7.4 Logarithmic and Exponential Equations ................................................................................. 726 7.5 Financial Models .................................................................................................................... 745 7.6 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models .................................................................................................................... 753 7.7 Building Exponential, Logarithmic, and Logistic Models from Data..................................... 763 Chapter Review.............................................................................................................................. 768 Chapter Test ................................................................................................................................... 777 Cumulative Review........................................................................................................................ 780 Chapter Projects ............................................................................................................................. 782 Appendix A Review A.1 Algebra Essentials .................................................................................................................. 784 A.2 Geometry Essentials ............................................................................................................... 789 A.3 Factoring Polynomials; Completing the Square..................................................................... 795 A.4 Solving Equations .................................................................................................................. 799 A.5 Complex Numbers; Quadratic Equations in the Complex Number System .......................... 813 A.6 Interval Notation; Solving Inequalities .................................................................................. 818 A.7 n th Roots; Rational Exponents ............................................................................................... 830 A.8 Lines....................................................................................................................................... 840 Appendix B Graphing Utilities B.1 The Viewing Rectangle .......................................................................................................... 857 B.2 Using a Graphing Utility to Graph Equations ........................................................................ 858 B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry .............................. 863 B.5 Square Screens ....................................................................................................................... 865 1 Chapter 1 Graphs and Functions Section 1.1 1. 0 2.   5 3 8 8     3. 2 2 3 4 25 5    4. 2 2 2 11 60 121 3600 3721 61      Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle. 5. 1 2 bh 6. true 7. x- coordinate or abscissa; y -coordinate or ordinate 8. quadrants 9. midpoint 10. False; the distance between two points is never negative. 11. False; points that lie in quadrant IV will have a positive x -coordinate and a negative y -coordinate. The point   1, 4  lies in quadrant II. 12. True; 1 2 1 2 , 2 2 x x y y M          13. b 14. a 15. (a) Quadrant II (b) x -axis (c) Quadrant III (d) Quadrant I (e) y -axis (f) Quadrant IV 16. (a) Quadrant I (b) Quadrant III (c) Quadrant II (d) Quadrant I (e) y -axis (f) x -axis 17. The points will be on a vertical line that is two units to the right of the y -axis. Chapter 1: Graphs and Functions 2 18. The points will be on a horizontal line that is three units above the x -axis. 19. 