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Study Guide and Review - Chapter 4 State whether each sentence is true or false . If false , replace the underlined term to make a true sentence. 1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. 5. The rate at which an object moves along a circular path is called its linear speed. SOLUTION: The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed. SOLUTION: The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg. 6. 0 , SOLUTION: 0 , 7. The period of the graph of y = 4 sin 3x is 4. The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio. 3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line. SOLUTION: The amplitude of the graph of y = 4 sin 3x is 4. The period is . The frequency is . 8. For f (x) = cos bx, as b increases, the frequency decreases. SOLUTION: SOLUTION: An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression is the angle formed by a horizontal line and an observer’s line of sight to an object below the line. are examples of quadrantal angles. , and Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles. 2. The secant ratio is the reciprocal of the sine ratio. SOLUTION: are examples of reference angles. , and For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequency and period are reciprocals. 9. The range of the arcsine function is [0, ]. SOLUTION: 4. The radian measure of an angle is equal to the ratio The range of the arccosine function is [0, ], while the range of the arcsine function is . of the length of its intercepted arc to the radius. SOLUTION: The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle. 10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. SOLUTION: The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS are known, the Law of Cosines will be needed as well. Find the exact values of the six trigonometric functions of θ. 5. The rate at which an object moves along a circular eSolutions Manual - Powered by Cognero Page 1 path is called its linear speed. SOLUTION: 11. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSAGuide are known. SSS and4SAS are Study andSometimes, Review when - Chapter known, the Law of Cosines will be needed as well. Find the exact values of the six trigonometric functions of θ. 11. Find the value of x. Round to the nearest tenth, if necessary. 13. SOLUTION: SOLUTION: The length of the side opposite is 36, the length of the side adjacent to is 15, and the length of the hypotenuse is 39. An acute angle measure and the length of the hypotenuse are given, so the sine function can be used to find the length of the side opposite . 14. SOLUTION: An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse. 12. SOLUTION: The length of the side opposite θ is 9, the length of the side adjacent to is 40, and the length of the hypotenuse is 41. Find the measure of angle θ. Round to the nearest degree, if necessary. 15. SOLUTION: Because the length of the side opposite θ and the hypotenuse are given, use the sine function. Find the value of x. Round to the nearest tenth, if necessary. eSolutions Manual - Powered by Cognero 13. Page 2 Study Guide and Review - Chapter 4 Find the measure of angle θ. Round to the nearest degree, if necessary. 18. 450 SOLUTION: To convert a degree measure to radians, multiply by 15. SOLUTION: Because the length of the side opposite θ and the hypotenuse are given, use the sine function. 19. SOLUTION: To convert a radian measure to degrees, multiply by 16. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, use the tangent function. 20. SOLUTION: To convert a radian measure to degrees, multiply by Write each degree measure in radians as a multiple of π and each radian measure in degrees. 17. 135 SOLUTION: To convert a degree measure to radians, multiply by Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle. 21. 342 SOLUTION: 18. 450 SOLUTION: To convert a degreebymeasure eSolutions Manual - Powered Cognero to radians, multiply by All angles measuring are coterminal with a 342 angle. Page 3 Sample answer: Let n = 1 and −1. Study Guide and Review - Chapter 4 Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle. 21. 342 22. SOLUTION: SOLUTION: All angles measuring All angles measuring are coterminal with a 342 angle. Sample answer: Let n = 1 and −1. are coterminal with a radian angle. Sample answer: Let n = 1 and −1. 