MAT-129 WA#2.docx - MAT-129 OL009 Written Assignment 2...

MAT-129
OL009
Written Assignment 2
Section 5.1:
EX: 24

20x+10) + (3x+9) = 180
20x + 19 = 180
23x=180-19
23x=161
X=
161
23
X=7
(20x+10) = (20 x 7 +10)
=(140+10)
=150°
(3x+9) = (3 x 7 + 9)
= (21+9)
= 30°
The two angles are 150° and 30°
EX: 32
(3x-5) + (6x-40) = 90°
9x-45=90
9x=90+45
9x=135
X=
135
9
X=15
(3x-5) = (3 x 15 – 5)
=(45-5)
=40°
(6x-40) = 6 x 15 – 40)
=(90-40)
= 50°
The two angles are 40° and 50°

EX: 48
180°-124°51’
179° 60’
-124° 51’
55° 09’
EX: 62
34°51’35”
= 34° +
51 °
60
+
35 °
3600
1’=
1 °
60
1= {1 °} over {3600
≈ 34° + 0.85° + 0.009722°
≈ 34.8597°
EX: 84
-203° 20’
-203° 20’ + 360° = 359° 60’ - 203° 20’
= 156° 40’
EX: 104
225°
225°+n.360
EX: 126
1 minute = 1000 times
60 seconds = 1000 times
2x30 seconds = 1000 times
2 seconds =
1000
30
times
2 seconds =
100
3
times
=
100
3
x 360 degrees
= 100 x 120 degrees
= 12,000 degrees in 2 seconds.

Section 5.2
EX: 14
EX: 44
3x+5y=0; x≥0
3(5)+5y=o
15+5y=0
5y=-15
y=-3
r=
√
x
2
+ y
2
=
√
( 5 )
2
+ ( − 3 )
2
=
√
25 + 9

=
√
34
sin θ =
y
r
=
− 3
√
34
=
− 3
√
34
X
√
34
√
34
=
− 3
√
34
34
cos θ =
x
r
=
5
√
34
=
5
√
34
X
√
34
√
34
=
5
√
34
34
tan θ =
y
x
=
− 3
5
csc θ =
r
y
=
√
34
− 3
=
−
√
34
3
sec θ =
r
x
=
√
34
5
cot θ =
x
y
=
5
− 3
=
− 5
3
EX: 90
csc θ, given that sin θ = -
8
43
csc θ =
1
sin θ
=
1
− 8
43
= 1 /
(
− 8
43
)
= 1 X
(
− 43
8
)
=
− 43
8
csc θ =
− 43
8
EX: 110
Since cos θ > 0 in quadrants I and IV, sec θ > 0 in quadrants I and IV. Both conditions are met in