2 2 1 2 2 2 ( , ) (2 0) (1 0) 2 1 4 1 5 d P P          20. 2 2 1 2 2 2 ( , ) ( 2 0) (1 0) ( 2) 1 4 1 5 d P P            21. 2 2 1 2 2 2 ( , ) ( 2 1) (2 1) ( 3) 1 9 1 10 d P P            22.   2 2 1 2 2 2 ( , ) 2 ( 1) (2 1) 3 1 9 1 10 d P P           23.       2 2 1 2 2 2 ( , ) (5 3) 4 4 2 8 4 64 68 2 17 d P P            24.         2 2 1 2 2 2 ( , ) 2 1 4 0 3 4 9 16 25 5 d P P            25.   2 2 1 2 2 2 ( , ) 4 ( 7) (0 3) 11 ( 3) 121 9 130            d P P 26.     2 2 1 2 2 2 ( , ) 4 2 2 ( 3) 2 5 4 25 29 d P P           27.   2 2 1 2 2 2 ( , ) (6 5) 1 ( 2) 1 3 1 9 10           d P P 28.     2 2 1 2 2 2 ( , ) 6 ( 4) 2 ( 3) 10 5 100 25 125 5 5 d P P             29.     2 2 1 2 2 2 ( , ) 2.3 ( 0.2) 1.1 (0.3) 2.5 0.8 6.25 0.64 6.89 2.62            d P P 30.     2 2 1 2 2 2 ( , ) 0.3 1.2 1.1 2.3 ( 1.5) ( 1.2) 2.25 1.44 3.69 1.92              d P P 31. 2 2 1 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) d P P a b a b a b           32. 2 2 1 2 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) 2 2 d P P a a a a a a a a             33. ( 2,5), (1,3), ( 1, 0) A B C            2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 1 ( 2) (3 5) 3 ( 2) 9 4 13 ( , ) 1 1 (0 3) ( 2) ( 3) 4 9 13 ( , ) 1 ( 2) (0 5) 1 ( 5) 1 25 26 d A B d B C d A C                                    Section 1.1: The Distance and Midpoint Formulas 3 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 13 13 26 13 13 26 26 26 d A B d B C d A C        The area of a triangle is 1 2 A bh   . In this problem,     1 ( , ) ( , ) 2 1 1 13 13 13 2 2 13 square units 2 A d A B d B C          34. ( 2, 5), (12, 3), (10, 11) A B C            2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 12 ( 2) (3 5) 14 ( 2) 196 4 200 10 2 ( , ) 10 12 ( 11 3) ( 2) ( 14) 4 196 200 10 2 ( , ) 10 ( 2) ( 11 5) 12 ( 16) 144 256 400 20 d A B d B C d A C                                       Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 10 2 10 2 20 200 200 400 400 400 d A B d B C d A C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 10 2 10 2 2 1 100 2 100 square units 2 A d A B d B C           35. ( 5,3), (6, 0), (5,5) A B C           2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 6 ( 5) (0 3) 11 ( 3) 121 9 130 ( , ) 5 6 (5 0) ( 1) 5 1 25 26 ( , ) 5 ( 5) (5 3) 10 2 100 4 104 2 26 d A B d B C d A C                                 Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: Chapter 1: Graphs and Functions 4             2 2 2 2 2 2 ( , ) ( , ) ( , ) 104 26 130 104 26 130 130 130 d A C d B C d A B        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 104 26 2 1 2 26 26 2 1 2 26 2 26 square units A d A C d B C              36. ( 6, 3), (3, 5), ( 1, 5) A B C             2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 3 ( 6) ( 5 3) 9 ( 8) 81 64 145 ( , ) 1 3 (5 ( 5)) ( 4) 10 16 100 116 2 29 ( , ) 1 ( 6) (5 3) 5 2 25 4 29 d A B d B C d A C                                     Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:             2 2 2 2 2 2 ( , ) ( , ) ( , ) 29 2 29 145 29 4 29 145 29 116 145 145 145 d A C d B C d A B           The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 29 2 29 2 1 2 29 2 29 square units A d A C d B C           37. (4, 3), (0, 3), (4, 2) A B C              2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (0 