22. Find the area of each sector. SOLUTION: All angles measuring are coterminal with a radian angle. Sample answer: Let n = 1 and −1. eSolutions Manual - Powered by Cognero 23. SOLUTION: The measure of the sector’s central angle is 31° and the radius is 14 inches. Convert the central angle measure to radians. Use the central angle and the radius to find the area Page 4 of the sector. Therefore, the area of the sector is about 246.1 square meters. Study Guide and Review - Chapter 4 Find the area of each sector. Sketch each angle. Then find its reference angle. 25. 240 SOLUTION: 23. The terminal side of 240º lies in Quadrant III. Therefore, its reference angle is ' = 240 – 180 or 60º. SOLUTION: The measure of the sector’s central angle is 31° and the radius is 14 inches. Convert the central angle measure to radians. Use the central angle and the radius to find the area of the sector. 26. 75 SOLUTION: The terminal side of 75 lies in Quadrant I. Therefore, its reference angle is ' = 75 – 0 or 75 . Therefore, the area of the sector is about 53.0 square inches. 24. SOLUTION: The measure of the sector’s central angle is and the radius is 10 meters. 27. SOLUTION: Use the central angle and the radius to find the area of the sector. The terminal side of lies in Quadrant III. Therefore, its reference angle is ' = . Therefore, the area of the sector is about 246.1 square meters. Sketch each angle. Then find its reference angle. 25. 240 eSolutions Manual - Powered by Cognero SOLUTION: The terminal side of 240º lies in Quadrant III. 28. Page 5 Study Guide and Review - Chapter 4 Find the exact values of the five remaining trigonometric functions of θ. 27. 29. cos SOLUTION: The terminal side of lies in Quadrant III. Therefore, its reference angle is ' = = , where sin > 0 and tan > 0 SOLUTION: To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ and tan are positive, must lie in Quadrant I. This means that both x and y are positive. . = Because cos or , x =2 and r = 5. Find y. 28. Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios. SOLUTION: The terminal side of lies in Quadrant II. Therefore, its reference angle is ' = . 30. where sin > 0 and cos < 0 SOLUTION: Find the exact values of the five remaining trigonometric functions of θ. 29. cos = , where sin > 0 and tan > 0 To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ is positive and cos is negative, must lie in Quadrant II. This means that x is negative and y is positive. Because tan = or – , use the point (–4, 3) to find r. SOLUTION: To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ and tan are positive, must lie in Quadrant I. This means that both x and y are positive. eSolutions Manualcos - Powered Cognero Because = by or ,x =2 and r = 5. Find y. Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios. Page 6 Study Guide and Review - Chapter 4 30. where sin > 0 and cos < 0 SOLUTION: = or – , use the point (–4, 3) to find r. To find the other function values, find the coordinates of a point on the terminal side of . Because cos θ is positive and cot θ is negative, must lie in Quadrant IV. This means that x is positive and y is negative. = or – , y = –5 and r = 13. Find x. Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios. where cos Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios. > 0 and cot < 0 32. cot θ = SOLUTION: where cos > 0 and cot < 0 To find the other function values, find the coordinates of a point on the terminal side of . Because cos θ is positive and cot θ is negative, must lie in Quadrant IV. This means that x is positive and y is negative. Because sin > 0 and cot < 0 where cos Because sin 31. > 0 and cot < 0 where cos SOLUTION: To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ is positive and cos is negative, must lie in Quadrant II. This means that x is negative and y is positive. Because tan 31. = or – , y = –5 and r = 13. Find x. , where sin < 0 and tan > 0 SOLUTION: To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ is negative and tan is positive, must lie in Quadrant III. This means that both x and y are negative. Because cot = or , use the point (–2, –3) to find r. Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios. eSolutions Manual - Powered by Cognero Use x = –2, y = –3, and r = remaining trigonometric ratios. to write the five Page 7 Study Guide and Review - Chapter 4 32. cot θ = , where sin < 0 and tan > 0 34. cot SOLUTION: SOLUTION: Because the terminal side of lies in Quadrant IV, the reference angle θ' of is To find the other function values, find the coordinates of a point on the terminal side of . Because sin θ is negative and tan is positive, must lie in Quadrant III. This means that both x and y are negative. . Because cot = or , use the point (–2, –3) to find r. In quadrant IV cot θ is negative. Use x = –2, y = –3, and r = remaining trigonometric ratios. to write the five Find the exact value of each expression. If undefined, write undefined. 