Download 3 Practice Trig Test

Name: ________________________ Class: ___________________ Date: __________
ID: A
Practice Trigonometry Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. If tan B = 1.0724 then the measure of ∠B, to the nearest degree, is
a. 2°
c. 43°
b. 18°
d. 47°
____
2. In
a.
b.
____
0.60
0.75
c.
d.
1.20
1.33
3. Determine the measure of ∠C, to the nearest degree.
a.
b.
____
ABC, AB = 15 cm and AC = 9 cm. Determine the tangent ratio of ∠B, to the nearest hundredth.
19°
20°
c.
d.
21°
22°
4. A school soccer field measures 45 m by 65 m. To get home more quickly, you decide to walk along the
diagonal of the field. What is the angle of your path, with respect to the 45-m side, to the nearest degree?
a. 55°
c. 2°
b. 34°
d. 1°
1
Name: ________________________
ID: A
Use the diagram to answer the following question(s).
____
5. Determine the length of y, to the nearest tenth of a metre.
a. 5.7 m
c. 7.0 m
b. 6.3 m
d. 8.2 m
____
6. A wheelchair ramp is being built for the entrance to a school. If the ramp makes an angle of 4° with the
ground and has a horizontal length of 6 m, determine the height of the ramp, to the nearest tenth of a metre.
a. 1.1 m
c. 0.4 m
b. 0.8 m
d. 0.3 m
____
7. Evaluate sin 67°, to four decimal places.
a. 2.3558
b. 0.9205
c.
d.
0.8554
0.3907
____
8. Determine the value of sin 28°, to four decimal places.
a. 0.4695
c. 0.8829
b. 0.5317
d. 0.9709
____
9. If cos A = 0.9903, then the measure of ∠A, to the nearest degree, is
a. 45°
c. 8°
b. 82°
d. 1°
____ 10. If cos B = 0.6561, then the measure of ∠B, to the nearest degree, is
a. 49°
c. 33°
b. 41°
d. 29°
____ 11. If sin A = 0.7986, determine the measure of ∠A, to the nearest degree.
a. 1°
c. 39°
b. 7°
d. 53°
2
Name: ________________________
____ 12. In
a.
b.
ID: A
ABC, AC = 8 cm and BC = 11 cm. Determine the sine ratio of ∠B, rounded to the nearest thousandth.
0.588
0.728
c.
d.
0.809
1.375
____ 13. Write the cosine ratio of ∠A.
a.
b.
AB
AC
BC
AB
____ 14. Determine the value of cos 0°.
a. –1
b. 0
____ 15. In
a.
b.
c.
d.
c.
d.
AC
BC
BC
AC
1
undefined
TUV, UV = 8 m, ∠U = 90°, and ∠T = 38°. Determine the measure of ∠V, to the nearest degree
42°
c. 90°
52°
d. 128°
3
Name: ________________________
ID: A
Use the diagram to answer the following question(s).
Kelly is flying a kite in a field. He lets out 40 m of his kite string, which makes an angle of 72° with the
ground.
____ 16. Suppose the sun is shining directly above the kite. How far is the kite’s shadow from Kelly, to the nearest
metre?
a. 12 m
c. 42 m
b. 38 m
d. 129 m
7
, what is the measure of ∠A, to the nearest degree?
8
62°
c. 60°
61°
d. 59°
____ 17. If sin A =
a.
b.
____ 18. Determine the length of x and the length of y, to the nearest tenth of a metre.
a.
b.
x = 7.2 m and y = 9.7 m
x = 9.6 m and y = 12.9 m
c.
d.
4
x = 9.6 m and y = 14.3 m
x = 7.2 m and y = 10.8 m
Name: ________________________
ID: A
____ 19. A window cleaner places a ladder that is 8 m long against a wall. The top of the ladder is 6 m above the
ground. Determine the angle between the base of the ladder and the ground, to the nearest tenth of a degree.
a.
b.
36.9°
41.4°
c.
d.
48.5°
51.2°
____ 20. Which statement is incorrect?
a. You can solve for the unknown side in any triangle, if you know the lengths of the other
two sides, by using the Pythagorean theorem.
b. The hypotenuse is the longest side in a right triangle.
c. The hypotenuse is always opposite the 90° angle in a right triangle.
d. The Pythagorean theorem applies to all right triangles.
