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PRECALCULUS B
TEST #1 – ACUTE ANGLES AND RIGHT TRIANGLES, PRACTICE
SECTION 2.1 – Trigonometric Functions of Acute Angles
1)
B
Find the value of sin A, cos A, tan A, csc A, sec A, and cot A using the
triangle pictured to the right. HINT – use the Pythagorean Theorem to
find the hypotenuse.
55
A
2)
C
48
Express each of the following functions in terms of its confuction:
A)
sin 52°
B)
sec 58°
C)
tan 23° 16'
3)
Solve the following equation, given that all angles are acute angles: cot ( 5θ + 2 ) = tan ( 2θ + 4 ) .
4)
Give the exact (NO decimals) value for each expression:
A)
5)
cos 45°
B)
tan 60°
C)
csc 30°
Find the exact value (NO decimals) of each
variable in the triangle pictured to the right.
p
s
6
60°
q
r
45°
SECTION 2.2 – Trigonometric Functions of Non-Acute Angles
6)
Give the measure of the reference angle for each of the following:
A)
7)
105°
B)
–120°
C)
310°
Give the exact (NO decimals) value for each expression:
A)
sin 120°
B)
cos 390°
C)
tan 225°
D)
csc –420°
E)
sec 405°
F)
cot –480°
2
, find all values of θ in the interval [ 0°, 360° ) that make the function true.
2
8)
If sin θ =
9)
If secθ = −2, find all values of θ in the interval [ 0°, 360° ) that make the function true.
SECTION 2.3 – Finding Trigonometric Function Values Using a Calculator
10)
Use a calculator to evaluate sin 27° 32'. Round the answer to three decimal places.
11)
Use a calculator to evaluate cot 3.45°. Round the answer to three decimal places.
12)
Find a value of θ in the interval [0°, 90°) such that sec θ = 1.1606249. Leave the answer in
decimal degrees rounded to three decimal places.
13)
Find a value of θ in the interval [0°, 90°) such that tan θ = 1.0548399. Leave the answer in
decimal degrees rounded to three decimal places.
14)
Which has a greater grade resistance: a 2200-lb car on a 2° uphill grade or a 2000-lb car on
a 2.2° uphill grade? Use the formula F = W sin θ.
SECTION 2.4 – Solving Right Triangles
15)
Solve the following right triangle, assuming that C = 90°: A = 31° 45' and a = 35.9.
HINT – Drawing a picture would be beneficial.
16)
Solve the following right triangle, assuming that C = 90°: B = 46.3° and c = 29.7.
HINT – Drawing a picture would be beneficial.
17)
From a point 250 feet from the base of a tower, then angle of elevation to the top of the tower is
18.3°. How tall is the tower to the nearest tenth of a foot?
18)
A 20-foot ladder is leaned against a building. The top of this ladder reaches a point 19.5 feet
above the ground. What is the measure of the angle formed by the ladder and the ground,
measured to the nearest tenth of a degree?
19)
A man sights the top of a 300 foot tall building using an angle of elevation of 6°. A woman
standing farther away sights the top of the same building using an angle of elevation of 4°.
What is the distance between the man and the woman, measured to the nearest foot?
SECTION 2.5 – Further Applications of Right Triangles
20)
A ship leaves a pier on a bearing of S 62° E and travels for 75 km. It then turns and continues
on a bearing of N 28° E for 53 km. How far is the ship from the pier?
21)
A plane is currently 120 miles west and 85 miles north of an airport. The pilot wants to fly
directly to the airport. What bearing should the pilot take, and how far will the plane have to
fly?
22)
A ship leaves port and travels 30 nautical miles due north. The ship then changes course to due
east and travels for 10 nautical miles. Find the ship’s bearing from its point of departure.
23)
Lighthouse A is located directly west of lighthouse B. Lighthouse A sights a ship 125 miles
away at a bearing of N 15° E. Lighthouse B sights the same ship at a bearing of N 75° W. How
far apart are the two lighthouses?
24)
A man sights the top of a building and determines that the angle of elevation to the top of the
building is 41.2°. He then walks 168 meters closer to the building and determines that the angle
of elevation is now 57.5°. Determine the height of the building, measured to the nearest foot.
25)
The bearing from point A to point C is 25°. The bearing from point C to point B is 115°. The
bearing from point A to point B is 61°. If the distance between point A and point C is 38.5
miles, what is the distance from point C to point B?
