1
, CHAPTER 1 dt
Section 1.1 Solutions --------------------------------------------------------------------------------
dt dt dt
1 x 1 x
d t d t d t dt d t d t d t
1. Solve for x:
d t dt dt d t dt 2. Solve for x:
d t dt dt d t dt
2 360∘ 4 360∘
360∘ 2x, so that x 180∘ .
dt dt d t dt d t dt dt dt 360∘ 4x, so that x 90∘ .
d t dt d t dt d t dt dt dt
1 x 2 x
3. Solve for x: 4. Solve for x:
d t d t d t d t d t d t
d t dt dt d t dt dt d t dt dt d t d t dt
3 360∘ 3 360∘
360∘ 3x, so that x 120∘ . (Note:
dt dt dt dt d t dt dt dt dt 720∘ 2(360∘ ) 3x, so that x 240∘ . (
dt dt dt dt dt dt dt d t dt dt dt dt
The angle has a negative measure sin
dt dt dt dt dt dt dt Note: The angle has a negative measur
d t dt dt dt dt d t
ce it is a clockwise rotation.)
dt dt dt dt dt e since it is a clockwise rotation.)
dt dt dt dt dt dt
5 x 7 x
d t d t d t dt dtd t d t d t
5. Solve for x:
d t dt dt d t dt 6. Solve for x:
d t dt dt d t dt
6 360∘ 12 360∘
1800∘ 5(360∘ ) 6x, so that x 300∘ .
dt dt dt dt dt dt dt d t dt dt dt 2520∘ 7(360∘ ) 12x, so that x 210∘ .
dt dt dt dt dt dt dt d t dt dt dt
4 x 5 x
7. Solve for x: 8. Solve for x:
dt d t d t d t d t d t d t
d t dt dt d t dt dt d t dt dt d t dt dt
5 360∘ 9 360∘
1440∘ 4(360∘ ) 5x, so that
dt dt dt dt dt dt dt 1800∘ 5(360∘ ) 9x, so that
dt dt dt dt dt dt dt
x 288∘ .
dt dt dt x 200∘ .
dt dt dt
(Note: The angle has a negative meas
d t dt dt dt dt dt (Note: The angle has a negative measur
d t dt dt dt dt dt
ure since it is a clockwise rotation.)
dt dt dt dt dt dt e since it is a clockwise rotation.)
dt dt dt dt dt dt
9. 10.
a) complement: 90∘ 18∘ 72∘ d t dt d t d t a) complement: 90∘ 39∘ 51∘ d t dt dt d t d t
b) supplement: 180∘ 18∘ 162∘ d t dt d t d t b) supplement: 180∘ 39∘ 141∘ d t dt dt d t d t
11. 12.
a) complement: 90∘ 42∘ 48∘ d t dt dt d t d t a) complement: 90∘ 57∘ 33∘ d t dt dt d t d t
b) supplement: 180∘ 42∘ 138∘ d t dt dt d t d t b) supplement: 180∘ 57∘ 123∘ d t dt dt d t d t
2
, Section 1.1 dt
13. 14.
a) complement: 90∘ 89∘ 1∘ d t dt dt d t d t a) complement: 90∘ 75∘ 15∘ d t dt dt d t d t
b) supplement: 180∘ 89∘ 91∘ d t dt dt d t d t b) supplement: 180∘ 75∘ 105∘ d t dt dt d t d t
15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complement
d t dt dt dt dt dt d t d t dt dt dt dt dt
ary, we know that 4x∘ 6x∘ 90∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
10x∘ 90∘, dt dt d
t d t so that x 9. So, the two angles have measures 36∘and 54∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt
16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplement
d t dt dt dt dt dt d t d t dt dt dt dt dt
ary, we know that 3x∘ 15x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
18x∘ 180∘, so that dt dt dt dt d t x 10. So, the two angles have measures 30∘ and 150∘ .
dt dt d t dt dt dt dt dt d t dt dt dt
17. Since the angles with measures 8x∘ and 4x∘ are assumed to be supplementar
d t dt dt dt dt d t dt d t dt dt dt dt dt
y, we know that 8x∘ 4x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
12x∘ 180∘, dt dt d t so that x 15. So, the two angles have measures 60∘ and 120∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt dt
18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be com
d t dt dt dt dt d t dt d
t d t dt d
t dt dt dt dt
plementary, we know that 3x 15∘ 10x 10∘ 90∘. Simplifying this yields dt dt dt dt dt dt dt dt dt dt d t dt dt
13x 25∘ 90∘, dt dt dt dt d t so that 13x∘ 65∘ and thus, x 5. So, the two angles have measur
dt dt dt dt d t dt d t dt dt d t dt dt dt dt dt
es 30∘and 60∘ .
d t dt dt
19. Since 180∘, we know tha
d t dt dt dt dt dt d t dt d t dt dt 20. Since 180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt
t 1 10∘ –45∘ 180∘ and so, 25∘ .
–
dt dt dt dt dt dt dt dt dt dt dt
1 17∘ –33∘ 180∘ and so, 30∘ .
dt
dt155∘
–
dt dt dt dt dt dt dt d t dt dt dt
dt
dt150∘
21. Since 180∘, we know th
d t dt dt dt dt dt d t dt d t dt dt 22. Since 180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt
at 3 180∘ and so, 36∘.
dt dt dt dt dt dt dt dt dt dt dt dt dt
4 180∘ and so, 30∘. –– ––
dt5
dt dt dt dt dt dt dt dt dt dt dt dt dt
–– ––
dt6dt
Thus, 3 108∘ and 36∘ .
d t dt dt d t dt d t dt d t dt d t dt dt
Thus, 4 d t d t dt d t 120∘ dt d t and dt d t dt d t 30∘ dt dt .
