Pre-Calculus Homework Solutions 228

# Pre-Calculus Homework Solutions 228 - 46. B; d2 ¬e2 f 2...

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Unformatted text preview: 46. B; d2 ¬e2 f 2 2ef cos D d2 ¬122 152 2(12)(15)cos75° d2 ¬369 360cos75° d ¬ 369 360cos 75° d ¬16.6 47. C; earnings base salary commission Let s represent her sales for the month. 4455 ¬1280 0.125s 3175 ¬0.125s 25,400 ¬s Her sales that month were \$25,400. 53. To show that AB CD, we must show that AB CD AB CD AE CE . AE 8 4 or 2, and CE 9 4 . Since the side lengths are not proportional, AB is not parallel to CD. 54. To show that AB CD, we must show that AB CD AB CD BE DE . BE 5.4 9 18 9 3 or 5 , and DE 10 or 5 . AB BE Thus, CD DE . Since the sides have proportional lengths, AB CD. Page 390 Maintain Your Skills sin X x sin 22° x 48. 55. Given: ¬ sin Y y ¬ sin 49° 4.7 4.7sin22° Prove: H ¬x sin49° 4.7 sin 22° sin 49° ¬x 2.3 ¬x sin X x sin 50° 14 49. D 10sin50° sin ¬sin Y 1 33° JF EF ¬Y x 1.55 m 100 m x ¬ 100 100tan23° ¬x 42.45 ¬x x 1.55 ¬42.45 1.55 or 44.0 The height of the building is about 44.0 meters. 51. To show that AB CD, we must show that D proportional lengths, AB CD. BD DE . AE DE BE So, BD DE AC 15 BD 22.5 7 or 8. 7 8. 10.5 or 12. AC 10.5 7 12 or 8 . Thus, CE BD DE . Since the sides have proportional lengths, AB CD. Chapter 7 C JF LF definition of similar triangles. EF GF by the Transitive Porperty of Equality and EFG EFG by the Reflexive Property of Congruence. Thus, JFL EFG by SAS Similarity and FJL FEG by the definition of similar triangles. JL EG because if two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel. 52. To show that AB CD, we must show that AC So, CE M A B Proof: Since JM EB and LM GB, then MJF BEF and FML FBG because if two parallel lines are cut by a transversal, corresponding angles are congruent. EFB EFB and BFG BFG by the Reflexive Property of Congruence. Then EFB JFM and FBG FML by AA JF LF MF MF Similarity. Then EF GF by the BF , BF AC BD CE DE . AC 8.4 7 BD 6.3 CE 6 or 5 , and DE 4.5 AC BD or 7 . Thus, CE . Since the sides have 5 DE CE MF MF BF and BF Reflexive Property of Congruence. Then, by SAS EFG. Similarity, JFL 56. Given: JM EB LM GB Prove: JL EG H G L J E F 23 AC CE C LF GF . By the Transitive JF LF Property of Equality, EF F by the GF . F ¬Y 50. tan23° F M A B Proof: Since JFM EFB and LFM GFB, then by the definition of similar triangles, ¬14sin Y 10sin 50° 14 10sin 50° 14 EFB GFB EFG G L J E ¬ sin Y y ¬ sin Y 10 JFM LFM JFL 230 ...
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## This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.

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