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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.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus

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