PreCalc 11 Chapter 6 Rev Pack v1 Answer Section
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1 PreCalc 11 Chapter 6 Rev Pack v1 Answer Section MULTIPLE CHOICE 1. ANS: A PTS: 0 DIF: Moderate 2. ANS: D PTS: 0 DIF: Easy REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T1. ANS: A PTS: 0 DIF: Moderate 4. ANS: D PTS: 0 DIF: Easy 5. ANS: D PTS: 0 DIF: Moderate 6. ANS: C PTS: 0 DIF: Moderate 7. ANS: A PTS: 0 DIF: Moderate 8. ANS: B PTS: 0 DIF: Easy REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T1 KEY: Conceptual Understanding 9. ANS: A PTS: 0 DIF: Easy REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T1 10. ANS: A PTS: 0 DIF: Moderate REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T2 11. ANS: D PTS: 0 DIF: Moderate REF: 6. Constructing Triangles KEY: Procedural Knowledge 12. ANS: B PTS: 0 DIF: Moderate REF: 6.4 The Sine Law 1. ANS: D PTS: 0 DIF: Moderate REF: 6. Constructing Triangles KEY: Procedural Knowledge 14. ANS: B PTS: 0 DIF: Moderate REF: 6.4 The Sine Law 15. ANS: C PTS: 0 DIF: Moderate REF: 6.4 The Sine Law 1
2 16. ANS: C PTS: 1 DIF: Moderate REF: 6.5 The Cosine Law 17. ANS: A PTS: 1 DIF: Moderate REF: 6.5 The Cosine Law 18. ANS: B PTS: 0 DIF: Easy REF: 6.4 The Sine Law SHORT ANSWER 19. ANS: tan 45 = 1 PTS: 0 DIF: Moderate REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T2 20. ANS: approximately 10.4 m PTS: 0 DIF: Moderate REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T2 21. ANS: approximately 7.6 m PTS: 0 DIF: Moderate REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T2 22. ANS: cos 0 2 PTS: 0 DIF: Moderate REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T2 2. ANS: DEF is scalene. PTS: 0 DIF: Moderate REF: 6. Constructing Triangles 2
3 24. ANS: No, the Sine Law cannot be used because only one angle measure is given and the angle is contained between the two given sides. PTS: 0 DIF: Easy REF: 6.4 The Sine Law KEY: Conceptual Understanding Communication 25. ANS: Two triangles can be constructed. PTS: 0 DIF: Easy REF: 6.4 The Sine Law 26. ANS: Yes, the Cosine Law can be used; AC = 8.6 cm. PTS: 1 DIF: Moderate REF: 6.5 The Cosine Law 27. ANS: G = 74 H = 55 I = 51 PTS: 1 DIF: Moderate REF: 6.5 The Cosine Law 28. ANS: C = 4 PTS: 1 DIF: Moderate REF: 6.5 The Cosine Law
4 PROBLEM 29. ANS: Label the right triangles ABC and DEF. In ABC, B represents the angle of inclination of the guy wire attached to the shorter building. In ABC, sin B opposite sin B AC AB sin B 4 47 Ê 4 ˆ B sin 1 Ë Á 47 B The angle of inclination of the guy wire attached to the shorter building is approximately 46.. In DEF, E represents the angle of inclination of the guy wire attached to the taller building. In DEF, cos E adjacent cos E EF DE cos E 8 51 Ê 8 ˆ E cos 1 Ë Á 51 E The angle of inclination of the guy wire attached to the taller building is approximately The student is not correct. The angles of inclination are different. PTS: 0 DIF: Difficult REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T2 KEY: Communication Problem-Solving Skills 4
5 0. ANS: a) In right BCO, BO is the, BC is opposite O, and CO is adjacent to O. To determine the length of BO, use the sine ratio. sin O opposite sin O BC BO sin BO Solve the equation for BO. sin BO BO sin1 8.0 BO sin sin 1 sin BO sin 1 BO The length of cord needed to reach corner B is approximately 15.5 m. b) In right BCO, to determine the length of CO, use the cosine ratio. cos O adjacent cos O CO BO CO cos Solve the equation for CO. cos 1 CO ( )cos 1 CO CO To determine the distance between the electrical outlet and corner N, subtract CO from CN. NO CN CO NO NO The distance between the electrical outlet and corner N is approximately 26.7 m. PTS: 0 DIF: Difficult REF: 6.1 Angles in Standard Position in Quadrant 1 LOC: 11.T2 KEY: Communication Problem-Solving Skills 5
6 1. ANS: a) Determine the distance r from the origin to P. x, y 4 r x 2 y 2 r () 2 ( 4) 2 r 25 cos x r cos sin y r 25 sin 4 25 tan y x tan 4 Ê ˆ b) The reference angle is: cos 1 Ë Á Since is in Quadrant 4, the angle is approximately: PTS: 0 DIF: Moderate REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T2 KEY: Procedural Knowledge Communication 2. ANS: Draw a labelled diagram to represent the problem. In right FLP, FP is the and FL is the side opposite P. So, use the sine ratio to determine the length of FL. sin P opposite sin P FL FP sin 48 FL 7.6 Solve the equation for FL. sin 48 FL sin 48 FL FL The distance between the fishing boat and the lighthouse is approximately 5.6 km. PTS: 0 DIF: Moderate REF: 6.2 Angles in Standard Position in All Quadrants LOC: 11.T2 KEY: Communication Problem-Solving Skills 6
7 . ANS: Possible solution: sin A sin Length of BC (cm) Value of BC AB How does BC AB compare with sin A? BC AB sina BC AB 1 sina BC AB 1 sina sina BC AB 1 Description of possible triangles No triangles are possible. 1 isosceles triangle 1 scalene triangle 2 scalene triangles PTS: 0 DIF: Moderate REF: 6. Constructing Triangles KEY: Conceptual Understanding Problem-Solving Skills 7
8 4. ANS: The treasure chest could be between the two divers or on one side of both divers. Case 1: The treasure chest C is between the two divers, A and B. C 180 ( A B) C 180 (8 51 ) C 91 a sin A c sin C a sin 8 60 a sin sin 8 sin 91 a Ö 7 b sin B c sin C b sin b sin sin 51 sin 91 b Ö 47 The treasure chest is approximately 7 m and 47 m from the divers. Case 2: The treasure chest C is on one side of both divers A and B. a sin A c sin C a sin 8 60 a sin 1 60 sin 8 sin 1 a Ö 164 In ABC, B C 180 (8 129 ) 1 b sin B c sin C b sin b sin 1 60 sin 129 sin 1 b Ö 207 The treasure chest is approximately 164 m and 207 m from the divers. PTS: 0 DIF: Difficult REF: 6.4 The Sine Law KEY: Communication Problem-Solving Skills 8
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