33. sin 180 35. sec 450 SOLUTION: Because the terminal side of lies on the positive yaxis, the reference angle ' is . SOLUTION: Because the terminal side of lies on the negative x-axis, the reference angle ' is 180°. 34. cot SOLUTION: Because the terminal side of lies in Quadrant IV, the reference angle θ' of is eSolutions Manual - Powered by Cognero . 36. SOLUTION: Page 8 Because the terminal side of θ lies in Quadrant III, Study Guide and Review - Chapter 4 36. SOLUTION: Because the terminal side of θ lies in Quadrant III, the reference angle θ' of is . Describe how the graphs of f (x)and g(x) are related. Then find the amplitude and period of g(x) , and sketch at least one period of both functions on the same coordinate axes. 37. f (x) = sin x; g(x) = 5 sin x SOLUTION: The graph of g(x) is the graph of f (x) expanded vertically. The amplitude of g(x) is |5| or 5, and the period is or 2 . Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph. Describe how the graphs of f (x)and g(x) are related. Then find the amplitude and period of g(x) , and sketch at least one period of both functions on the same coordinate axes. 37. f (x) = sin x; g(x) = 5 sin x Functions f (x) = sin x g(x) = 5 sin x x-int (0, 0) (0, 0) (π, 0) (π, 0) (2π, 0) (2π, 0) Max x-int SOLUTION: Min The graph of g(x) is the graph of f (x) expanded vertically. The amplitude of g(x) is |5| or 5, and the x-int period is or 2 . Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph. Sketch the curve through the indicated points for each function. Then repeat the pattern to complete a second period. Functions f (x) = sin x g(x) = 5 sin x x-int (0, 0) (0, 0) Max 38. f (x) = cos x; g(x) = cos 2x x-int (π, 0) eSolutions Manual - Powered by Cognero Min (π, 0) SOLUTION: Page 9 The graph of g(x) is the graph of f (x) compressed horizontally. The amplitude of g(x) is |1| or 1, and the Study Guide and Review - Chapter 4 38. f (x) = cos x; g(x) = cos 2x 39. f (x) = sin x; g(x) = sin x SOLUTION: The graph of g(x) is the graph of f (x) compressed horizontally. The amplitude of g(x) is |1| or 1, and the SOLUTION: period is vertically. The amplitude of g(x) is or . Create a table listing the coordinates of the xintercepts and extrema for f (x) = cos x for one period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph. Functions f (x) = cos x g(x) = cos 2x Max (0, 1) (0, 1) x-int Min (π, –1) x-int Max (2π, 1) (π, 1) The graph of g(x) is the graph of f (x) compressed the period is or , and or 2 . Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph. g(x) = Functions f (x) = sin x x-int (0, 0) (0, 0) (π, 0) (π, 0) (2π, 0) (2π, 0) sin x Max x-int Min Sketch the curve through the indicated points for each function. Then repeat the pattern to complete a second period. x-int Sketch the curve through the indicated points for each function. Then repeat the pattern to complete a second period. 39. f (x) = sin x; g(x) = sin x SOLUTION: The graph of g(x) is the graph of f (x) compressed vertically. The amplitude of g(x) is or , and the period or by 2 Cognero . eSolutions Manual -isPowered Create a table listing the coordinates of the x- 40. f (x) = cos x; g(x) = −cos SOLUTION: x Page 10 The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The Study Guide and Review - Chapter 4 40. f (x) = cos x; g(x) = −cos State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 41. y = 2 cos (x – ) x SOLUTION: The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The SOLUTION: amplitude of g(x) is |–1| or 1, and the period is or In this function, a = 2, b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. 6 . Create a table listing the coordinates of the xintercepts and extrema for f (x) = cos x for one period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph. g(x) = –cos Functions f (x) = cos x x Max (0, 1) (0, −1) (π, –1) (3π, 1) (2π, 1) (6π, −1) x-int Min x-int Max Graph y = 2 cos x shifted units to the right. Sketch the curve through the indicated points for each function. Then repeat the pattern to complete a second period. 42. y = −sin 2x + 1 SOLUTION: In this function, a = , b = 2, c = 0, and d = 1. State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 41. y = 2 cos (x – ) SOLUTION: eSolutions Manual - Powered by Cognero In this function, a = 2, b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. Graph y = – sin 2x shifted 1 unit up. Page 11 Study Guide and Review - Chapter 4 42. y = −sin 2x + 1 43. SOLUTION: In this function, a = , b = 2, c = 0, and d = 1. SOLUTION: In this function, a = , b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. Graph y = – sin 2x shifted 1 unit up. Graph y = cos x shifted units to the left. 43. SOLUTION: In this function, a = , b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. 