Completion
Complete each statement.
1. The hypotenuse is the side _________________________ the right angle in a right triangle.
2. The tangent ratio of ∠A can be written as _________________________ .
3. The sine ratio of ∠A compares the length of the _________________________ side to the length of the
hypotenuse.
4. The angle of elevation is the angle between the horizontal and the line of sight looking
_________________________ to an object.
5. The angle between the horizontal and the line of sight looking up to an object is called the angle of
_________________________ .
5
Name: ________________________
ID: A
Matching
Match the correct side or angle of PQR to each of the following definitions, descriptions, or expressions. A
term may be used more than once or not at all.
a.
b.
c.
p
p
q
p
r
f.
q
r
g.
–1
h.
0
d.
Pythagorean theorem
i.
e.
q
j.
____
1. the side opposite ∠Q
____
2. r 2 = q 2 + p 2
1
pq
2
1
Match each term to the correct definition, description, or expression. A term may be used more than once or
not at all.
a. angle of elevation
e. Pythagorean theorem
b. angle of depression
f. sine ratio
c. cosine ratio
g. tangent ratio
d. hypotenuse
h. trigonometry
____
3. the angle between the horizontal and the line of sight looking down to an object
____
4. in a right triangle with side lengths a, b, and c, where c is the hypotenuse, a 2 + b 2 = c 2
____
5.
____
6. the side opposite the right angle in a right triangle
____
7.
length of side opposite an acute angle
length of side adjacent to an acute angle
length of side adjacent to an acute angle
length of hypotenuse
6
Name: ________________________
ID: A
Short Answer
1. A flight of stairs has a ratio of vertical distance to horizontal distance of 3 to 5. What angle does the flight of
stairs make with the ground, to the nearest degree?
2. Denis is building a playhouse for his younger brother. The roof of the playhouse is shaped like an isosceles
triangle. The diagonal trusses of the roof make an angle of 24° with the horizontal. The height of the roof is
3.5 m. How long are the diagonal trusses AB and AC, to the nearest tenth of a metre?
3. In XYZ, XY and XZ have equal lengths of 6 cm. YZ is 10 cm. Determine the measure of ∠X, to the
nearest degree.
4. Triangle ABC is a right triangle. Side AB is 16 cm, and side AC is 18 cm. Using trigonometry, determine the
length of BC, to the nearest tenth of a centimetre.
5. A telephone pole is secured with a guy-wire as shown in the diagram. The guy-wire makes an angle of 75°
with the ground and is secured to the ground 6 m from the bottom of the pole. Determine the length of the
wire, to the nearest tenth of a metre.
6. A ski jumper begins at the top of a ski jump hill that is 90 m high. The hill makes an angle of 30° with the
horizontal. Determine the length of the hill, to the nearest metre.
7
Name: ________________________
ID: A
7. Khalil decides to go parasailing while on vacation. The flyer advertises that the maximum height reached
during the trip will be 68 m. If the parasailing cable is 91 m long, what angle will the cable make with the
horizontal when Khalil reaches the maximum height? Answer to the nearest degree.
Problem
1. A submarine is travelling parallel to the surface of the ocean at a depth of 626 m. It begins a constant ascent
in order to reach the surface after travelling a distance of 4420 m. The ascent takes 35 min.
a) What angle of ascent, or angle of elevation, would the submarine need to make to reach the surface in
4420 m? Answer to the nearest degree.
b) How far did the submarine travel horizontally during its ascent to the surface? Answer to the nearest
metre.
c) Determine the horizontal speed of the submarine during its ascent. Give your answer to the nearest metre
per second.
2. The percent grade of a road is the ratio of the vertical rise of the road to the horizontal distance, expressed as
a percent. A road in the mountains has a vertical rise of 40 m over 1 km of horizontal distance.
a) Determine the percent grade of the road, to the nearest percent.
b) What is the angle of elevation of the road, to the nearest degree?