4) 3 ( 3) ( 4) 0 16 0 16 4 ( , ) 4 0 2 ( 3) 4 5 16 25 41 ( , ) (4 4) 2 ( 3) 0 5 0 25 25 5 d A B d B C d A C                                   Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: Section 1.1: The Distance and Midpoint Formulas 5         2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 5 41 16 25 41 41 41 d A B d A C d B C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 4 5 2 10 square units A d A B d A C        38. (4, 3), (4, 1), (2, 1) A B C             2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (4 4) 1 ( 3) 0 4 0 16 16 4 ( , ) 2 4 1 1 ( 2) 0 4 0 4 2 ( , ) (2 4) 1 ( 3) ( 2) 4 4 16 20 2 5 d A B d B C d A C                                   Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:         2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 2 2 5 16 4 20 20 20 d A B d B C d A C        The area of a triangle is 1 2 A bh  . In this problem,     1 ( , ) ( , ) 2 1 4 2 2 4 square units A d A B d B C        39. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 4 3 5 , 2 2 8 0 , 2 2 (4, 0) x x y y x y                            40. The coordinates of the midpoint are:   1 2 1 2 ( , ) , 2 2 2 2 0 4 , 2 2 0 4 , 2 2 0, 2 x x y y x y                            41. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 1 8 4 0 , 2 2 7 4 , 2 2 7 , 2 2                                  x x y y x y Chapter 1: Graphs and Functions 6 42. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 2 4 3 2 , 2 2 6 1 , 2 2 1 3, 2 x x y y x y                                    43. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 7 9 5 1 , 2 2 4 16 , 2 2 (8, 2)                              x x y y x y 44. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 2 3 2 , 2 2 2 1 , 2 2 1 1, 2 x x y y x y                                       45. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a b a b                          46. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a a a a                          47. The x coordinate would be 2 3 5   and the y coordinate would be 5 2 3   . Thus the new point would be   5,3 . 48. The new x coordinate would be 1 2 3     and the new y coordinate would be 6 4 10   . Thus the new point would be   3,10  49. a. If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length. One of the legs of the triangle will be 2+3=5. Thus the other leg will be: 2 2 2 2 2 5 13 25 169 144 12 b b b b       Thus the coordinates will have an y value of 1 12 13     and 1 12 11    . So the points are   3,11 and   3, 13  . b. Consider points of the form   3, y that are a distance of 13 units from the point   2, 1   .             2 2 2 1 2 1 2 2 2 2 2 2 3 ( 2) 1 5 1 25 1 2 2 26 d x x y y y y y y y y                           2 2 2 2 2 2 13 2 26 13 2 26 169 2 26 0 2 143 0 11 13 y y y y y y y y y y                11 0 11 y y    or 13 0 13 y y     Thus, the points   3,11 and   3, 13  are a distance of 13 units from the point   2, 1   . Section 1.1: The Distance and Midpoint Formulas 7 50. a. If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length. One of the legs of the triangle will be 2+6=8. Thus the other leg will be: 2 2 2 2 2 8 17 64 289 225 15 b b b b       Thus the coordinates will have an x value of 1 15 14    and 1 15 16   . So the points are   14, 6   and   16, 6  . b. Consider points of the form   , 6 x  that are a distance of 17 units from the point   1, 2 .             