******************************ANSWERS*******************************
1)
2
2
5329 = AB → 5329 =
2)
2
Calculate AB by using the Pythagorean Theorem → 482 + 552 = AB → 2304 + 3025 = AB →
2
AB → 73 = AB
sin A =
opp BC 55
=
=
hyp AB 73
cos A =
adj AC 48
=
=
hyp AB 73
tan A =
opp BC 55
=
=
adj AC 48
csc A =
hyp AB 73
=
=
opp BC 55
sec A =
hyp AB 73
=
=
adj AC 48
cot A =
adj AC 48
=
=
opp BC 55
A) sin 52° = cos ( 90 − 52 ) ° → cos 38°
B) sec 58° = csc ( 90 − 58 ) ° → csc 32°
C) tan 23° 16 ' = cot ( 90° − 23° 16 ') → cot 66° 44 '
3)
cot ( 5θ + 2 ) = tan ( 2θ + 4 ) → cot ( 5θ + 2 ) = cot ⎡⎣90 − ( 2θ + 4 ) ⎤⎦ → 5θ + 2 = 90 − 2θ − 4 →
5θ + 2 = 86 − 2θ → 7θ = 84 → θ = 12
4)
A) cos 45° =
1
2
→ Multiply top & bottom by 2 →
1⋅ 2
2⋅ 2
=
2
2
→ cos 45° =
2
2
3
→ tan 60° = 3
1
1
1
1
C) csc 30° =
→ sin 30° = → csc 30° =
→ Multiply top & bottom by 2 →
sin 30°
2
1/2
1⋅ 2
2
= = 2 → csc 30° = 2
1/2 ⋅ 2 1
B) tan 60° =
2
60°
2
1
1
45°
1
30°
3
5)
q=
6
3
→ Multiply top & bottom by 3 →
6⋅ 3
3⋅ 3
6 3
=2 3→q=2 3
3
=
p = q⋅2 → p = 2 3 ⋅2 → p = 4 3
r = 6 → Legs in a 45°-45°-90° triangles are equal
s = r ⋅ 2 → s = 6⋅ 2 → s = 6 2
6)
A) 105° → 180° − 105° = 75° → Reference angle = 75°
B) − 120° → −120° + 180° = 60° → Reference angle = 60°
C) 310° → 360° − 310° = 50° → Reference angle = 50°
7)
SEE 45°-45°-90° TRIANGLE AND 30°-60°-90° TRIANGLE FROM PROBLEM #4
3
A) sin120° → Reference angle = 60° → sin 60° =
→ 120° is in Quadrant II →
2
3
sine is positive in Quadrant II → sin120° =
2
3
B) cos 390° → Reference angle = 30° → cos 30° =
→ 390° is in Quadrant I →
2
3
cosine is positive in Quadrant I → cos 390° =
2
C) tan 225° → Reference angle = 45° → tan 45° = 1 → 225° is in Quadrant III →
tangent is positive in Quadrant III → tan 225° = 1
D) csc ( −420° ) → Reference angle = 60° → csc 60° =
Multiply top & bottom by 2 →
(
1⋅ 2
)
3/2 ⋅ 2
=
1
3
1
→ sin 60° =
→ csc 60° =
→
sin 60°
2
3/2
2
2⋅ 3
2 3
→ Multiply top & bottom by 3 →
=
→
3
3
3⋅ 3
−420° is in Quadrant IV → cosecant is negative in Quadrant IV → csc ( −420° ) = −
E) sec 405° → Reference angle = 45° → sec 45° =
1⋅ 2
Multiply top & bottom by 2 →
Multiply top & bottom by 2 →
(
2⋅ 2
1⋅ 2
)
2 /2 ⋅ 2
=
=
2 3
3
1
1
→ cos 45° =
→
cos 45°
2
2
2
→ cos 45° =
→ sec 45° =
2
2
1
2 /2
→
2
2⋅ 2
2 2
→ Multiply top & bottom by 2 →
=
= 2→
2
2
2⋅ 2
405° is in Quadrant I → secant is positive in Quadrant I → sec 405° = 2
F) cot ( −480° ) → Reference angle = 60° → cot 60° =
Multiply top & bottom by 3 →
1
1
→ tan 60° = 3 → cot 60° =
→
tan 60°
3
1⋅ 3
3
3
=
→ cot 60° =
→ −480° is in Quadrant III →
3
3
3⋅ 3
cotangent is positive in Quadrant III → cot ( −480° ) =
8)
3
3
2
