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Practice Test - Chapter 4 Find the value of x. Round to the nearest tenth, if necessary. Find the measure of angle θ. Round to the nearest degree, if necessary. 1. SOLUTION: 3. SOLUTION: 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 . Because the length of the side opposite and adjacent to θ are given, use the tangent function. 2. SOLUTION: An acute angle measure and the length of the side opposite the angle are given, so the tangent function can be used to find the length of the side adjacent to the angle. 4. SOLUTION: Because the length of the side opposite and the hypotenuse are given, use the cosine function. 5. MULTIPLE CHOICE What is the linear speed Find the measure of angle θ. Round to the nearest degree, if necessary. 3. SOLUTION: Because the length of the side opposite and adjacent to θ are given, use the tangent function. of a point rotating at an angular speed of 36 radians per second at a distance of 12 inches from the center of the rotation? A 420 in./s B 432 in./s C 439 in./s D 444 in./s SOLUTION: The formula for linear speed is v = where s is the arc length traveled. Arc length is equal to rθ, where r is the radius and θ is the central angle. Thus, eSolutions Manual - Powered by Cognero . The distance from the center of rotation, also known as the radius, is 12 inches, so . Page 1 The correct choice is B. Practice Test - Chapter 4 5. MULTIPLE CHOICE What is the linear speed of a point rotating at an angular speed of 36 radians per second at a distance of 12 inches from the center of the rotation? A 420 in./s B 432 in./s C 439 in./s D 444 in./s Write each degree measure in radians as a multiple of π and each radian measure in degrees. 6. 200 SOLUTION: To convert a degree measure to radians, multiply by SOLUTION: The formula for linear speed is v = where s is the arc length traveled. Arc length is equal to rθ, where r is the radius and θ is the central angle. Thus, . The distance from the center of rotation, also known as the radius, is 12 inches, so . 7. SOLUTION: To convert a radian measure to degrees, multiply by The formula for angular speed is speed is 36 radians per second, so . The angular . Therefore, 8. Find the area of the sector of the circle shown. The correct choice is B. Write each degree measure in radians as a multiple of π and each radian measure in degrees. 6. 200 SOLUTION: SOLUTION: The area of a sector A = r 2 , where r is the radius and θ is the central angle. To convert a degree measure to radians, multiply by eSolutions Manual - Powered by Cognero 7. Page 2 The area of the sector is about 209.4 square inches. Practice Test - Chapter 4 8. Find the area of the sector of the circle shown. 10. SOLUTION: The terminal side of lies in Quadrant IV. Therefore, its reference angle is SOLUTION: The area of a sector A = r 2 ' = . , where r is the radius and θ is the central angle. Find the exact value of each expression. The area of the sector is about 209.4 square inches. Sketch each angle. Then find its reference angle. 9. 165 11. sec SOLUTION: Because the terminal side of θ lies in Quadrant III, the reference angle θ´ of SOLUTION: is . The terminal side of 240 lies in Quadrant II. Therefore, its reference angle is ' = 180 –165 or 15 . 10. SOLUTION: The terminal side of lies in Quadrant IV. Therefore, its reference angle is . eSolutions Manual - Powered by Cognero ' = 12. cos (−240 ) SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 ) Because the terminal side of lies in Quadrant II, the reference angle ' is . Page 3 this additional restriction, must lie in the 3rd quadrant. With sin < 0 and cos θ < 0, cot must be > 0. Practice Test - Chapter 4 The correct choice is H. State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 12. cos (−240 ) SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 ) Because the terminal side of lies in Quadrant II, the reference angle ' is . 14. y = 4 cos – 5 SOLUTION: In this function, a = 4, b = , c = 0, and d = –5. 13. MULTIPLE CHOICE An angle satisfies the following inequalities: csc θ < 0, cot θ > 0, and sec θ < 0. In which quadrant does lie? FI G II H III J IV Graph y = 4 cos shifted 5 units down. SOLUTION: If csc < 0, then sin < 0. So, must lie in the 3rd or 4th quadrant. If sec < 0, then cos < 0. With this additional restriction, must lie in the 3rd quadrant. With sin < 0 and cos θ < 0, cot must be > 0. The correct choice is H. State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function. 14. y = 4 cos – 5 SOLUTION: In this function, a = 4, b = 15. , c = 0, and d = –5. SOLUTION: In this function, a = –1, b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. eSolutions Manual - Powered by Cognero Page 4 Practice Test - Chapter 4 16. TIDES The table gives the approximate times that 15. the high and low tides occurred in San Azalea Bay over a 2-day period. SOLUTION: In this function, a = –1, b = 1, c = , and d = 0. Because d = 0, there is no vertical shift. a. The tides can be modeled with a trigonometric function. Approximately what is the period of this function? b. The difference in height between the high and low tides is 7 feet. What is the amplitude of this function? c. Write a function that models the tides where t is measured in hours. Assume the function has no phase shift or vertical shift. SOLUTION: a. The period lasts from peak to peak, or from high tide to high tide. 3:04 PM is 12 hours and 30 minutes after 2:35 AM. b. The amplitude is one half of the difference between the maximum (high tide) and the minimum (low tide). One half of 7 is 3.5. c. The period is 12 hours and 30 minutes, or 12.5 hours. Graph y = – sin x shifted units to the left. 16. TIDES The table gives the approximate times that the high and low tides occurred in San Azalea Bay over a 2-day period. a. The tides can be modeled with a trigonometric function. Approximately what is the period of this function? b. The difference in height between the high and low tides is 7 feet. What is the amplitude of this function? c. Write a function that models the tides where t is measured in hours. Assume the function has no phase shift or vertical shift. SOLUTION: a. The period lasts from peak to peak, or from high tide to high tide. 3:04 PM is 12 hours and 30 minutes after 2:35 AM. eSolutions Manual - Powered by Cognero b. The amplitude is one half of the difference between the maximum (high tide) and the minimum Locate the vertical asymptotes, and sketch the graph of each function. 