3
,
, CHAPTER 1 dt
Section 1.1 Solutions --------------------------------------------------------------------------------
dt dt dt
1 x 1 x
d t d t d t dt d t d t d t
1. Solve for x:
d t dt dt d t dt 2. Solve for x:
d t dt dt d t dt
2 360∘ 4 360∘
360∘ 2x, so that x 180∘ .
dt dt d t dt d t dt dt dt 360∘ 4x, so that x 90∘ .
d t dt d t dt d t dt dt dt
1 x 2 x
3. Solve for x: 4. Solve for x:
d t d t d t d t d t d t
d t dt dt d t dt dt d t dt dt d t d t dt
3 360∘ 3 360∘
360∘ 3x, so that x 120∘ . (Note:
dt dt dt dt d t dt dt dt dt 720∘ 2(360∘ ) 3x, so that x 240∘ . (
dt dt dt dt dt dt dt d t dt dt dt dt
The angle has a negative measure sin
dt dt dt dt dt dt dt Note: The angle has a negative measur
d t dt dt dt dt d t
ce it is a clockwise rotation.)
dt dt dt dt dt e since it is a clockwise rotation.)
dt dt dt dt dt dt
5 x 7 x
d t d t d t dt dtd t d t d t
5. Solve for x:
d t dt dt d t dt 6. Solve for x:
d t dt dt d t dt
6 360∘ 12 360∘
1800∘ 5(360∘ ) 6x, so that x 300∘ .
dt dt dt dt dt dt dt d t dt dt dt 2520∘ 7(360∘ ) 12x, so that x 210∘ .
dt dt dt dt dt dt dt d t dt dt dt
4 x 5 x
7. Solve for x: 8. Solve for x:
dt d t d t d t d t d t d t
d t dt dt d t dt dt d t dt dt d t dt dt
5 360∘ 9 360∘
1440∘ 4(360∘ ) 5x, so that
dt dt dt dt dt dt dt 1800∘ 5(360∘ ) 9x, so that
dt dt dt dt dt dt dt
x 288∘ .
dt dt dt x 200∘ .
dt dt dt
(Note: The angle has a negative meas
d t dt dt dt dt dt (Note: The angle has a negative measur
d t dt dt dt dt dt
ure since it is a clockwise rotation.)
dt dt dt dt dt dt e since it is a clockwise rotation.)
dt dt dt dt dt dt
9. 10.
a) complement: 90∘ 18∘ 72∘ d t dt d t d t a) complement: 90∘ 39∘ 51∘ d t dt dt d t d t
b) supplement: 180∘ 18∘ 162∘ d t dt d t d t b) supplement: 180∘ 39∘ 141∘ d t dt dt d t d t
11. 12.
a) complement: 90∘ 42∘ 48∘ d t dt dt d t d t a) complement: 90∘ 57∘ 33∘ d t dt dt d t d t
b) supplement: 180∘ 42∘ 138∘ d t dt dt d t d t b) supplement: 180∘ 57∘ 123∘ d t dt dt d t d t
2
, Section 1.1 dt
13. 14.
a) complement: 90∘ 89∘ 1∘ d t dt dt d t d t a) complement: 90∘ 75∘ 15∘ d t dt dt d t d t
b) supplement: 180∘ 89∘ 91∘ d t dt dt d t d t b) supplement: 180∘ 75∘ 105∘ d t dt dt d t d t
15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complement
d t dt dt dt dt dt d t d t dt dt dt dt dt
ary, we know that 4x∘ 6x∘ 90∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
10x∘ 90∘, dt dt d
t d t so that x 9. So, the two angles have measures 36∘and 54∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt
16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplement
d t dt dt dt dt dt d t d t dt dt dt dt dt
ary, we know that 3x∘ 15x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
18x∘ 180∘, so that dt dt dt dt d t x 10. So, the two angles have measures 30∘ and 150∘ .
dt dt d t dt dt dt dt dt d t dt dt dt
17. Since the angles with measures 8x∘ and 4x∘ are assumed to be supplementar
d t dt dt dt dt d t dt d t dt dt dt dt dt
y, we know that 8x∘ 4x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt
12x∘ 180∘, dt dt d t so that x 15. So, the two angles have measures 60∘ and 120∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt dt
18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be com
d t dt dt dt dt d t dt d
t d t dt d
t dt dt dt dt
plementary, we know that 3x 15∘ 10x 10∘ 90∘. Simplifying this yields dt dt dt dt dt dt dt dt dt dt d t dt dt
13x 25∘ 90∘, dt dt dt dt d t so that 13x∘ 65∘ and thus, x 5. So, the two angles have measur
dt dt dt dt d t dt d t dt dt d t dt dt dt dt dt
es 30∘and 60∘ .
d t dt dt
19. Since 180∘, we know tha
d t dt dt dt dt dt d t dt d t dt dt 20. Since 180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt
t 1 10∘ –45∘ 180∘ and so, 25∘ .
–
dt dt dt dt dt dt dt dt dt dt dt
1 17∘ –33∘ 180∘ and so, 30∘ .
dt
dt155∘
–
dt dt dt dt dt dt dt d t dt dt dt
dt
dt150∘
21. Since 180∘, we know th
d t dt dt dt dt dt d t dt d t dt dt 22. Since 180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt
at 3 180∘ and so, 36∘.
dt dt dt dt dt dt dt dt dt dt dt dt dt
4 180∘ and so, 30∘. –– ––
dt5
dt dt dt dt dt dt dt dt dt dt dt dt dt
–– ––
dt6dt
Thus, 3 108∘ and 36∘ .
d t dt dt d t dt d t dt d t dt d t dt dt
Thus, 4 d t d t dt d t 120∘ dt d t and dt d t dt d t 30∘ dt dt .
3
,
