Download Know the following:

Practice Test
Chapter 5
For Credit show all work
Know the following:
Trigonometric Ratios
Sine (sin  = ratio)
Cosine (cos  = ratio)
= opposite side /hypotenuse
= adjacent side / hypotenuse
= y/r
= x/r
= 1/csc 
= 1/ sec 
Inverse: sin-1(ratio) = angle
Inverse: cos-1(ratio) = angle
Cosecant (csc )
Secant (sec )
= r/y
= r/x
= 1/ csc 
= 1/ cos 
Values Quadrant I

sin  cos  tan  csc  sec  cot 
30 0
1
2
3
2
3
3
2
2 3
3
45 0
2
2
2
2
1
2
2
1
2 3
3
2
3
3
60 0
1
3
3
2
2
Formulas
Pythagorean Theorem
Use: for all right triangles.
c2  a2  b2
Area: A 
1
bh
2
Page 1 of 11
Tangent (tan  = ratio)
= opposite side/ adjacent side
= y/x
= 1/ cot 
Inverse: tan-1 (ratio) = angle
Cotangent (cot )
= x/y
= 1/ tan 
3
Law of Sines
Use: Angle (must be less than 90o) and
opposite side, and another side or
angle;
a
b
c


sin A sin B sin C
1
bc sin A
2
1
Area: K  ac sin B
2
1
K  ab sin C
2
K
Law of Cosines
Use: 3 sides or
Angle and 2 non-opposing
sides.
c 2  a 2  b 2  2ab cos C
b 2  a 2  c 2  2ac cos B
a 2  b 2  c 2  2bc cos A
1 sin B sin C
K  a2
2
sin A
1 sin A sin C
Area: K  b 2
2
sin B
1 sin A sin B
K  c2
2
sin C
Practice Test
Chapter 5
For Credit show all work
Law of Sines
Ambiguous Case
(A copy of the Ambiguous
Case will be supplied
during the test)
Case I: A <900Case I: A <900
a<b
a≥b
No solution
One Solution
Two Solutions
One solution
Case II: A ≥900
a≤b
a>b
No solution
One Solution
Use the figure on the
right for questions 1 3.
a< b sin a
a = b sin a
a > b sin a
B
9
a
C
4
1. Find the value of the sine for A
2. Find the value of the cotangent of A
3. Find the value of the secant of A
4. If the tan  = -3 find the cot 
5. Find the tan 1800
6. Find the exact value of cos 3300
For exercises 7 and 8, refer to
River to waterfall
the figure on the right.
The angle of elevation from
Height
of the
the far side of the pool to the
0
waterfall
top of the waterfall is 68 and
the distance is 200 feet.
A
1. ___________
2. __________
3. __________
4. _________
5. _________
6. _______________
pool of water
7. Find the height of the waterfall to the nearest foot.
8. Find the width across the pool to the nearest foot.
3
9. If 00≥ x ≤ 3600, solve sin x  
2
Page 2 of 11
7. _____
8. __________
9. ___________
Practice Test
Chapter 5
For Credit show all work
10. Given the triangle on the
right, find B to the nearest
tenth of a degree if b = 12, and
c = 18.
10. ________
B
c
a
C
b
A
For exercise 11 and 12, round the numbers to the nearest tenth.
11. If ∆ABC, A= 470, B=580, b=23. Find a
11. ______________
0
12. If C=37.2 , a =17.9 and b = 22.3, find the area of ∆ABC
12. _______________
13. Determine the number of possible solutions if A=470, a = 2
13. ____________
and b=4.
For exercise 14 and 15, round the numbers to the nearest tenth.
14. In ∆ABC, A=670, b = 10 and c= 5, Find a
14. _________
15. In ∆ABC, a=8, b = 6 and c= 12, Find C
15. __________
16. If a=18, b = 22 and c =30, find the area of ∆ABC
16. _________
17. Aviation A traffic helicopter is flying 1000 feet above the
17. __________
downtown area. To the right, the pilot sees the baseball stadium
at an angle of depression of 630. To the left the pilot sees the
football stadium at an angle of depression of 180. Find the
distance between the two stadiums.
18. Railways: The steepest railway in the world is the
18 A) __________
Katoomba Scenic Railway in Australia. The passenger car is
18 B) __________
pulled up the mountain by twin steel cables. It travels along the
track 1020 feet to obtain a change of altitude of 647 feet.
a) Find the angle of elevation
b) How far does the train travel horizontally.
19. Engineering The escalator of St. Petersburg Metro in
19. ________
Russia has a vertical rise of 195.8 feet. If the angle of elevation
of the escalator is 100 21’ 36”, find the length of the escalator?
20. Architecture: A flagpole 40 feet high stands on top of the
20. _________
Wentworth Building. From the point in front of Bailey’s
Drugstore, the angle of elevation for the top of the pole is 540
54’ and the angle to the bottom of the pole is 470 30’. How high
is the building?
Page 3 of 11
Practice Test
Chapter 5
For Credit show all work
Solutions
Use the figure on the
right for questions 1 3.
B
9
a
C
4
A
1. Find the value of the sine for A
Sin A = opposite side / hypotenuse = a/c.
Hypotenuse = 9
Find opposite side a
a2  b2  c2
1. ___________
65
9
a 2  42  92
a 2  9 2  4 2  81  16  65
a 2  75
a 2  65
a  65
Find sin A
a
65
sin A  
9
9
2. Find the value of the cotangent of A
Cot A = adjacent side / opposite side = b/a
From the prior problem we know that a  65
B =4
4
Cot A =
which is improper ¾ credit for this answer.
65
Complete answer
4  65  4 65