3. A cell phone tower is supported by two guy wires, attached on opposite sides of the tower. One guy wire is
attached to the top of the base of the tower at point A. The other is attached to the base at point E, at a height
of 70 m above the ground.
a) Demonstrate that ABC and EBD are similar triangles.
b) Determine the height of the base of the tower, to the nearest metre.
c) Determine the length of each guy-wire, to the nearest metre.
8
Name: ________________________
ID: A
4. José is sitting in a tree, so that his eyes are 3.2 m above the ground. When he looks down at an angle of
depression of 43°, he can see his cat sitting in the yard.
a) Draw a diagram of the situation.
b) Determine the horizontal distance, to the nearest tenth of a metre, from the base of the tree to José’s cat.
5. Myriam is participating in a water-skiing competition. She goes over a water-ski ramp that is 4.0 m long.
When she leaves the ramp, she is 1.7 m above the surface of the water.
a) What is the horizontal length of the ramp along the surface of the water, to the nearest tenth of a metre?
b) Determine the angle of elevation of the ramp, to the nearest degree.
6. A 100-m cable is attached to the top of a cell phone tower and is secured to the ground 45 m from the base of
the tower. Determine the angle that the cable makes with the ground, to the nearest degree.
7. The string on Yuri’s kite is 45 m long and makes an angle of 55° with the ground. Yuri’s friend, Abdul, is
standing directly below the kite.
a) How far apart are Abdul and Yuri now, to the nearest tenth of a metre?
b) Abdul runs away from Yuri, so that the angle of elevation between Abdul and the kite is 15°. How far
apart are Abdul and Yuri, to the nearest tenth of a metre?
9
ID: A
Practice Trigonometry Test
Answer Section
MULTIPLE CHOICE
1. ANS:
NAT:
KEY:
2. ANS:
NAT:
KEY:
3. ANS:
NAT:
KEY:
4. ANS:
NAT:
KEY:
5. ANS:
NAT:
KEY:
6. ANS:
NAT:
KEY:
7. ANS:
NAT:
8. ANS:
NAT:
9. ANS:
NAT:
KEY:
10. ANS:
NAT:
KEY:
11. ANS:
NAT:
KEY:
12. ANS:
NAT:
KEY:
13. ANS:
NAT:
KEY:
14. ANS:
NAT:
15. ANS:
NAT:
KEY:
D
PTS: 1
DIF: B
M4
TOP: The Tangent Ratio
tangent ratio | determine an angle measure
B
PTS: 1
DIF: B
M4
TOP: The Tangent Ratio
tangent ratio | calculate a tangent ratio | right triangle
C
PTS: 1
DIF: B
M4
TOP: The Tangent Ratio
tangent ratio | determine an angle measure | right triangle
A
PTS: 1
DIF: C
M4
TOP: Solving Right Triangles
tangent ratio | determine an angle measure
B
PTS: 1
DIF: C
M4
TOP: The Tangent Ratio
tangent ratio | determine a distance using trigonometry
C
PTS: 1
DIF: B
M4
TOP: Solving Right Triangles
tangent ratio | determine a distance using trigonometry
B
PTS: 1
DIF: A
M4
TOP: The Sine and Cosine Ratios
A
PTS: 1
DIF: A
M4
TOP: The Sine and Cosine Ratios
C
PTS: 1
DIF: B
M4
TOP: The Sine and Cosine Ratios
cosine ratio | determine an angle measure
A
PTS: 1
DIF: B
M4
TOP: The Sine and Cosine Ratios
cosine ratio | determine an angle measure
D
PTS: 1
DIF: B
M4
TOP: The Sine and Cosine Ratios
sine ratio | determine an angle measure
A
PTS: 1
DIF: B
