Section 6.1 Solutions
1. The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.
3. When the known values are the side opposite the missing angle and another side and its opposite angle.
5. A triangle with two given sides and a non-included angle.
7. β=72∘,a≈12.0,b≈19.9
9. γ=20∘,b≈4.5,c≈1.6
11. b≈3.78
13. c≈13.70
15. one triangle, α≈50.3∘,β≈16.7∘,a≈26.7
17. two triangles, γ≈54.3∘,β≈90.7∘,b≈20.9 or γ′≈125.7∘,β′≈19.3∘,b′≈6.9
19. two triangles, β≈75.7∘,γ≈61.3∘,b≈9.9 or β′≈18.3∘,γ′≈118.7∘,b′≈3.2
21. two triangles, α≈143.2∘,β≈26.8∘,a≈17.3 or α′≈16.8∘,β′≈153.2∘,a′≈8.3
23. no triangle possible
25. A≈47.8∘ or A′≈132.2∘
27. 8.6
29. 370.9
31. 12.3
33. 12.2
35. 16.0
37. 29.7∘
39. x=76.9∘or x=103.1∘
41. 110.6∘
43. A≈39.4, C≈47.6, BC≈20.7
45. 57.1
47. 42.0
49. 430.2
51. 10.1
53. AD≈ 13.8
55. AB≈2.8
57. L≈49.7, N≈56.3, LN≈5.8
59. 51.4 feet
61. The distance from the satellite to station A is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.
63. 2.6 ft
65. 5.6 km
67. 371 ft
69. 5936 ft
71. 24.1 ft
73. 19,056 ft2
75. 445,624 square miles
77. 8.65 ft2
Section 6.2 Solutions
1. two sides and the angle opposite the missing side
3. s is the semi-perimeter, which is half the perimeter of the triangle.
5. The Law of Cosines must be used for any oblique (non-right) triangle.
7. 11.3
9. 34.7
11. 26.7
13. 257.4
15. not possible
17. 95.5°
19. 26.9°
21. B≈45.9∘,C≈99.1∘,a≈6.4
23. A≈20.6∘,B≈38.4∘,c≈51.1
25. A≈37.8∘,B≈43.8,C≈98.4∘
27. 177.56 in2
29. 0.04 m2
31. 0.91 yd2
33. 3.0
35. 29.1
37. 0.5
39. 70.7°
41. 77.4°
43. 25.0
45. 9.3
47. 43.52
49. 1.41
51. 0.14
53. 18.3
55. 48.98
57.
59. 7.62
61. 85.1
63. 24.0 km
65. 99.9 ft
67. 37.3 miles
69. 2371 miles
71.
73. 599.8 miles
75. 65.4 cm2
77. 468 ft2
Section 6.3 Solutions
1. For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.
3. Determine θ for the point, then move r units from the pole to plot the point. If r is negative, move r units from the pole in the opposite direction but along the same angle. The point is a distance of r away from the origin at an angle of θ from the polar axis.
5. The point (−3,π2) has a positive angle but a negative radius and is plotted by moving to an angle of π2 and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point (3,−π2) has a negative angle and a positive radius and is plotted by first moving to an angle of −π2 and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y-axis.
7.
a) (5,−4π3)
b) (−5,5π3)
c) (5,8π3)
9.
a) (3,−5π4)
b) (−3,7π4)
c) (3,11π4)
11.
a) (4,−135∘)
b) (−4,45∘)
c) (4,585∘)
13.
a) (5,−60∘)
b) (−5,120∘)
c) (5,480∘)
15. (−5,0)
17. (−3√32,−32)
19. (2√5,0.464)
21. (√34,5.253)
23. (8√2,π4)
25. r=4cscθ
27. r=3√sinθ2cos4θ
29. r=3cosθ
31. r=3sinθcos(2θ)
33. r=9sinθcos2θ
35. r=√19cosθsinθ
37. x2+y2=4x or (x−2)24+y24=1; circle
39. 3y+x=6; line
41. y=3; line
43. xy=4; hyperbola
45. x2+y2=4; circle
47. x−5y=3; line
49. (3,3π4)
51. (5,π)
53.
55.
57.
59.
61.
63. r=65cosθ−sinθ
65. r=2sinθ
67. r=2cosθ
69. r=3cosθ
71. x2+y2=16
73. y=x
75. x2+(y+5)2=25
77. A vertical line with a units left of the y-axis.
79. A horizontal line with a units below the x-axis.
Section 6.4 Solutions
1. Symmetry with respect to the polar axis is similar to symmetry about the x -axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line θ=π2 is similar to symmetry about the y -axis.
3. Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at θ=0,π2,πand 3π2, and sketch the graph.