65
65  65 
3. Find the value of the secant of A
Sec A = hypotenuse / adjacent side = c/b {inverse of the cos}
Adjacent = 4
Hypotenuse = 9
Sec A = 9/4 or 2.25
Page 4 of 11
2. __________
4 65
65
3. __________
9/4 or 2.25
Practice Test
Chapter 5
For Credit show all work
4. If the tan  = -3 find the cot 
Cot  = 1/ tan  tan  = -3
Cot  = 1/(-3) = -1/3
5. Find the tan 1800
Tan  = y/x, use calculator or coordinates (-1,0)
Tan 180 = 0/(-1)=0
6. Find the exact value of cos 3300
Since the question asks for exact value use the radical
3300 is in Quadrant IV
3300 has a reference angle in Quadrant I 300
Cosine = x/r and x is positive in both Quadrant I and IV
3
Cos 3300 = Cos 300 =
2
For exercises 7 and 8, refer to
River to waterfall
the figure on the right.
The angle of elevation from
Height
of the
the far side of the pool to the
waterfall
top of the waterfall is 680 and
the distance is 200 feet.
4. _________
-1/3
5. _________
0
6. _______________
3
2
pool of water
7. Find the height of the waterfall to the nearest foot.
x
200
feet
680
The angle of elevation is 680
The height is the opposite side =x
The hypotenuse is 200 feet
x
sin 680 
200
200 sin 68 0  x  185.43
Page 5 of 11
7. _____
185 feet
Practice Test
Chapter 5
For Credit show all work
8. Find the width across the pool to the nearest foot.
200 feet
680
Adjacent = x
8. __________
75 feet
The angle of elevation is 680
The pool is the adjacent side = x
The hypotenuse is 200 feet
Cos = adjacent side / hypotenuse
x
cos 680 
200
200 cos 680  x  74.9
9. If 00≥ x ≤ 3600, solve sin x  
9. ___________
3
2
Sin  = y/r
Y is negative in Quadrants III and IV
3
Sin 2400 and sin 3000  
2
10. Given the triangle on the
right, find B to the nearest
B
tenth of a degree if b = 12, and
a
c = 18.
C
Sin 2400 and 3000
10. ________
c
b
A
For angle B b is the opposite side
41.80
For a right triangle c is the hypotenuse
Sin-1(opposite/hypotenuse) = angle
Sin-1(12/18)= 41.810
For exercise 11 and 12, round the numbers to the nearest tenth.
11. If ∆ABC, A= 470, B=580, b=23. Find a
11. ______________
Law of Sines: we have an acute triangle and a side angle pair.
19.8 feet
a
b
c