44. SOLUTION: In this function, a = 3, b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. eSolutions Manual - Powered by Cognero Graph y = cos x shifted Page 12 units to the left. Study Guide and Review - Chapter 4 44. Locate the vertical asymptotes, and sketch the graph of each function. 45. y = 3 tan x SOLUTION: The graph of y = 3 tan x is the graph of y = tan x SOLUTION: In this function, a = 3, b = 1, c = expanded vertically. The period is , and d = 0. or . Find the location of two consecutive vertical asymptotes. Because d = 0, there is no vertical shift. and Create a table listing the coordinates of key points for y = 3 tan x for one period on . Function Graph y = 3 sin x shifted Vertical Asymptote Intermediate Point x-int units to the left. y = tan x y = 3 tan x (0, 0) (0, 0) Intermediate Point Vertical Asymptote Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Locate the vertical asymptotes, and sketch the graph of each function. 45. y = 3 tan x SOLUTION: The graph of y = 3 tan x is the graph of y = tan x expanded vertically. The period is or . Find the location of two consecutive vertical asymptotes. and Create a table listing the coordinates of key points for y = 3 tan x for one period on eSolutions Manual - Powered by Cognero Function . 46. Page 13 SOLUTION: Asymptote Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Study Guide and Review - Chapter 4 46. SOLUTION: is the graph of y = The graph of tan x compressed vertically and translated units or . Find the location of right. The period is two consecutive vertical asymptotes. 47. and SOLUTION: Create a table listing the coordinates of key points for for one period on [0, ]. x shifted Function Vertical Asymptote Intermediate Point x-int units to the left. The period is or . Find the location of two consecutive vertical asymptotes. y = tan x x =0 (0, 0) Intermediate Point Vertical Asymptote is the graph of y = cot The graph of and Create a table listing the coordinates of key points for x=π for one period on . Function Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Vertical Asymptote Intermediate Point x-int y = cot x x =0 Intermediate Point Vertical Asymptote x=π eSolutions Manual - Powered by Cognero 47. Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Page 14 Study Guide and Review - Chapter 4 48. y = −cot (x – 47. ) SOLUTION: The graph of is the graph of y = cot x reflected in the x-axis and shifted π units to the SOLUTION: is the graph of y = cot The graph of or . Find the location of two left. The period is x shifted units to the left. The period is or . consecutive vertical asymptotes. Find the location of two consecutive vertical asymptotes. and Create a table listing the coordinates of key points for for one period on [ , 2 ]. and Function Create a table listing the coordinates of key points for for one period on Vertical Asymptote Intermediate Point x-int . Function y = cot x Vertical Asymptote Intermediate Point x-int x =0 Intermediate Point Vertical Asymptote x =0 x=π x = 2π Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Intermediate Point Vertical Asymptote y = cot x x=π Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. 49. SOLUTION: 48. y = −cot (x – ) SOLUTION: eSolutions Manual - Powered by Cognero The graph of is the graph of y = cot x reflected in the x-axis and shifted π units to the The graph of is the graph of y = sec x expanded vertically and expanded horizontally. Page The 15 period is or 4 . Find the location of two Study Guide and Review - Chapter 4 50. y = –csc (2x) 49. SOLUTION: The graph of is the graph of y = csc x compressed horizontally and reflected in the x-axis. SOLUTION: is the graph of y = sec x The graph of or . Find the location of two The period is expanded vertically and expanded horizontally. The consecutive vertical asymptotes. or 4 . Find the location of two period is and consecutive vertical asymptotes. Create a table listing the coordinates of key points for and for one period on . Create a table listing the coordinates of key points for Function for one period on [−π, 3π]. Vertical Asymptote Intermediate Point x-int Function Vertical Asymptote Intermediate Point x-int y = csc x y = sec x (0, 1) Intermediate Point Vertical Asymptote Intermediate Point Vertical Asymptote (0, 1) x=π y = –csc 2x x = −π x =0 x =0 x=π x = 3π Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. 