Download MATH 130 FINAL REVIEW version2

MATH 130 FINAL REVIEW version2
Problems 1 – 3 refer to triangle ABC, with = °. Find the remaining angle(s) and side(s).
1. = 50°, = 25
a) = 40°, = 32.6, = 21.0
b) = 50°, = 21.0, = 32.6
c) = 40°, = 21.0, = 32.6
d) = 50°, = 32.6, = 21.0
2. = 9, = 12
a) = 48.6°, = 41.4°, = 3√7
b) = 41.4°, = 48.6°, = 3√7
c) = 53.1°, = 36.9°, = 15
d) = 36.9°, = 53.1°, = 15
3. = 55°42 , = 8.85ℎ !
a) = 34°58 , = 4.98ℎ !, = 7.29ℎ !
b) = 34°18 , = 4.99ℎ !, = 7.31ℎ !
c) = 34°58 , = 4.99ℎ !, = 7.31ℎ !
d) = 34°18 , = 5.08ℎ !, = 7.25ℎ !
For problems 4 – 5, find the remaining five trig functions of ϴ if:
'(
4. sin % = − ') and ϴ terminates in Quadrant III.
a) csc % = −
')
'(
,
c) csc % = −
cos % = − ')
')
,
tan % = −
')
'(
,
cos % = ')
sec % = b) csc % = sec % = −
sec % = −
'(
tan % = ,
cot % = '(
tan % = − ,
,
cot % = − '(
cot % = − '(
,
cos % = − ')
')
,
'(
,
')
'(
')
,
d) csc % = ')
'(
,
cos % = ')
sec % = ')
,
'(
,
tan % = '(
,
,
cot % = '(
,
'
5. tan % = − 0 and ϴ terminates in Quadrant II.
a) cot % = −3
sin % = b) cot % = −3
√'1
'1
sin % = −
csc % = √10
c) cot % = −3
√'1
'1
sin % =
csc % = −√10
d) cot % = −3
0√'1
'1
sin % = −
0√'1
'1
√'1
0
csc % = −
√'1
0
csc % = √'1
'1
cos % = −
0√'1
'1
cosθ = 0√'1
'1
cos % = −
sec % = −
√'1
0
sec % = √'1
0
sec % = −√10
cos % = √'1
'1
sec % = √10
For problems 6 - 7, find the requested trig function:
(
6. If cotθ = − 0 and sin % > 0, then sec % =
a)
√'0
(
b)
(
√'0
c) −
√'0
(
d) −
(
√'0
7. If sec % = √5 and tan % < 0 , then csc % =
a) −
(√)
)
b)
√)
(
c) −√6
d) −
√)
(
For problems 8 – 9, simplify after making the given substitution
8. √16 − 165 ( , 5 = cos %
a) 16 sin %
b) 4 sin %
c) 4|sin %|
d) 16|sin %|
b) 3 csc %
c) 9 csc %
d) 9|csc %|
9. √5 ( + 9 , 5 = 3 cot %
a) 3|csc %|
For problems 10 – 11, perform the operation and simplify. Answers should be in terms of
89: ;and/or <=8 ; .
10.
>?@A BCDE@A B
'C>FGA B
=
a) cos ( %
b) sin( %
c) − sin( %
d) − cos ( %
11. sec 0 % − tan( % sec % =
'
a)
GH> B
b) cos %
c)
>?@A B
GH>A B
d)
'
GH>A B
For problems 12 – 13, write in terms of 89: ;and/or <=8 ; and simplify.
12.
'
G>G B DE@ B
a) cos %
b)
>?@A B
c)
b)
'
c)
GH> B
'
d)
'
d)
GH> B
GH> B
>?@A B
13. tan % + cot %
a) sin % cos %
>?@ B GH> B
>?@ B
>?@A B
GH>A B
For problems 14 – 15, give an angle between °and IJ° coterminal with the given angle.
14. −155°
a) 155°
b) 205°
c) 25°
d) 115°
a) 15°
b) 215°
c) 115°
d) 205°
15. 475°
For problems 16 - 20, evaluate without using a calculator.
16. arcsin L−
a)
√(
M
(
N
O
b)
)N PN
c)
PN
b)
N
c)
N
O
,
O
O
d) −
Q
4
17. tanC'R√3S
a)
N
0
0
,
ON
0
T
d)
N
T
,
PN
T
'
18. sin LcosC' L− (MM
a)
'
b) -
(
'
c) -
(
√0
(
d)
√0
(
19. sinC'Utan 135°V
a) −90°
b) 90°
c) 270°
d) 0°
'
(W
20. sec LtanC' L MM
a) √45 ( + 1
b)
√OW A X'
(W
c)
√(W A X'
(W
d)
(W
√OW A X'
For problems 21 and 22, ϴ is a central angle in a circle of radius r. Find the requested value.