sin A sin B sin C
a
23

sin 47 sin 58
sin 47  a  23 sin 47 
sin 47 sin 58
23 sin 47   19.83
a
sin 58
Page 6 of 11
Practice Test
Chapter 5
For Credit show all work
12. If C=37.20, a =17.9 and b = 22.3, find the area of ∆ABC
12. _______________
120.7 square units
1
K  ab sin C
2
K  .517.922.3sin 37.2  120.66
13. Determine the number of possible solutions if A=470, a = 2
13. ____________
and b=4.
Law of Sines Ambiguous Case
no solution
0
0
Case I because 47 < 90
a < b true 2 <4
a ? b sin a
2 ? 4 sin 470
2 < 2.9 using rules no solution
For exercise 14 and 15, round the numbers to the nearest tenth.
14. In ∆ABC, A=670, b = 10 and c= 5, Find a
14. _________
Law of Cosines
a= 9.3
2
2
2
a = b + c -2bc cosA
a2 = 102 + 52 -2(10)(5)cos 670
a2 = 100 + 25 – 100 (cos 67) = 85.92
a = 85.92 = 9.26
15. In ∆ABC, a=8, b = 6 and c= 12, Find C
15. __________
Law of Cosines
117.30
c2 = a2+ b2 – 2ab cos C
122 = 82 + 62 -2(8)(6) cos C
144 = 64 + 36 -96 cos C
144- 64-36 = -96 cos C
(144-64-36) / (-96) = cos C
-0.45833 = cos C
cos-1(-0.45833) = C = 117.27
16. If a=18, b = 22 and c =30, find the area of ∆ABC
16. _________
Using the law of cosines find an angle,
 196.7 square units
Then using the area formulas we can find the area.
1
a 2  b 2  c 2  2ab sin A
K  bc sin A
2
182 = 222 + 302—2(22)(30) cos A
(182 - 222 - 302) = -1320 Cos A
-1060 = -1320 Cos A
Cos-1(-1060/-1320) = A = 36.57  36.60
Area = (.5)(22)(30)(sin 36.58) = 196.66
Page 7 of 11
Practice Test
Chapter 5
For Credit show all work
17. Aviation A traffic helicopter is flying 1000 feet above the
downtown area. To the right, the pilot sees the baseball stadium
at an angle of depression of 630. To the left the pilot sees the
football stadium at an angle of depression of 180. Find the
distance between the two stadiums.
Step 1. Draw the picture
Step 2 Name Variables
H = helicopter
F = Football Stadium
B = baseball stadium
D = downtown
Step 3 Write in constants
Angle of depression is look down from the helicopter.
17. __________
H
180
630
1000 Feet
F
D
B
H
180
630
1000 Feet
180
F
630
D
B
Step 4 Write the question in terms of the drawing.
Find the distance of FD plus DB
There are two right triangle ∆FDH and ∆BDH
FD is adjacent, DH is opposite B and
DB is adjacent, DH is opposite B
1000
1000
tan 63 
tan 18 
DB
FD
1000
1000
DB 
 509.52
FD 
 3077.68
tan 63
tan 18
3077.68 + 509.52 = 3587.2
Page 8 of 11
3587 feet
Practice Test
Chapter 5
For Credit show all work
18. Railways: The steepest railway in the world is the
Katoomba Scenic Railway in Australia. The passenger car is
pulled up the mountain by twin steel cables. It travels along the
track 1020 feet to obtain a change of altitude of 647 feet.
a) Find the angle of elevation
b) How far does the train travel horizontally.
Step 1. Draw the picture
Step 2 Name Variables
a= altitude b = horizontal
Step 3 Write in constants
18 A) __________
18 B) __________
B
647
feet
C
1020 feet
b
A
Step 4 Write the question in terms of the drawing
Angle of elevation is looking up or A.
 647 
0
A  sin 1 
  39.36
 1020 
2
2
647  b  1020 2
b 2  1020 2  647 2  621791
b  621791  788.53
Page 9 of 11
a) 39.40
b) 788 feet
Practice Test
Chapter 5
For Credit show all work
19. Engineering The escalator of St. Petersburg Metro in
Russia has a vertical rise of 195.8 feet. If the angle of elevation
of the escalator is 100 21’ 36”, find the length of the escalator?
Step 1. Draw the picture
Step 2 Name Variables
c = length of the escalator
Step 3 Write in constants
19. ________
B
195.8
feet
x
C
Covert to decimal
10  21 / 60  36 / 3600  10.36
195.8
sin( 10.36) 
x
195.8
x
 1088.79
sin 10.36
Page 10 of 11
A = 100 21’ 36”
 1088.8 feet
Practice Test
Chapter 5
For Credit show all work
20. Architecture: A flagpole 40 feet high stands on top of the
Wentworth Building. From the point in front of Bailey’s
Drugstore, the angle of elevation for the top of the pole is 540
54’ and the angle to the bottom of the pole is 470 30’. How high
is the building?
Step 1. Draw the picture
Step 2 Name Variables
x = height of the building.
d = distance to the building
Step 3 Write in constants
F
50 feet
20. _________
FAC= 540 54’ =54.9
BAC= 470 30’ =47.50
B
x
C
Wentworth
Building
d
A
Bailey’s
Drugstore
50  x   50  x 
50  x
d 
d
tan FAC  tan 54.9
x
tan( BAC ) 
d
x
tan 47.5 
d
d tan 47.5  x
tan( FAC) 
 50  x  

tan 47.5  x
 tan 54.9 
50 tan 47.5  x tan 47.5  x tan 54.9
50 tan 47.5  x tan 54.9  x tan 47.5
50 tan 47.5  xtan 54.9  tan 47.5
50 tan 47.5
 x  164.57
tan 54.9  tan 47.5
Page 11 of 11
 165.6 feet