5. The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.
7. symmetric with respect to the polar axis
9. symmetric with respect to the polar axis, symmetric with respect to the line θ=π2, symmetric with respect to the pole
11. no symmetry
13. no symmetry
15. symmetric with respect to the pole
17. circle
19. cardioid
21. cardioid
23. one-loop/dimpled limaçon
25. one-loop/dimpled limaçon
27. inner loop/two-loop limaçon
29. inner loop/two-loop limaçon
31. inner loop/two-loop limaçon
33. lemniscate
35. lemniscate
37. rose curve
39. rose curve
41. Archimedes’ spiral
43. Archimedes’ spiral
45.
47.
49.
51.
53.
55. They are both spirals, but not quite the same.
57. Both graphs are curves with 2 loops. The equation with a coefficient of θ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to 4π to get a better picture.
59. When the width of the domain is increased, more petals of the flower are visible.
61. The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.
63. The graphs are spirals. The smaller the coefficient, the tighter the spiral.
65. (4,π3),(4,5π3)
67. (32,π3),(32,5π3)
69. (0,π2),(0,π),(0,3π2),(0,2π)
71. (4√82,π4),(4√82,5π4)
and at θ=3π4,7π4 since r is squared
Section 6.5 Solutions
1. a is the real part, b is the imaginary part, and i=√−1
3. Polar form converts the real and imaginary part of the complex number in polar form using x=rcosθ and y=rsinθ
5. zn=rn(cos(nθ)+isin(nθ)). It is used to simplify polar form when a number has been raised to a power.
7. 5√2
9. √38
11. √14.45
13. 4√5cis(333.4∘)
15. 2cis(π6)
17. 7√32+i72
19. −2√3−2i
21. −1.5−i3√32
23. 4√3cis(198∘)
25. 34cis(180∘)
27. 5√3cis(17π24)
29. 7cis(70∘)
31. 5cis(80∘)
33. 5cis(π3)
35. 125cis(135∘)
37. 9cis(240∘)
39. cis(3π4)
41. 3cis(80∘), 3cis(200∘), 3cis(320∘)
43. 23√4cis(2π9), 23√4cis(8π9), 23√4cis(14π9)
45. 2√2cis(7π8), 2√2cis(15π8)
47.
49.
51.
53.
55.
57. 3.61e−0.59i
59. −2+3.46i
61. −4.33−2.50i
Section 6.6 Solutions
1. lowercase, bold letter, usually u, v, w
3. They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.
5. The first number always represents the coefficient of the i, and the second represents the j.
7. ⟨7,−5⟩
9. not equal
11. equal
13. equal
15. 7i−3j
17. −6i−2j
19. u+v=⟨−5,5⟩,u−v=⟨−1,3⟩,2u−3v=⟨0,5⟩
21. −10\boldsymbol{i}–4\boldsymbol{j}
23. −\frac{2\sqrt{29}}{29}\boldsymbol{i}+\frac{5\sqrt{29}}{29}\boldsymbol{j}
25. –\frac{2\sqrt{229}}{229}\boldsymbol{i}+\frac{15\sqrt{229}}{229}\boldsymbol{j}
27. –\frac{7\sqrt{2}}{10}\boldsymbol{i}+\frac{\sqrt{2}}{10}\boldsymbol{j}
29. |\boldsymbol{v}|=7.810,\theta=39.806^{\circ}
31. |\boldsymbol{v}|=7.211,\theta=236.310^{\circ}
33. −6
35. −12
37.
39.
41.
43.
45.
47. \langle 4,1\rangle
49. \boldsymbol{v}=−7\boldsymbol{i}+3\boldsymbol{j}
51. 3\sqrt{2}\boldsymbol{i}+3\sqrt{2}\boldsymbol{j}
53. \boldsymbol{i}−\sqrt{3}\boldsymbol{j}
55. Magnitude: 29.05 pounds, Direction: 130.44 degrees
57. Magnitude: 8.29 pounds, Direction: -44.56 degrees
59. a. 58.7; b. 12.5
61. x=7.13 pounds, y=3.63 pounds
63. x=2.87 pounds, y=4.10 pounds
65. 4.635 miles, 17.764^{\circ} N of E
67. 17 miles. 10.318 miles
69. Distance: 2.868. Direction: 86.474^{\circ} North of West, or 3.526^{\circ} West of North
71. 4.924^{\circ}. 659 km/hr
73. 4.424^{\circ}
75. (0.081, 8.602)
77. 21.801^{\circ}, relative to the car’s forward direction
79. parallel: 16.28, perpendicular: 47.28 pounds
81. 19.35 pounds, 231.54^{\circ} from the horizontal
83. 5.1583 pounds, 75.8^{\circ} from the horizontal