TEST BANK FOR Trigonometry 5th Edition by Cynthia Y. Young ISBN:978-1119742623 COMPLETE GUIDE ALL CHAPTERS COVERED 100% VERIFIED A+ GRADE ASSURED!!!!NEW LATEST UPDATE!!!!

1

, CHAPTER 1 zl




Section 1.1 Solutions --------------------------------------------------------------------------------
zl zl zl




1 x 1 x
 
z l z l z l zl z l z l z l

1. Solve for x:
z l zl zl z l zl 2. Solve for x:
z l zl zl z l zl



2 360∘ 4 360∘
360∘  2x, so that x 180∘ .
zl zl z l zl z l zl zl zl 360∘  4x, so that x  90∘ .
zl zl z l zl z l zl zl zl




1 x 2 x
3. Solve for x:   4. Solve for x:  
z l z l z l z l z l z l

z l zl zl z l zl zl z l zl zl z l z l zl



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Note:
zl zl zl zl z l zl zl zl zl zl 720∘  2(360∘)  3x, so that x  240∘ . (
zl zl lz zl zl zl zl z l zl zl zl zl



The angle has a negative measure sinc
zl zl zl zl zl zl Note: The angle has a negative measur
z l zl zl zl zl z l



e it is a clockwise rotation.)
zl zl zl zl zl e since it is a clockwise rotation.)
zl zl zl zl zl zl




5 x 7 x
 
z l z l z l zl zlz l z l z l

5. Solve for x:
z l zl zl z l zl 6. Solve for x:
z l zl zl z l zl



6 360∘ 12 360∘
1800∘  5(360∘)  6x, so that x  300∘ .
zl zl zl zl zl zl zl z l zl zl zl 2520∘  7(360∘) 12x, so that x  210∘ .
zl zl zl zl zl zl zl z l zl zl zl




4 x 5 x
7. Solve for x:   8. Solve for x:  
zl z l z l z l z l z l z l

z l zl zl z l zl zl z l zl zl z l zl zl



5 360∘ 9 360∘
1440∘  4(360∘)  5x, so that zl zl zl zl zl zl zl 1800∘  5(360∘)  9x, so that zl zl zl zl zl zl zl




x  288∘ .
zl zl zl x  200∘ .
zl zl zl




(Note: The angle has a negative meas z l zl zl zl zl zl (Note: The angle has a negative measure
z l zl zl zl zl zl



ure since it is a clockwise rotation.)
zl zl zl zl zl zl since it is a clockwise rotation.)
zl zl zl zl zl zl




9. 10.
a) complement: 90∘ 18∘  72∘ z l zl z l z l a) complement: 90∘ 39∘  51∘ z l zl zl z l z l




b) supplement: 180∘ 18∘  162∘ z l zl z l z l b) supplement: 180∘ 39∘  141∘ z l zl zl z l z l




11. 12.
a) complement: 90∘  42∘  48∘ z l zl zl z l z l a) complement: 90∘ 57∘  33∘ z l zl zl z l z l




b) supplement: 180∘  42∘  138∘ z l zl zl z l z l b) supplement: 180∘ 57∘  123∘ z l zl zl z l z l




2

, Section 1.1 zl




13. 14.
a) complement: 90∘ 89∘  1∘ z l zl zl z l z l a) complement: 90∘ 75∘  15∘ z l zl zl z l z l




b) supplement: 180∘ 89∘  91∘ z l zl zl z l z l b) supplement: 180∘ 75∘  105∘ z l zl zl z l z l




15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complementa
z l zl zl zl zl zl z l z l zl zl zl zl zl




ry, we know that 4x∘ 6x∘  90∘. Simplifying this yields
zl zl zl zl zl zl zl zl z l zl zl




10x∘  90∘, zl zl lz z l so that x  9. So, the two angles have measures 36∘and 54∘ .
zl z l zl zl z l zl zl zl zl zl z l zl zl




16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplementa
z l zl zl zl zl zl z l z l zl zl zl zl zl




ry, we know that 3x∘ 15x∘ 180∘. Simplifying this yields
zl zl zl zl zl zl zl zl z l zl zl




18x∘ 180∘, so that zl zl zl zl z l x 10. So, the two angles have measures 30∘ and 150∘ .
zl zl z l zl zl zl zl zl z l zl zl zl




17. Since the angles with measures 8x∘and 4x∘are assumed to be supplementary
z l zl zl zl zl z l lz z l zl zl zl zl zl




, we know that 8x∘ 4x∘ 180∘. Simplifying this yields
zl zl zl zl zl zl zl zl z l zl zl




12x∘ 180∘, zl zl z l so that x 15. So, the two angles have measures 60∘ and 120∘ .
zl z l zl zl z l zl zl zl zl zl z l zl zl zl




18. Since the angles with measures 3x 15∘and 10x10∘are assumed to be comp
z l zl zl zl zl z l zl lz z l zl lz zl zl zl zl




lementary, we know that 3x15∘ 10x10∘  90∘. Simplifying this yields
zl zl zl zl zl zl zl zl zl zl z l zl zl




13x25∘  90∘, zl zl zl zl z l so that 13x∘  65∘ and thus, x  5. So, the two angles have measur
zl zl zl zl z l zl z l zl zl z l zl zl zl zl zl




es 30∘and 60∘ .
z l zl zl




19. Since     180∘, we know tha
z l zl zl zl zl zl z l zl z l zl zl 20. Since     180∘, we know that
z l zl zl zl zl zl z l zl z l zl zl




t 1 10∘ –45∘  180∘ and so,   25∘ .
–
zl zl zl zl zl zl zl zl zl zl zl




1 17∘ –33∘  180∘ and so,  30∘ .
zl


zl155∘
–
zl zl zl zl zl zl zl z l zl zl zl

zl


zl150∘



21. Since     180∘, we know tha
z l zl zl zl zl zl z l zl z l zl zl 22. Since     180∘, we know that
z l zl zl zl zl zl z l zl z l zl zl




t 3    180∘ and so,   36∘.
zl zl zl zl zl zl zl zl zl zl zl zl zl




 4      180∘ and so,   30∘. –– ––
zl5
zl zl zl zl zl zl zl zl zl zl zl zl zl


–– ––
zl6z
l 
Thus,   3 108∘ and     36∘ .
z l zl zl z l zl z l zl z l zl z l zl zl




Thus,   4 z l zl zl z l 120∘ zl z l and    zl z l zl z l  30∘ zl zl .


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