M4
TOP: The Sine and Cosine Ratios
sine ratio | calculate a sine ratio | right triangle
A
PTS: 1
DIF: A
M4
TOP: The Sine and Cosine Ratios
cosine ratio | write a cosine ratio | right triangle
C
PTS: 1
DIF: A
M4
TOP: The Sine and Cosine Ratios
B
PTS: 1
DIF: A
M4
TOP: The Sine and Cosine Ratios
sine ratio | determine an angle measure
1
OBJ: Section 3.1
OBJ: Section 3.1
OBJ: Section 3.1
OBJ: Section 3.3
OBJ: Section 3.1
OBJ: Section 3.3
OBJ:
KEY:
OBJ:
KEY:
OBJ:
Section 3.2
sine ratio | calculate a sine ratio
Section 3.2
sine ratio | calculate a sine ratio
Section 3.2
OBJ: Section 3.2
OBJ: Section 3.2
OBJ: Section 3.2
OBJ: Section 3.2
OBJ: Section 3.2
KEY: cosine ratio | calculate a cosine ratio
OBJ: Section 3.2
ID: A
16. ANS:
NAT:
KEY:
17. ANS:
NAT:
KEY:
18. ANS:
NAT:
KEY:
19. ANS:
NAT:
KEY:
20. ANS:
NAT:
A
PTS: 1
DIF: C
OBJ: Section 3.3
M4
TOP: Solving Right Triangles
cosine ratio | determine a distance using trigonometry | solve a right triangle
B
PTS: 1
DIF: A
OBJ: Section 3.2
M4
TOP: The Sine and Cosine Ratios
sine ratio | determine an angle measure
D
PTS: 1
DIF: C
OBJ: Section 3.2
M4
TOP: The Sine and Cosine Ratios
sine ratio | determine a distance using trigonometry | right triangle
C
PTS: 1
DIF: B
OBJ: Section 3.3
M4
TOP: Solving Right Triangles
sine ratio | determine a distance using trigonometry | solve a right triangle
A
PTS: 1
DIF: B
OBJ: Section 3.3
M4
TOP: Solving Right Triangles
KEY: Pythagorean theorem | hypotenuse
COMPLETION
1. ANS: opposite
PTS: 1
DIF: A
TOP: The Tangent Ratio
2. ANS:
Example:
length of side opposite ∠A
tanA =
length of side adjacent to ∠A
OBJ: Section 3.1 NAT: M4
KEY: right triangle | hypotenuse
PTS:
TOP:
KEY:
3. ANS:
1
DIF: B
OBJ: Section 3.1 NAT: M4
The Tangent Ratio
tangent ratio | write a tangent ratio | primary trigonometric ratios
opposite
PTS:
TOP:
KEY:
4. ANS:
1
DIF: A
OBJ: Section 3.2 NAT: M4
The Sine and Cosine Ratios
sine ratio | define the sine ratio | opposite side | primary trigonometric ratios
up
PTS: 1
DIF: A
TOP: Solving Right Triangles
5. ANS: elevation
OBJ: Section 3.3 NAT: M4
KEY: angle of elevation
PTS: 1
DIF: B
TOP: Solving Right Triangles
OBJ: Section 3.3 NAT: M4
KEY: angle of elevation
MATCHING
1. ANS: E
NAT: M4
PTS: 1
DIF:
TOP: The Tangent Ratio
2
A
OBJ: Section 3.1
KEY: label a right triangle | opposite side
ID: A
2. ANS: D
NAT: M4
3. ANS:
NAT:
4. ANS:
NAT:
5. ANS:
NAT:
KEY:
6. ANS:
NAT:
7. ANS:
NAT:
KEY:
PTS: 1
DIF:
TOP: The Tangent Ratio
A
B
PTS: 1
DIF: B
M4
TOP: Solving Right Triangles
E
PTS: 1
DIF: A
M4
TOP: Solving Right Triangles
G
PTS: 1
DIF: B
M4
TOP: The Tangent Ratio
tangent ratio | primary trigonometric ratios
D
PTS: 1
DIF: A
M4
TOP: The Tangent Ratio
C
PTS: 1
DIF: B
M4
TOP: The Sine and Cosine Ratios
cosine ratio | primary trigonometric ratios
OBJ: Section 3.1
KEY: Pythagorean theorem
OBJ:
KEY:
OBJ:
KEY:
OBJ:
Section 3.3
angle of depression
Section 3.3
Pythagorean theorem
Section 3.1
OBJ: Section 3.1
KEY: hypotenuse
OBJ: Section 3.2
SHORT ANSWER
1. ANS:
Let x be the angle the stairs make with the horizontal, in degrees.
