Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter1: The Six Trigonometric Functions
Section1.2: The Rectangular Coordinate System
Problem 92PS: Draw an angle in standard position whose terminal side contains the point (2, –3). Find the...
Related questions
Question
![-
Name: Vanesa Gomzalez
Date: 11-28-23
Bell: 4th
** This is a 2-page document! **
Directions: Find each value or measure. Assume that segments that appear to be tangent are tangent.
258
2. MAD
1. mZYZV
144
+114
161°
2
258
161+x= 93
2
161+x=186
961 -161
X=25
3. mQS
128
x 2
250
5. mZEFG
-39
7. mCF
180
53
129
127
366
ما25 -
34+x=121
2
104
391+x=242
-39
my
mQS=104
x=203
37
P
9. Solve for x.
5x-7+84+27:/8C
5x+190-120
(x = 8)
119
144°
127-53-37
2
121
2
R
C
128°
T
Z
D.
Ĉ
V
W
D
127
X
39°
41%
S
114°
E
G
E
53°
Unit 10: Circles
Homework 6: Arc & Angle Measures
(5x - 7)X\27°
p
4. m/JLK
298/2 = 149
<JLK = 31
6. m/TUV
360
-108
252
8. mHJ
252/2=2
126
402
2606
102
18G
- 149
31
298
180
29=1/2(96-X)
58-96-X
38=x
10. Solve for x.
360 23x-3=204
+3 +3
U
B
93°
E
C
N
23x=20723x-3)
X = 9
T
108
96°
G
X80
K
204
K
DAYAN
L
360
29°
H
M
25/
78°
torat
Sy)
S
1
951
Gina Wilson (All Things Algebra), 2018](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd145341-759a-4e89-8c83-6d087e71e7ba%2F3b875c65-bdda-49e3-a306-53e660aa72a6%2Fb5nkqhs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:-
Name: Vanesa Gomzalez
Date: 11-28-23
Bell: 4th
** This is a 2-page document! **
Directions: Find each value or measure. Assume that segments that appear to be tangent are tangent.
258
2. MAD
1. mZYZV
144
+114
161°
2
258
161+x= 93
2
161+x=186
961 -161
X=25
3. mQS
128
x 2
250
5. mZEFG
-39
7. mCF
180
53
129
127
366
ما25 -
34+x=121
2
104
391+x=242
-39
my
mQS=104
x=203
37
P
9. Solve for x.
5x-7+84+27:/8C
5x+190-120
(x = 8)
119
144°
127-53-37
2
121
2
R
C
128°
T
Z
D.
Ĉ
V
W
D
127
X
39°
41%
S
114°
E
G
E
53°
Unit 10: Circles
Homework 6: Arc & Angle Measures
(5x - 7)X\27°
p
4. m/JLK
298/2 = 149
<JLK = 31
6. m/TUV
360
-108
252
8. mHJ
252/2=2
126
402
2606
102
18G
- 149
31
298
180
29=1/2(96-X)
58-96-X
38=x
10. Solve for x.
360 23x-3=204
+3 +3
U
B
93°
E
C
N
23x=20723x-3)
X = 9
T
108
96°
G
X80
K
204
K
DAYAN
L
360
29°
H
M
25/
78°
torat
Sy)
S
1
951
Gina Wilson (All Things Algebra), 2018
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