Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Math 2201 Chapter 3 Review Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. Solve for the unknown side length. Round your answer to one decimal place. a. b. c. d. ____ 2. Solve for the unknown angle measure. Round your answer to the nearest degree. a. b. c. d. ____ 4.1 5.1 4.7 5.6 77° 76° 75° 74° 3. Which expression describes the ratios of side-angle pairs in ∆QRS? a. q(sin R) = r(sin S) = s(sin Q) b. c. d. q(sin Q) = r(sin R) = s(sin S) ____ 4. What information do you need to know about an acute triangle to use the sine law? a. b. c. d. ____ two angles and any side two sides and any angle all the sides all the angles 5. Determine the length of c to the nearest tenth of a centimetre. a. b. c. d. 26.1 cm 24.0 cm 25.0 cm 23.0 cm June 2014 1 Math 2201 ____ 30 cm 27 cm 29 cm 28 cm 8. Determine the measure of ∠R to the nearest degree. a. b. c. d. ____ 78.6 cm 78.2 cm 79.4 cm 79.0 cm 7. Determine the length of k to the nearest centimetre. a. b. c. d. ____ Review 6. Determine the length of f to the nearest tenth of a centimetre. a. b. c. d. ____ Chapter 3 56° 54° 52° 50° 9. In ∆DEF, ∠D = 61°, d = 23.9 cm, and ∠E = 38°. Determine the length of side e to the nearest tenth of a centimetre. a. b. c. d. 18.4 cm 17.6 cm 16.0 cm 16.8 cm ____ 10. In ∆DEF, d = 10.0 cm, e = 8.6 cm, and ∠E = 45°. Determine the measure of ∠D to the nearest degree. a. b. c. d. 65° 35° 55° 45° June 2014 2 Math 2201 Chapter 3 Review ____ 11. What information do you need to know about an acute triangle to use the cosine law? a. b. c. d. two angles and any side two sides and any angle all the sides all the angles ____ 12. Determine the length of EF to the nearest centimetre. a. b. c. d. 88 cm 84 cm 86 cm 82 cm ____ 13. Determine the length of AC to the nearest tenth of a centimetre. a. b. c. d. 30.1 cm 30.2 cm 31.1 cm 31.0 cm ____ 14. Determine the measure of θ to the nearest degree. a. b. c. d. 50° 40° 30° 60° June 2014 3 Math 2201 Chapter 3 Review ____ 15. Determine the measure of θ to the nearest degree. a. b. c. d. 60° 62° 59° 61° ____ 16. In ∆DEF, d = 13.5 cm, e = 18.2 cm, and ∠F = 60°. Determine the measure of f to the nearest tenth of a centimetre. a. b. c. d. 17.0 cm 16.8 cm 16.4 cm 16.6 cm ____ 17. In ∆DEF, d = 23.9 cm, e = 16.8 cm, and f = 27.0 cm. Determine the measure of ∠D to the nearest degree. a. b. c. d. 61° 64° 58° 54° ____ 18. How you would determine the indicated angle measure, if it is possible? a. b. c. d. the cosine law primary trigonometric ratios the sine law not possible ____ 19. How you would determine the indicated angle measure, if it is possible? a. b. c. d. primary trigonometric ratios the sine law the cosine law not possible June 2014 4 Math 2201 Chapter 3 Review ____ 20. How you would determine the indicated angle measure, if it is possible? a. b. c. d. primary trigonometric ratios the cosine law the sine law not possible ____ 21. A radar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 120 m directly below the ship. The length of the vessel is a clue to which wreck has been found. The radar operator measures the angles of depression to the front and back of the sunken vessel to be 55° and 46°. How long, to the nearest tenth of a metre, is the sunken vessel? a. b. c. d. 199.9 m 201.8 m 203.7 m 198.0 m Short Answer 22. Solve for the unknown side length. Round your answer to one decimal place. 