Algebra 2: Unit 10: Trigonometry
NAME: ____________________
Review
(1) Given the triangle below, determine the value of each trig function. Try not to use the calculator.
sinθ =
5
13
cscθ =
cosθ =
12
13
secθ =
5
12
cotθ =
tanθ =
13
5
13
12
θ
12
12
5
30
Find the missing sides and angles:
(2)
(3)
45 5√10
2
5√5
45
5√10
2
Find the missing angles and sides:
(11)
9
11)
64o
3.94
12)
sin 
3√3
30
3√2
6
60
√6
(12)
8.09
36
√7
2
(4)
2√6
26o
(14) If sec θ =
13
5
(13)
5
2.77
32
4.35
41
49
3.27
6.63
58
, find the values of the other 5 trig functions.
21
7
cos 
2 7
3
tan 
7
2
csc  
21
2 3
cot  
3
3
Solve for the missing angles and sides of the triangles with the following properties.
(15) B = 53o, b = 24.5, A = 102.1o
(16) b = 250, c = 125.9, A = 45o
A = 102.1o
a = 30.00
A = 45o
a = 183.95
o
o
B = 53
b = 24.5
B = 106.06
b = 250
o
o
C = 24.9
c = 12.92
C = 28.94
c = 125.9
a
b
c
b


sin A sin B
sin C sin B
sin102.1
sin24.9
a  24.5
c  24.5
sin53
sin53
a  30.00
c  12.92
a  2502  125.92  2(250)(125.9)cos(45 )
a  183.95
sin C sin A

c
a
125.9 

C  sin1  sin(45)
183.95 

C  28.94
60
3
(17) B = 35o, b = 22, a = 35
Triangle #1
A = 65.85o
a = 35
o
B = 35
b = 22
o
C = 79.15
c = 37.67
(18) b = 14, a = 5, A = 100.4o
No Triangle possible
A = 100.4o
a=5
o
B=
b = 14
o
C=
c=
c
b
a
b


sin C sin B
sin A sin B
sin  79.15 
35 

A  sin1  sin(35)  c  22
sin  35 
22 

A  65.85
c  37.67
Triangle #2
A = 114.15o
B = 35o
C = 30.85 o
a = 35
b = 22
c = 19.67
c
b

sin C sin B
sin  30.85 
c  22
sin  35 
c  19.67
(19) a = 240, b = 180, c = 222
A = 72.46o
B = 45.65o
C = 61.89o
a = 240
b = 180
c = 222
 2402  180 2  2222 
A  cos 

 2(180)(222) 
A  72.46
(20) a = 10, b = 15, A = 30o
Triangle #1
A = 30o
a = 10
o
B = 48.59
b = 15
o
C = 101.41
c = 19.60
1
sin B sin A

b
a
180 

B  sin1  sin(72.46)
240 

B  45.65
c
a
sin B sin A


sin C sin A
b
a
sin 101.41
15 

B  sin1  sin(30)  c  10
sin  30 
10 

B  48.59
c  19.60
Triangle #2
A = 30o
B = 131.41o
C = 18.59o
a = 10
b = 15
c = 6.38
c
a

sin C sin A
sin 18.59 
c  10
sin  30 
c  6.38
Determine if there is No triangle, One triangle, or Two triangles possible for each given set of values.
a. A = 76.4◦; a = 176; b = 189
No triangle
A is acute. a < b
bsinA = 183.7 > a = 176
b. A = 48.2◦; a = 15; b = 20
Two triangles
A is acute. a < b
bsinA = 14.91 < a = 15
c. A = 20◦; a = 10; c = 11
Two triangles
A is acute. a < c
csinA = 3.76 < a = 10
d. C = 95◦; a = 8; c = 9
One triangle
A is obtuse. c > a
1. Surveying Two sides of a triangular plot of land have lengths of 400 feet and 600 feet. The measure of the angle
between those sides is 46.3o. Find the missing side and angles of the plot of land.
400‘ 46.3o
o
600‘
a  4002  6002  2(400)(600)cos(46.3 )
a  434.02'
41.78o
91.92
434.02‘
sin C sin A

c
a
400 

C  sin1  sin(46.3)
434.02 

C  41.78
2. Aviation A pilot takes off from the Newport News Airport in Newport News, Virginia. The pilot flies out towards the
Atlantic Ocean and after reaching point C, the plane develops mechanical difficulties, and the pilot needs to return
to either the Newport News or Norfolk Airport. How far is it to the nearer airport?
58o
24.2miles 96o
C
a
c

sin A sin C
sin58
a  24.2
sin26
C  46.82 miles
3. Emergency Medicine A medical rescue helicopter has flown 45 miles from its home base to pick up an accident
victim and 35 miles from there to the hospital. The angle between the two legs of the trip was 125o. The pilot needs
to know how far he is now from his home base so he can decide whether to refuel before returning. How far is the
hospital from the helicopter’s base?
Base
a  452  352  2(45)(35)cos(125 )
a  71.11miles
45m
i
125o
Accident
35m
i
Hospital