51. y = sec (x – ) SOLUTION: The graph of y = sec (x − 50. y = –csc (2x) eSolutions Manual - Powered by Cognero SOLUTION: The graph of is the graph of y = csc x compressed horizontally and reflected in the x-axis. ) is the graph of y = sec x translated π units to the right. The period is or Page 16 2 . Find the location of two vertical asymptotes. 51. y = sec (x – ) Study Guide and Review - Chapter 4 SOLUTION: The graph of y = sec (x − 52. ) is the graph of y = sec x translated π units to the right. The period is or 2 . Find the location of two vertical asymptotes. SOLUTION: is the graph of y = The graph of csc x compressed vertically and translated the left. The period is and units to or 2 . Find the location of two consecutive vertical asymptotes. Create a table listing the coordinates of key points for y = sec (x − π) for one period on and . Function Vertical Asymptote Intermediate Point x-int y = sec x y = sec (x − π) Create a table listing the coordinates of key points for for one period on . (0, 1) Function Vertical Asymptote Intermediate Point x-int Intermediate Point Vertical Asymptote Intermediate Point Vertical Asymptote Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. y= csc x x =0 x=π Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve. 52. eSolutions Manual - Powered by Cognero Find the exact value of each expression, if Page it 17 exists. SOLUTION: The graph of is the graph of y = 53. sin−1 (−1) When t = Study Guide and Review - Chapter 4 –1 , cos t = . Therefore, cos = . Find the exact value of each expression, if it exists. 55. 53. sin−1 (−1) SOLUTION: SOLUTION: Find a point on the unit circle on the interval Find a point on the unit circle on the interval such that with a y-coordinate of –1. = When t = , sin t = –1. Therefore, sin –1 When t = –1 = . , tan t = = . Therefore, tan –1 . 56. arcsin 0 54. SOLUTION: SOLUTION: Find a point on the unit circle on the interval Find a point on the unit circle on the interval with a x-coordinate of with a y-coordinate of 0. . When t = 0, sin t = 0. Therefore, arcsin 0 = 0. When t = , cos t = –1 . Therefore, cos . = 57. arctan −1 SOLUTION: Find a point on the unit circle on the interval such that = 55. SOLUTION: eSolutions Manual - Powered by Cognero Find a point on the unit circle on the interval Page 18 When t = Study Guide and Review - Chapter 4 , cos t = . Therefore, arccos = . When t = 0, sin t = 0. Therefore, arcsin 0 = 0. 57. arctan −1 59. SOLUTION: Find a point on the unit circle on the interval such that = SOLUTION: The inverse property applies, because lies on the interval [–1, 1]. Therefore, = . 60. SOLUTION: The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore, When t = = , tan t = . Therefore, arctan Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree. 61. a = 11, b = 6, A = 22 . 58. arccos SOLUTION: Draw a diagram of a triangle with the given dimensions. SOLUTION: Find a point on the unit circle on the interval with a x-coordinate of . Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B. When t = , cos t = . Therefore, arccos = . 59. SOLUTION: The inverse property applies, because eSolutions Manual - Powered by Cognero the interval [–1, 1]. Therefore, lies on = Because two angles are now known, C 180 – (22 + 12 ) or about 146 . Apply the Law of Sines to find c. Page 19 SOLUTION: The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore, Study Guide and Review - Chapter 4 62. a = 9, b = 10, A = 42 SOLUTION: Draw a diagram of a triangle with the given dimensions. Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree. 61. a = 11, b = 6, A = 22 SOLUTION: Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h. Draw a diagram of a triangle with the given dimensions. Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B. 9 > 6.69, so two solutions exist. Apply the Law of Sines to find B. Because two angles are now known, C 180 – (22 + 12 ) or about 146 . Apply the Law of Sines to find c. Because two angles are now known, C 180 – (42 + 48 ) or about 90 . Apply the Law of Sines to find c. Therefore, the remaining measures of B 12 , C 146 , and c 16.4. are 62. a = 9, b = 10, A = 42 SOLUTION: Draw a diagram of a triangle with the given dimensions. When, then C 180 – (42 + 132 ) or about 6º. Apply the Law of Sines to find c. eSolutions Manual - Powered by Cognero Page 20 Study Guide and Review - Chapter 4 When, then C 180 – (42 + 132 ) or about 6º. Apply the Law of Sines to find c. Therefore, the remaining measures of B 29 , C 73 , and c 19.5. are 64. a = 2, b = 9, A = 88 SOLUTION: Notice that A is acute and a < b because 2 < 9. Find h. Therefore, the remaining measures of are B 48 , C 90 , and c 13.5 or B = 132 , C = 6 , and c = 1.4. Because a < b and a < h, no triangle can be formed with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution. Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. 65. a = 13, b = 12, c = 8 63. a = 20, b = 10, A = 78 SOLUTION: Draw a diagram of a triangle with the given dimensions. SOLUTION: Use the Law of Cosines to find an angle measure. Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B. Use the Law of Sines to find a missing angle measure. Because two angles are now known, C 180 – (78 + 29 ) or about 73 . Apply the Law of Sines to find c. Find the measure of the remaining angle. Therefore, A 78 , B 65 , and C 37 . 66. a = 4, b = 5, C = 96 Therefore, the remaining measures of B 29 , C 73 , and c 19.5. are 64. a = Manual 2, b = -9,Powered A = 88by Cognero eSolutions SOLUTION: Notice that A is acute and a < b because 2 < 9. Find SOLUTION: Use the Law of Cosines to find the missing side measure. Page 21 Find the measure of the remaining angle. Study GuideA and Therefore, , B 65 - ,Chapter and C 374 . 78 Review 66. a = 4, b = 5, C = 96 SOLUTION: Use the Law of Cosines to find the missing side measure. Find the measure of the remaining angle. Therefore, c = 6.7, A = 36 , and B = 48 . 67. CONSTRUCTION A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp? SOLUTION: a. We are given the angle and the height opposite the angle. Draw a diagram. Use the Law of Sines to find a missing angle measure. We need to find the length of the ramp, so we can use the tangent function. Find the measure of the remaining angle. Therefore, c = 6.7, A = 36 , and B = 48 . 67. CONSTRUCTION A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp? b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3). SOLUTION: a. We are given the angle and the height opposite the angle. Draw a diagram. 68. NATURE For a photography project, Maria is photographing deer from a tree stand. From her sight 30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first? We need to find the length of the ramp, so we can use the tangent function. SOLUTION: eSolutions Manual - Powered by Cognero Draw a diagram of the situation. Page 22 Study Guide and Review - Chapter 4 68. NATURE For a photography project, Maria is photographing deer from a tree stand. From her sight 30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first? 69. FIGURE SKATING An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute. SOLUTION: a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given SOLUTION: Draw a diagram of the situation. by . Each rotation requires 2 6 . The angular speed is 2.5 The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles. radians, so θ = radians per second. b. Convert the seconds to minutes then convert the radians to degrees. We can use trigonometry to find x and y and then find y − x, which is the distance between the deer. The angular speed is about 27,000 degrees per minute. 70. TIMEPIECES The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? SOLUTION: After 40 minutes, the minute hand has traveled of a revolution. Multiply by 360 to determine the central angle θ. 69. FIGURE SKATING An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. eSolutions Manual - the Powered by Cognero b. Express angular speed of the figure skater in degrees per minute. SOLUTION: The area of a sector is determined by A = r 2 , where r is the radius and is the measure of the central angle in radians. We need to convert thePage 23 degrees to radians. TheGuide angular speed about 27,000 degrees per Study and is Review - Chapter 4 minute. The area is about 4.71 square inches. 70. TIMEPIECES The length of the minute hand of a 71. WORLD’S FAIR The first Ferris wheel had a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet? SOLUTION: SOLUTION: After 40 minutes, the minute hand has traveled a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds. of a revolution. Multiply by 360 to determine the central angle θ. The area of a sector is determined by A = r 2 , where r is the radius and is the measure of the central angle in radians. We need to convert the degrees to radians. One sixth of a revolution is equal to degrees. b. If someone has been on it for 7 minutes, then they have made Substitute the known values of r and A. to determine or 60 of a revolution. The wheel has a diameter of 250 feet, so the circumference is 250 feet. After 7 minutes, of the circumference has been traveled. They have traveled about 550 feet. c. It takes 10 minutes to travel the circumference, or 250 feet. Use a proportion to solve for t. The area is about 4.71 square inches. 