3
tan x =
5
ÊÁ 3 ˆ˜
x = tan −1 ÁÁÁÁ ˜˜˜˜
Ë 5¯
x = 30.9638. . .
The stairs make an angle of approximately 31° with the stairs.
PTS: 1
DIF:
TOP: The Tangent Ratio
2. ANS:
AD
sin B =
AB
sin 24° =
AB =
B
OBJ: Section 3.1 NAT: M4
KEY: tangent ratio | determine an angle measure
3.5
AB
3.5
sin 24°
AB = 8.6051. . .
The trusses are each approximately 8.6 m long.
PTS: 1
DIF: B
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: sine ratio | determine a distance using trigonometry | isosceles triangle
3
ID: A
3. ANS:
cos Z =
side adjacent to ∠Z
hypotenuse
cos Z =
ZW
XZ
cos Z =
5
6
cos Z = 0.8333
Z = cos −1(0.8333)
Z = 33.5573. . .
∠Z measures approximately 34°. ∠Y is equal to ∠Z.
180 = ∠X + ∠Y + ∠Z
180 = ∠X + 34 + 34
180 = ∠X + 68
112 = ∠X
∠X measures approximately 112°.
PTS: 1
DIF: C
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: cosine ratio | determine an angle measure | isosceles triangle
4. ANS:
side opposite ∠C
sinC =
hypotenuse
sinC =
AB
AC
sinC =
16
18
C = sin −1(0.8888)
C = 62.7340. . .
C = 63°
BC
cos 63° =
18
18(cos 63°) = BC
8.1718. . . = BC
BC is approximately 8.2 cm.
PTS: 1
DIF: C
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: sine ratio | cosine ratio | determine a distance using trigonometry
4
ID: A
5. ANS:
Let x represent the length of the wire, in metres.
distance of wire from base
cos 75° =
length of wire
cos 75° =
x=
6
x
6
cos 75°
x = 23.1822. . .
The length of the wire is approximately 23.2 m.
PTS: 1
DIF: B
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: cosine ratio | determine a distance using trigonometry | right triangle
6. ANS:
Let h represent the length of the hill, in metres.
90
sin30° =
h
h=
90
sin30°
h = 180
The length of the ski jump hill is 180 m.
PTS: 1
DIF: B
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: sine ratio | determine a distance using trigonometry
7. ANS:
Let x represent Khalil’s angle above the water, in degrees.
height of Khalil above the water
sinx =
length of cable
sinx =
68
91
sinx = 0.7473
x = sin −1 (0.7473)
x = 48.3570. . .
When Khalil is at the maximum height, the cable makes an angle of approximately 48°.
PTS: 1
DIF: B
TOP: The Sine and Cosine Ratios
OBJ: Section 3.2 NAT: M4
KEY: sine ratio | determine an angle measure
5
ID: A
PROBLEM
1. ANS:
a) sin x =
sin x =
depth of submarine below surface
distance travelled during ascent
626
4420
sin x = 0.1416
x = sin−1 (0.1416)
x = 8.1421. . .
The submarine’s angle of ascent is approximately 8°.
b) Let d represent the distance travelled by the submarine, in metres..
d
cos 8° =
4420
4420(cos 8°) = d
4376.9849. . . = d
The submarine traveled approximately 4377 m horizontally.
1min = 60 s
c)
35 min ×
60 s
= 2100 s
1min
4420 m
= 2.1
2100 s
During the ascent, the submarine travels at a horizontal speed of 2.1 m/s.
PTS: 1
DIF: D
OBJ: Section 3.3 NAT: M4
TOP: Solving Right Triangles
KEY: sine ratio | cosine ratio | solve a right triangle | determine a distance using an angle of elevation | speed
6
ID: A
2. ANS:
a)
1 km = 1000 m
ÊÁ
ˆ˜
vertical rise
˜˜ × 100
percent grade = ÁÁÁÁ
˜˜
horizontal
distance
Ë
¯
ÊÁ 40 ˆ˜
˜˜ × 100
percent grade = ÁÁÁÁ
˜˜
1000
Ë
¯
percent grade = 0.04 × 100
percent grade = 4
The percent grade of the road is 4%.
b) Let θ represent the angle of elevation, in degrees.
vertical rise
tan θ =
horizontal distance
tan θ =
40
1000
θ = tan−1 (0.04)
θ = 2.2906. . .