2 2 2 1 2 1 2 2 2 2 2 2 1 2 6 2 1 8 2 1 64 2 65 d x x y y x x x x x x x                          2 2 2 2 2 2 17 2 65 17 2 65 289 2 65 0 2 224 0 14 16 x x x x x x x x x x                14 0 14 x x     or 16 0 16 x x    Thus, the points   14, 6   and   16, 6  are a distance of 13 units from the point   1, 2 . 51. Points on the x -axis have a y -coordinate of 0. Thus, we consider points of the form   , 0 x that are a distance of 6 units from the point   4, 3  .           2 2 2 1 2 1 2 2 2 2 2 2 4 3 0 16 8 3 16 8 9 8 25 d x x y y x x x x x x x                        2 2 2 2 2 2 2 6 8 25 6 8 25 36 8 25 0 8 11 ( 8) ( 8) 4(1)( 11) 2(1) 8 64 44 8 108 2 2 8 6 3 4 3 3 2 x x x x x x x x x                             4 3 3 x   or 4 3 3 x   Thus, the points   4 3 3, 0  and   4 3 3, 0  are on the x -axis and a distance of 6 units from the point   4, 3  . 52. Points on the y -axis have an x -coordinate of 0. Thus, we consider points of the form   0, y that are a distance of 6 units from the point   4, 3  .         2 2 2 1 2 1 2 2 2 2 2 2 4 0 3 4 9 6 16 9 6 6 25 d x x y y y y y y y y y                     Chapter 1: Graphs and Functions 8   2 2 2 2 2 2 2 6 6 25 6 6 25 36 6 25 0 6 11 ( 6) (6) 4(1)( 11) 2(1) 6 36 44 6 80 2 2 6 4 5 3 2 5 2 y y y y y y y y y                               3 2 5 y    or 3 2 5 y    Thus, the points   0, 3 2 5   and   0, 3 2 5   are on the y- axis and a distance of 6 units from the point   4, 3  . 53. a. To shift 3 units left and 4 units down, we subtract 3 from the x -coordinate and subtract 4 from the y -coordinate.     2 3,5 4 1,1     b. To shift left 2 units and up 8 units, we subtract 2 from the x -coordinate and add 8 to the y -coordinate.     2 2,5 8 0,13    54. Let the coordinates of point B be   , x y . Using the midpoint formula, we can write   1 8 2,3 , 2 2 x y           . This leads to two equations we can solve. 1 2 2 1 4 5 x x x        8 3 2 8 6 2 y y y       Point B has coordinates   5, 2  . 55.   1 2 1 2 , , 2 2 x x y y M x y           .   1 1 1 , ( 3, 6) P x y    and ( , ) ( 1, 4) x y   , so 1 2 2 2 2 2 3 1 2 2 3 1 x x x x x x            and 1 2 2 2 2 2 6 4 2 8 6 2 y y y y y y        Thus, 2 (1, 2) P  . 56.   1 2 1 2 , , 2 2 x x y y M x y           .   2 2 2 , (7, 2) P x y    and ( , ) (5, 4) x y   , so 1 2 1 1 1 2 7 5 2 10 7 3 x x x x x x        and 1 2 1 1 1 2 ( 2) 4 2 8 ( 2) 6 y y y y y y             Thus, 1 (3, 6) P   . 57. The midpoint of AB is:   0 6 0 0 , 2 2 3, 0 D           The midpoint of AC is:   0 4 0 4 , 2 2 2, 2 E           The midpoint of BC is:   6 4 0 4 , 2 2 5, 2 F             2 2 2 2 ( , ) 0 4 (3 4) ( 4) ( 1) 16 1 17 d C D              2 2 2 2 ( , ) 2 6 (2 0) ( 4) 2 16 4 20 2 5 d B E            2 2 2 2 ( , ) (2 0) (5 0) 2 5 4 25 29 d A F          Section 1.1: The Distance and Midpoint Formulas 9 58. Let 1 2 (0, 0), (0, 4), ( , ) P P P x y          2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 , (0 0) (4 0) 16 4 , ( 0) ( 0) 4 16 , ( 0) ( 4) ( 4) 4 ( 4) 16 d P P d P P x y x y x y d P P x y x y x y                             Therefore,   2 2 2 2 4 8 16 8 16 2 y y y y y y y        which gives 2 2 2 2 16 12 2 3 x x x      Two triangles are possible. The third vertex is     2 3, 2 or 2 3, 2  . 59. 2 2 1 2 2 2 ( , ) ( 4 2) (1 1) ( 6) 0 36 6 d P P             2 2 2 3 2 2 ( , ) 4 ( 4) ( 3 1) 0 ( 4) 16 4 d P P             2 2 1 3 2 2 ( , ) ( 4 2) ( 3 1) ( 6) ( 4) 36 16 52 2 13 d P P               Since       2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. 60.   