71. WORLD’S FAIR The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet? SOLUTION: a. One complete revolution is done in 10 minutes, so eSolutions Manual - Powered by Cognero divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds. It would take about 2.5 minutes to travel 200 feet. 72. AIR CONDITIONING An air-conditioning unit turns on and off to maintain the desired temperature. On one summer day, the air conditioner turns on at Page 24 8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. Study Guide and Review - Chapter 4 It would take about 2.5 minutes to travel 200 feet. 72. AIR CONDITIONING An air-conditioning unit that, or 12 hours and 30 minutes. b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening. 74. MUSIC When plucked, a bass string is displaced turns on and off to maintain the desired temperature. On one summer day, the air conditioner turns on at 8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going to use a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning. 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s motion to be dampened so that −0.1 < y < 0.1. SOLUTION: SOLUTION: First, write a trigonometric function that models that motion of the string. The maximum displacement of –ct the string occurs when t = 0, so y = k e cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0. a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes. The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in. You can use the value of the frequency to find b. Sample answer: The function appears to not be periodic because it might take more or less time for the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will get hotter much quicker between 10 in the morning and 2 in the afternoon. 73. TIDES In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur? SOLUTION: a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes. b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening. 74. MUSIC When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s motion to be dampened so that −0.1 < y < 0.1. SOLUTION: First, write a trigonometric function that models that eSolutions Manual - Powered Cognero motion of the string.byThe maximum displacement of –ct the string occurs when t = 0, so y = k e cos t can be used to model the motion of the string . Write a function using the values of k, , and c. –1.9t y = 1.5e t is one model that describes the motion of the string. –1. Use a graphing calculator to graph y = 1.5e t, y = –0.1, and y = 0.1. Select the intersect feature under the TRACE menu to find the point –1.9t where the graphs of y = 1.5e t and y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y ≤ 0.1. From the graph, you can see that it takes approximately 1.41 seconds for the graph of y = –1.9t 1.5e t to oscillate within the interval – 0.1 ≤ y ≤ 0.1. 75. PAINTING A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder makes with the ground is less than 65 , it will slide Page 25 out from under him. What is the greatest distance that the bottom of the ladder can be from the side of From the graph, you can see that it takes approximately 1.41 seconds for the graph of y = –1.9t 1.5eGuide andt Review to oscillate -within the interval Study Chapter 4 – 0.1 ≤ y ≤ 0.1. 75. PAINTING A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter? At the minimum angle, the ladder is about 6.3 feet away from the side of the house. 76. NAVIGATION A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer. SOLUTION: Draw a diagram of the situation. SOLUTION: The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle. Use the cosine function. From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions. 77. GEOMETRY Consider quadrilateral ABCD. At the minimum angle, the ladder is about 6.3 feet away from the side of the house. 76. NAVIGATION A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer. a. Find C. b. Find the area of ABCD. SOLUTION: a. Use SAS to find the length of BD. SOLUTION: Draw a diagram of the situation. Use SSS to find C. From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous casebyofCognero the Law of Sines. 10 < 15 < eSolutions Manual - Powered 20, so there are two possible triangles with the given dimensions. b. Use SAS to find the area of each triangle formed Page 26 by BD. From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are twoReview possible triangles with the Study Guide and - Chapter 4 given dimensions. 77. GEOMETRY Consider quadrilateral ABCD. a. Find C. b. Find the area of ABCD. SOLUTION: a. Use SAS to find the length of BD. Use SSS to find C. b. Use SAS to find the area of each triangle formed by BD. eSolutions Manual - Powered by Cognero Page 27