The angle of elevation of the road is approximately 2°.
PTS: 1
DIF: D
TOP: Solving Right Triangles
OBJ: Section 3.3 NAT: M4
KEY: tangent ratio | angle of elevation
7
ID: A
3. ANS:
a) ∠C = ∠D = 70°; ∠ABC = ∠EBD = 90°; ∠A = ∠E = 20°
AC corresponds to ED, BC corresponds to BD, and AB corresponds to EB, so the triangles are similar.
height of tower
tanC =
b)
distance of wire to base of tower
tan 70° =
AB
42
42(tan70°) = AB
115.3941... = AB
The base of the tower is approximately 115 m tall.
c)
cos C =
cos 70° =
AC =
distance from wire to base of tower
length of wire
42
AC
42
cos 70°
AC = 122.7998. . .
The guy wire attached to the top of the base of the tower is approximately 123 m long.
height of tower
sinD =
length of wire
sin 70° =
ED =
70
ED
70
sin 70°
ED = 74.4924. . .
The second guy wire is approximately 74 m long.
PTS: 1
DIF: C
OBJ: Section 3.3 NAT: M4
TOP: Solving Right Triangles
KEY: tangent ratio | sine ratio | cosine ratio | solve a right triangle | similar triangles
8
ID: A
4. ANS:
a) Example:
b) Let d represent the distance from the cat to the base of the tree, in metres.
distance from base of tree to cat
tan43° =
height of Jose above the ground
tan43° =
d
3.2
3.2(tan43°) = d
2.9840. . . = d
The distance from the base of the tree to the cat is about 3.0 m.
PTS: 1
DIF: A
OBJ: Section 3.3 NAT: M4
TOP: Solving Right Triangles
KEY: tangent ratio | determine a distance using an angle of depression
9
ID: A
5. ANS:
a) Let x represent the length of the ramp along the surface of the water, in metres.
Draw a diagram of the situation.
x 2 + 1.7 2 = 4.0 2
x 2 = 16 − 2.89
x 2 = 13.11
x ≈ 3.62
The horizontal length of the ramp along the surface of the water is about 3.6 m.
b) sinθ =
1.7
4.0
sinθ = 0.425
θ = 25.1507. . .
θ ≈ 25
The angle of elevation of the ramp is about 25°.
PTS: 1
DIF: B
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: sine ratio | angle of elevation | Pythagorean theorem
6. ANS:
Let x represent the angle, in degrees, that the cable makes with the ground.
cos x =
distance from cable to base of tower
length of cable
cos x =
45
100
x = cos −1 (0.45)
x = 63.2563. . .
The angle that the cable makes with the ground is approximately 63°.
PTS: 1
DIF: A
TOP: Solving Right Triangles
OBJ: Section 3.3 NAT: M4
KEY: cosine ratio | determine an angle measure
10
ID: A
7. ANS:
a) Let x represent the distance between Abdul and Yuri, in metres.
distance between Abdul and Yuri
cos 55° =
length of kite string
cos 55° =
x
45
45(cos 55°) = x
25.8109. . . = x
Abdul and Yuri are approximately 25.8 m apart.
b) Let h represent the height of the kite above the ground, in metres.
height of the kite above the ground
sin55° =
length of kite string
sin55° =
h
45
45(sin55°) = h
36.8618. . . = h
Let x represent the horizontal distance between Abdul and the kite, in metres.
height of the kite above the ground
tan15° =
horizontal distance between Abdul and the kite
tan15° =
x=
36.9
x
36.9
tan15°
x = 137.713. . .
Distance between Abdul and Yuri = 25.8 + 137.7
= 163.5
The distance between Abdul and Yuri is now approximately 163.5 m.
PTS: 1
DIF: D
OBJ: Section 3.2 NAT: M4
TOP: The Sine and Cosine Ratios
KEY: cosine ratio | determine a distance using trigonometry | angle of elevation | solve a right triangle
11