21. % = 0N
O
, r = 4 inches. Find the arc length s.
a)
0N
'T
inches
b) 12Q inches
c) 3Q inches
d) 6Q inches
c) 900Q(
d)
22. % = 72°, Y = 5. Find the area of the sector.
a) 5Q(
b) 10Q(
N
)
(
For problems 23 – 25, Identify the amplitude, period, phase (horizontal) shift, and vertical shift.
23. Z = 1 − 3 sinU25 + QV
a) Amp = 3
Per = 2Q
N
b) Amp = 3
c) Amp = -3
d) Amp = 3
Per = Q
Per = Q
Per = Q
N
(
HS = - (
HS =
VS = 1
VS = 1
N
N
HS = -(
HS = − (
VS = 1
VS = 1
N
24. Z = 2 + 2 sec L5 − O M
a) Amp = none
b) Amp = none
Per = 2Q
Per = 2Q
N
O
HS =
HS = −
VS = 2
N
O
VS = 2
c) Amp = 2
Per = 2Q
d) Amp = 2
Per = Q
N
O
HS = -
VS = 2
VS = 2
HS =
N
O
N
25. Z = tan L25 − ( M
a) Amp = 1
Per =
HS =
b) Amp = none
N
(
HS =
VS = 0
Per = Q
N
O
HS = − O
HS = (
VS = 0
VS = 0
VS = 0
26.
NW
M
(
Z = 2 + 2[! L
c)
Z = 4 + cos L
NW
M
(
Per = Q
N
For problem 26, find the equation that matches the graph.
a)
d) Amp = none
N
Per = (
N
O
c) Amp = 1
b) Z = 2 + 2 cos 2Q5
W
d) Z = 2 + 2 cos L(M
N
,
)
For problems 27 - 31, let sin = − ') with A in QIII and cos = '0 with B in QIV. Evaluate the
following.
27. sinU + V
a)
T0
T)
b) - T0
T)
c) - 00
d)
00
T)
T)
28. cosU − V
a)
)T
b)
'T
c) -
'T
d) -
)T
'(1
b)
(O
c) -
(O
d) -
'(1
c)
P
T)
T)
T)
T)
29. sinU2V
a)
'T,
()
()
'T,
30. cosU2V
a) -
P
()
b) 1
57
()
d) 25
(O
d) -
31. tanU2V
a)
'(1
'',
b)
(O
P
c) -
)
'(1
'',
O
For problems 32 – 34, use Half-Angle formulas to evaluate given that tan = − 0 and A is in QII.
\
32. tan L ( M =
a)
'
(
b) - 2
c) -
'
(
d) 2
\
33. sin L ( M =
a)
√)
)
b)
(√)
)
c) -
√)
)
d) -
(√)
)
√)
)
b)
(√)
)
c) -
√)
)
d) -
(√)
)
\
34. cos L ( M =
a)
For problems 35 – 36, find all solutions in the interval ° ≤ ; < IJ°.
35. 2 sin % − 2 = 0
a) 0°, 180°
b) 270°
c) 90°
d) 30°, 150°
36. tan( % + tan % = 0
a) 0°, 180°
b) 135°, 315°
c) 45°, 90°, 225°270°
d) 0°, 135°, 180°, 315°
For problems 37 – 38, find all solutions in the interval ≤ ^ < _`.
37. 4 cos( 5 − 4 sin 5 − 5 = 0
a)
N
T
,
)N
T
b)
PN
b)
N ''N
T
,
''N
T
c)
N
c)
N
0
,
(N
0
d)
ON
d)
N
0
)N
,
0
'
38. cos 25 cos 5 + sin 25 sin 5 = (
a)
N (N
0
,
0
T
,
T
0
0
,
)N
0
For problem 39, find all solutions. Write the answer in radians using exact values.
39. Ucos 45V( = 1
a)
N
(
k
b) 2Qa
c)
N
O
k
d)
N
b
N
+ k
(
For problems 40 - 42, refer to triangle ABC which is not necessarily a right triangle. Find the requested
value(s).
40. If = 110°, c = 40°, d = 18.0ℎ !, find .
a) 9.6 inches
b) 12.3 inches
c) not enough
information
d) 33.8 inches
c) 5.7 m
d) 11.1 m
41. If = 3.7. = 6.4, d = 23°, find .
a) 3.3 m
b) 4.1 m
42. If = 51, = 24, d = 31,find the smallest angle.
a) 15°
b) 19°
c) 25°
d) 38°
For problem 43, the diagonals of a parallelogram are 56.0 cm and 34.0 cm. They meet at an angle of
120°.