23. Solve for the unknown angle measure. Round your answer to the nearest degree. 24. Determine the length of c to the nearest tenth of a centimetre. 25. Determine the measure of θ to the nearest degree. 26. In ∆ABC, ∠A = 65°, a = 23.5 cm, and ∠C = 71°. Determine the length of side c to the nearest tenth of a centimetre. 27. In ∆QRS, r = 4.1 cm, s = 2.7 cm, and ∠R = 88°. Determine the measure of ∠S to the nearest degree. June 2014 5 Math 2201 Chapter 3 Review 28. Determine the length of d to the nearest tenth of a centimetre. 29. Determine the measure of α to the nearest degree. 30. In ∆PQR, p = 18.0 cm, q = 15.0 cm, and ∠R = 45°. Determine the measure of r to the nearest tenth of a centimetre. 31. In ∆VWX, v = 52.5 cm, w = 48.0 cm, and x = 61.7 cm. Determine the measure of ∠V to the nearest degree. Problem 32. In ∆PQR, ∠P = 55°, ∠Q = 77°, and p = 4.5 cm. Solve the triangle. Round angles to the nearest degree and sides to the nearest tenth of a centimetre. Show your work. 33. A radio tower is supported by two wires on opposite sides. On the ground, the ends of the wire are 280 m apart. One wire makes a 60° angle with the ground. The other makes a 66° angle with the ground. Draw a diagram of the situation. Then, determine the length of each wire to the nearest metre. Show your work. 34. Two Jasper National Park rangers in their fire towers spot a fire. Determine the distances, to the nearest tenth of a kilometre, from each tower to the fire. Show your work. 35. Determine, to the nearest centimetre, the perimeter of the triangle. Show your work. June 2014 6 Math 2201 Chapter 3 Review 36. In ∆TUV, t = 6.0 m, u = 7.6 m, and v = 8.0 m. Solve the triangle. Round angles to the nearest degree and sides to the nearest tenth of a metre. Show your work. 37. In ∆ABC, a = 12.0 cm, b = 14.0 cm, and ∠C = 62°. Solve the triangle. Round angles to the nearest degree and sides to the nearest tenth of a centimetre. 38. Determine the perimeter of the triangle to the nearest tenth of a centimetre. 39. A radio tower is supported by two wires on opposite sides. On the ground, the ends of the wire are 84 m apart. One wire makes a 52° angle with the ground. The other makes a 74° angle with the ground. Draw a diagram of the situation. Then, determine the height of the tower to the nearest tenth of a metre. 