17. SOLUTION: The graph of x translated is the graph of y = tan units to the left. The period is . Find the location of two consecutive vertical asymptotes. Page or 5 17. SOLUTION: 18. y = sec (2x) Practice Test - Chapter 4 is the graph of y = tan The graph of x translated units to the left. The period is or . Find the location of two consecutive vertical asymptotes. SOLUTION: sec 2x is the graph of y = sec x The graph of y = compressed horizontally and compressed vertically. The period is or π. Find the location of two vertical asymptotes. and and Create a table listing the coordinates of key points for for one period on . Create a table listing the coordinates of key points Function Vertical Asymptote Intermediate Point x-int . Function Vertical Asymptote Intermediate Point x-int (0, 0) Intermediate Point Vertical Asymptote (0, 1) 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 = sec x y= sec 2x (0, 1) Intermediate Point Vertical Asymptote 18. y = sec 2x for one period on for y = y = tan x 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. sec (2x) SOLUTION: eSolutions Manual - Powered by Cognero The graph of y = sec 2x is the graph of y = sec x compressed horizontally and compressed vertically. 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.Page 6 19. a = 8, b = 16, A = 22 Practice Test - Chapter 4 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. 19. a = 8, b = 16, A = 22 When B 131 , then C 180 – (22 + 131 ) or about 27 . Apply the Law of Sines to find c. SOLUTION: Draw a diagram of a triangle with the given dimensions. Notice that A is acute and a < b because 8< 16. Therefore, two solutions may exist. Find h. Therefore, the remaining measures of B 49 , C 109 , c 20.1 and B 27 , c 9.7. are 131 , C 20. a = 9, b = 7, A = 84 SOLUTION: Draw a diagram of a triangle with the given dimensions. 8 > 6, so two solutions exist. Apply the Law of Sines to find B. Notice that A is acute and a > b because 9 > 7. Therefore, one solution exists. Apply the Law of Sines to find B. Because two angles are now known, C 180 – (22 + 49 ) or about 109 . Apply the Law of Sines to find c. Because two angles are now known, C ≈ 180 – (84 + 51 ) or about 45 . Apply the Law of Sines to find c. eSolutions Manual Powered by Cognero When B - 131 , then C 180 Page 7 – (22 + 131 ) or about 27 . Apply the Law of Sines to find c. Therefore, the remaining measures of are Therefore, the remaining measures of B 49 , C 109 , c 20.1 and B Practice 27 , Test c 9.7.- Chapter 4 are 131 , C Therefore, the remaining measures of B 51 , C 45 , and c 6.4. are 20. a = 9, b = 7, A = 84 21. a = 3, b = 5, c = 7 SOLUTION: SOLUTION: Draw a diagram of a triangle with the given dimensions. Use the Law of Cosines to find an angle measure. Notice that A is acute and a > b because 9 > 7. 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 – (84 + 51 ) or about 45 . Apply the Law of Sines to find c. Find the measure of the remaining angle. Therefore, A 22 , B 38 , and C 120 . 22. a = 8, b = 10, C = 46 SOLUTION: Use the Law of Cosines to find the missing side measure. Therefore, the remaining measures of B 51 , C 45 , and c 6.4. are 21. a = 3, b = 5, c = 7 SOLUTION: Use the Law of Cosines to find an angle measure. eSolutions Manual - Powered by Cognero Use the Law of Sines to find a missing angle Use the Law of Sines to find a missing angle measure. Page 8 Find the measure of the remaining angle. Practice Test - Chapter 4 22 , B Therefore, A 38 , and C Find the measure of the remaining angle. 120 . 7.3, A Therefore, c 52 , and B 82 . Find the exact value of each expression, if it exists. 22. a = 8, b = 10, C = 46 SOLUTION: Use the Law of Cosines to find the missing side measure. 23. SOLUTION: Find a point on the unit circle on the interval with a x-coordinate of . Use the Law of Sines to find a missing angle measure. When t = = , cos t = –1 . Therefore, cos . Find the measure of the remaining angle. 7.3, A Therefore, c 52 , and B 82 . Find the exact value of each expression, if it exists. 24. SOLUTION: Find a point on the unit circle on the interval 23. with a y-coordinate of . SOLUTION: Find a point on the unit circle on the interval with a x-coordinate of . When t = , sin t = . Therefore, sin –1 = . When t = = , cos t = . eSolutions Manual - Powered by Cognero 24. –1 . Therefore, cos 25. NAVIGATION A boat leaves a dock and travels 45º north of west averaging 30 knots for 2 hours. The boat then travels directly west averaging 40 knots for 3 hours. Page 9 25. NAVIGATION A boat leaves a dock and travels 45º north of west averaging 30 Practice Test - Chapter 4 knots for 2 hours. The boat then travels directly west averaging 40 knots for 3 hours. a. How many nautical miles is the boat from the dock after 5 hours? b. How many degrees south of east is the dock from the boat’s present position? SOLUTION: a. During the first leg of the trip the boat traveled 30 knots for 2 hours, and therefore traveled a distance of 30 · 2 or 60 nautical miles. During the second leg of the trip, the boat traveled 40 knots for 3 hours, so the distance the boat traveled is 40 · 3 or 120 nautical miles. Draw a diagram to model the situation. Let x represent the distance the boat has traveled from the dock after 5 hours. Use the Law of Cosines to find b. Therefore, the boat is 167.9 nautical miles from the dock after 5 hours. b. Use the Law of Sines to find C. Therefore, the dock is about 15 south of east from the boat's current position. eSolutions Manual - Powered by Cognero Page 10