43. Find the length of the shorter side.
a) 39.5 cm
b) 22.8 cm
c) 24.6 cm
d) 16.1 cm
For problems 44 – 45, refer to triangle ABC, which is not necessarily a right triangle. Find the area of
the triangle. Round to the nearest tenth.
44. = 57°, c = 31°, = 7.3
a) 43.4(
b) 23.0(
c) 46.0(
d) 11.5(
c) 45.9(
d) 18.0(
45. a = 4.8 cm, b = 6.3 cm, c = 7.5 cm
a) 4.9(
b) 15.0(
For problems 46 - 47, multiply or divide as indicated. Leave answers in standard form.
46. U2 + 3V ∙ U4 − 5V
a) 6 − 2
47.
b) 8 − 13
c) −7 + 2
d) 23 + 2
U0XOfV
U0COfV
a) -
P
(O
+ () ()
b) 1 − (O
P
c)
P
()
(O
+ () d) 1 − For problem 48, convert the complex number from standard form to trigonometric form.
48. g = 2 − 2√3
a) 4!60°
b) 4!330°
c) 4!300°
d) 4!30°
For problems 49 – 51, multiply or divide as indicated. Leave answers in Trigonometric form.
49. 6Ucos 120° + sin 120°V ∙ 3Ucos 40° + sin 40°V
a) 18!80°
b) 18!160°
c) 9!160°
d) 2!80°
50.
')UGH> (()°Xf >?@ (()°V
)UGH> O)°Xf >?@ O)°V
a) 3!270°
b) 5!180°
c) 75!270°
d) 3!180°
b) 6√3!300°
c) 27!300°
d) 27!360°
T
51. R√3!110°S
a) 6√3!660°
For problem 52, use DeMoivre’s Theorem and convert the answer back to standard form.
52. U1 − VO
a) −4
b) 4
c) 4
d) −4
For problems 53 – 54, find the indicated roots. Leave the answers in trigonometric form.
53. Find two square roots of g = 81Ucos 120° + sin 120°V .
a) 3!60°, 3!300°
b) 9!60°, 9!240°
c) 3!60°, 3!240°
d) 9!120°, 9!300°
54. Find three cube roots of g = −4√3 + 4.
a) 4!50°, 4!170°, 4!290°
b) 2!75°, 2!150°, 2!300°
c) 2!40°, 2!160°, 2!280°
d) 2!50°, 2!170°, 2!290°
Word Problems 55. Distance - Two straight wires are strung on opposite sides of a tent pole and anchored to the
ground by two stakes. One of the wires is 56 feet long and makes an angle of 47° with the ground. The
other wire is 65 feet long and makes an angle of 37° with the ground. How far apart are the stakes that
hold the wires to the ground?
a) 97 feet
b) 86 feet
c) 80 feet
d) 90 feet
56. Angle of Elevation/Height – To estimate the height of a tree, two people position themselves 25
feet apart. From the first person, the bearing of the tree is N 48° E and the angle of elevation to the top
of the tree is 73°. If the bearing of the tree from the second person is N 38° W, find the height of the
tree to the nearest foot.
a) 50 feet
b) 60 feet
c) 65 feet
d) 70 feet
57. Bearing & Distance - A boy is riding his motorcycle on a road that runs east and west. He leaves
the road at a service station and rides 5.25 miles in the direction N 15.5° E. Then he turns toward his
right (do NOT assume 90°) and rides 6.50 miles back to the road, where his motorcycle breaks down.
How far will he have to walk to get back to the service station? What direction will he be walking in?
a) 5.48 miles, W
b) 8.36 miles W
c) 8.36 miles, E
d) 5.48 miles, E
58. True Course and Speed – A helicopter is flying at 90 mph on a heading of 40°. A 20 mph wind is
blowing from the NE on a heading of 190°. What is the true course and speed of the helicopter relative
to the ground?
a) 73hℎ, Yi7.8°
b) 70hℎ, Yi150°
c) 73hℎ, Yi47.8°
d) 70hℎ, Yi50°
ANSWER KEY
1.
c
21.
c
41.
a
2.
d
22.
a
42.
b
3.
b
23.
d
43.
c
4.
c
24.
a
44.
d
5.
a
25.
b
45.
b
6.
c
26.
a
46.
d
7.
d
27.
d
47.
a
8.
c
28.
b
48.
c
9.
a
29.
b
49.
b
10.
b
30.
c
50.
d
11.
a
31.
a
51.
c
12.
a
32.
d
52.
a
13.
b
33.
b
53.
b
14.
b
34.
a
54.
d
15.
c
35.
c
55.
d
16.
d
36.
d
56.
c
17.
a
37.
b
57.
a
18.
d
38.
d
58.
c
19.
a
39.
c
20.
b
40.
a