40. Determine the perimeter of this quadrilateral to the nearest tenth of a centimetre. June 2014 7 Math 2201 Chapter 3 Review MULTIPLE CHOICE 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: ANS: A B C A B B A C D C C D C A B C A A D A A SHORT ANSWER 22. ANS: 23.2 23. ANS: 46° 24. ANS: c = 42.7 cm 25. ANS: θ = 57° 26. ANS: c = 24.5 cm 27. ANS: ∠S = 41° 28. ANS: d = 10.0 cm 29. ANS: α = 67° 30. ANS: r = 12.9 cm 31. ANS: ∠V = 56° PROBLEM 32. ANS: ∠P + ∠Q + ∠R = 180° 55° + 77° + ∠R = 180° ∠R = 48° The length of r is 4.1 cm. The length of q is 5.4 cm. June 2014 8 Math 2201 33. Chapter 3 Review ANS: Let the x and y be the length of the wires. The third angle is 180° – 66° – 60° = 54°. Use the sine law to determine the length of each wire: The wires are 316 m and 300 m long. 34. ANS: Let ∠C represent the measure of the remaining unknown angle. ∠A + ∠B + ∠C = 180° 64° + 48° + ∠C = 180° ∠C = 68° Let b represent the distance from tower A to the fire. The distance from tower A to the fire is 3.4 km. Let a represent the distance from tower B to the fire. The distance from tower B to the fire is 4.1 km. 35. ANS: Determine the measure of ∠N. ∠L + ∠M + ∠N = 180° 80° + 50° + ∠N = 180° ∠N = 50° ∆LMN is isosceles because ∠M = ∠N. So, m = 14 cm because it is also opposite a 50° angle. Determine the length of l. Perimeter = l + m + n Perimeter = 17.998... + 14 + 14 Perimeter = 45.998... The perimeter is 46 cm. June 2014 9 Math 2201 36. Chapter 3 Review ANS: t2 6.02 36.0 –85.76 = u2 + v2 – 2uv cos T = 7.62 + 8.02 – 2(7.6)(8.0) cos T = 57.76 + 64.00 – 121.60 cos T = –121.60 cos T = cos T ∠T = cos–1(0.7052...) ∠T = 45.149...° The measure of ∠T is 45°. u2 7.62 57.76 –42.24 = t2 + v2 – 2tv cos U = 6.02 + 8.02 – 2(6.0)(8.0) cos U = 36.00 + 64.00 – 96.00 cos U = –96.00 cos U = cos U ∠U = cos–1(0.44) ∠U = 63.896...° The measure of ∠U is 64°. v2 8.02 64.0 –29.76 = t2 + u2 – 2tu cos V = 6.02 + 7.62 – 2(6.0)(7.6) cos V = 36.0 + 57.76 – 91.2 cos V = –91.2 cos V = cos V ∠V = cos–1(0.3263...) ∠V = 70.954...° The measure of ∠V is 71°. PTS: 37. 1 DIF: Grade 11 REF: Lesson 3.3 ANS: c2 = a2 + b2 – 2ab cos C c2 = 12.02 + 14.02 – 2(12.0)(14.0) cos 62° c2 = 144.0 + 196.0 – 336.0(0.4694...) c2 = 182.257... c = 13.500... The length of c is 13.5 cm. a2 12.02 144.0 –234.25 = b2 + c2 – 2bc cos A = 14.02 + 13.52 – 2(14.0)(13.5) cos A = 196.00 + 182.25 – 378.00 cos A = –378.00 cos A = cos A ∠A = cos–1(0.6197...) ∠A = 51.705...° The measure of ∠A is 52°. b2 14.02 196.0 –130.25 = a2 + c2 – 2ac cos B = 12.02 + 13.52 – 2(12.0)(13.5) cos B = 144.00 + 182.25 – 324.00 cos B = –324.00 cos B = cos B ∠B = cos–1(0.4020...) ∠B = 66.296...° The measure of ∠B is 66°. 38. ANS: a2 = b2 + c2 – 2bc cos A a2 = 3.32 + 3.52 – 2(3.3)(3.5) cos 43° a2 = 10.89 + 12.25 – 23.10(0.7313...) a2 = 6.245... a = 2.499... Perimeter = a + b + c Perimeter = 2.499... + 3.3 + 3.5 Perimeter = 9.299... The perimeter of the triangle is 9.3 cm. June 2014 10 Math 2201 39. Chapter 3 Review ANS: Let the x and y be the lengths of the wires and h be the height of the tower. The third angle is 180° – 52° – 74° = 54°. Use the sine law to determine the length of one of the wires: Use the sine ratio to determine the height of the tower: The tower is 78.6 m tall. 40. ANS: AD2 AD2 AD2 AD2 AD = AB2 + BD2 – 2AB·BD cos ABD = 6.42 + 7.02 – 2(6.4)(7.0) cos 50° = 40.96 + 49.00 – 89.60(0.6427…) = 32.366… = 5.689… ∠BDC = 180° – 48° – 73° ∠BDC = 59° Perimeter = AB + BC + CD + DA Perimeter = 6.4 + 6.3438… + 5.5 + 5.689… Perimeter = 23.932… The perimeter of ABCD is 23.9 cm. June 2014 11