2 2 1 2 2 2 ( , ) 6 ( 1) (2 4) 7 ( 2) 49 4 53 d P P              2 2 2 3 2 2 ( , ) 4 6 ( 5 2) ( 2) ( 7) 4 49 53 d P P               2 2 1 3 2 2 ( , ) 4 ( 1) ( 5 4) 5 ( 9) 25 81 106 d P P             Since       2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. Since     1 2 2 3 , , d P P d P P  , the triangle is isosceles. Therefore, the triangle is an isosceles right triangle. 61.     2 2 1 2 2 2 ( , ) 0 ( 2) 7 ( 1) 2 8 4 64 68 2 17 d P P               2 2 2 3 2 2 ( , ) 3 0 (2 7) 3 ( 5) 9 25 34 d P P               2 2 1 3 2 2 ( , ) 3 ( 2) 2 ( 1) 5 3 25 9 34 d P P            Since 2 3 1 3 ( , ) ( , ) d P P d P P  , the triangle is isosceles. Since       2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle. Chapter 1: Graphs and Functions 10 62.     2 2 1 2 2 2 ( , ) 4 7 0 2 ( 11) ( 2) 121 4 125 5 5 d P P                2 2 2 3 2 2 ( , ) 4 ( 4) (6 0) 8 6 64 36 100 10 d P P                2 2 1 3 2 2 ( , ) 4 7 6 2 ( 3) 4 9 16 25 5 d P P            Since       2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P   , the triangle is a right triangle. 63. Using the Pythagorean Theorem: 2 2 2 2 2 90 90 8100 8100 16200 16200 90 2 127.28 feet d d d d         90 90 90 90 d 64. Using the Pythagorean Theorem: 2 2 2 2 2 60 60 3600 3600 7200 7200 60 2 84.85 feet d d d d          60 60 60 60 d 65. a. First: (90, 0), Second: (90, 90), Third: (0, 90) (0,0) (0,90) (90,0) (90,90) X Y b. Using the distance formula: 2 2 2 2 (310 90) (15 90) 220 ( 75) 54025 5 2161 232.43 feet d           c. Using the distance formula: 2 2 2 2 (300 0) (300 90) 300 210 134100 30 149 366.20 feet d          66. a. First: (60, 0), Second: (60, 60) Third: (0, 60) (0,0) (0,60) (60,0) (60,60) x y b. Using the distance formula: 2 2 2 2 (180 60) (20 60) 120 ( 40) 16000 40 10 126.49 feet d           c. Using the distance formula: 2 2 2 2 (220 0) (220 60) 220 160 74000 20 185 272.03 feet d          Section 1.1: The Distance and Midpoint Formulas 11 67. The Focus heading east moves a distance 60 t after t hours. The truck heading south moves a distance 40 t after t hours. Their distance apart after t hours is: 2 2 2 2 2 (60 ) (45 ) 3600 2025 5625 75 miles       d t t t t t t 68. 15 miles 5280 ft 1 hr 22 ft/sec 1 hr 1 mile 3600 sec      2 2 2 100 22 10000 484 feet d t t     100 22 t d 69. a. The shortest side is between 1 (2.6, 1.5) P  and 2 (2.7, 1.7) P  . The estimate for the desired intersection point is:   1 2 1 2 2.6 2.7 1.5 1.7 , , 2 2 2 2 5.3 3.2 , 2 2 2.65, 1.6 x x y y                          b. Using the distance formula: 2 2 2 2 (2.65 1.4) (1.6 1.3) (1.25) (0.3) 1.5625 0.09 1.6525 1.285 units d           70. Let 1 (2013, 102.87)  P and 2 (2017, 126.17)  P . The midpoint is:     1 2 1 2 , , 2 2 2013 2017 102.87 126.17 , 2 2 4030 229.04 , 2 2 2015, 114.52                           x x y y x y The estimate for 2010 is $114.52 billion. The estimate net sales of Costco Wholesale Corporation in 2015 is $0.85 billion off from the reported value of $113.67 billion. 71. For 2009 we have the ordered pair   2009, 21756 and for 2017 we have the ordered pair   2017, 24858 . The midpoint is     2009 2017 21756 24858 year, $ , 2 2 4026 46614 , 2 2 2013, 23307                  Using the midpoint, we estimate the poverty level in 2013 to be $23,307. This is lower than the actual value. 72. Let   1 0, 0 P  ,   2 , 0 P a  , and 3 3 , 2 2 a a P          . Then           2 2 1 2 2 1 2 1 2 2 2 , 0 0 0 d P P x x y y a a a                 2 2 2 3 2 1 2 1 2 2 2 2 2 2 , 3 0 2 2 3 4 4 4 4 d P P x x y y a a a a a a a a                            d 45t 60t

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