→ Other angles with reference angles of 45° are 135°, 225°, and 315° →
2
sine is positive in Quadrants I and II → ANSWERS: 45° and 135°
sin 45° =
9)
1
1
→ cos 60° = → Other angles with a reference angle of 60° are
2
2
120°, 240°, and 300° → secant is negative in Quadrants II and III → ANSWERS: 120° and 240°
If sec θ = −2, then cos θ = −
10)
27° 32 ' = 27 + 32/60 = 27.53333333° → sin 27.53333333° = 0.462264577 → sin 27 ' 32 ' ≈ 0.462
11)
cot 3.45° =
12)
sec θ =
13)
tan θ = 1.0548399 → θ = tan −1 (1.0548399 ) = 46.52875615° → θ ≈ 46.529°
1
1
=
= 16.58739618 → cot 3.45° ≈ 16.587
tan 3.45° 0.0602867375
1
1
→ sec θ = 1.1606249 →
= 0.8616048131 → cos −1 ( 0.8616048131) =
cos θ
1.1606249
30.50274845° → θ ≈ 30.503°
14)
F = W sin θ → F = 2200 ⋅ sin 2° → F = 2200 ⋅ 0.03488994967 → F = 76.77889275 →
F ≈ 76.779 pounds
F = W sin θ → F = 2000 ⋅ sin 2.2° → F = 2000 ⋅ 0.0383878091 → F = 76.77561818 →
F ≈ 76.776 pounds
15)
B = 180° − 90° − 31° 45' = 58° 15'
B
35.9
35.9
31° 45' = 31.75° → tan 31.75° =
→ 0.6188187767 =
→
c
b
b
a = 35.9
31° 45'
35.9
b=
= 58.01375354 → b ≈ 58.0
b
A
C
0.6188187767
35.9
35.9
35.9
sin 31.75° =
→ 0.5262139237 =
→c=
= 68.22320426 → c ≈ 68.2
c
c
0.5262139237
16)
A = 180° − 90° − 46.3° = 43.7°
b
b
sin 46.3° =
→ 0.7229671459 =
→ 0.7229671459 ⋅ 29.7 = b →
29.7
29.7
21.47212423 = b → 21.5 ≈ b
a
a
sin 43.7° =
→ 0.6908824111 =
→ 0.6908824111 ⋅ 29.7 = a →
29.7
29.7
20.51920761 = a → 20.5 ≈ a
17)
h
h
→ 0.3307183801 =
→ 0.3307183801 ⋅ 250 = h →
250
250
82.67959502 = h → h ≈ 82.7 feet
NOTE – diagram not drawn to scale
B
c = 29.7
A
b
46.3°
a
C
tan18.3° =
h
18.3°
250 feet
18)
19.5
→ sin x° = 0.975 → x = sin −1 ( 0.975 ) →
20
x = 77.16143186° → x ≈ 77.2°
NOTE – diagram not drawn to scale
sin x° =
20 feet
19.5 feet
x°
19)
20)
300
300
300
→ 0.0699268119 =
→a=
→
a
a
0.0699268119
a = 4290.199877 → a ≈ 4290.20 feet
300
300
300
tan 6° =
→ 0.1051042353 =
→b=
→
b
b
0.1051042353
b = 2854.309336 → b ≈ 2854.31 feet
Distance between woman and man = 4290.20 feet − 2854.31 feet =
1435.89 feet ≈ 1436 feet
NOTE – diagrams not drawn to scale
tan 4° =
300 feet
4°
a
300 feet
6°
b
∠ ABC = 62° + 28° = 90° → Use the Pythagorean Theorem to find x →
752 + 532 = x 2 → 5625 + 2809 = x 2 → 8434 = x 2 →
8434 = x 2 → 91.83681179 = x →
x ≈ 91.8 km
A
x
C
62°
28°
75 km
53 km
62°
B
21)
Use the Pythagorean Theorem to find d → 852 + 1202 = d 2 →
7225 + 14400 = d 2 → 21625 = d 2 → 21625 = d 2 →
147.0544117 = d → d ≈ 147.1 miles
Xθ
120
→ tan θ = 1.411764706 → θ = tan −1 (1.411764706 ) →
tan θ =
85
θ = 54.68878656° → θ ≈ 54.7°
85 miles
Bearing = 180° − 54.7° = 125.3°
Y
d
120 miles
Z
22)
10
→ tan θ = 0.3333333333 →
30
θ = tan −1 ( 0.3333333333) → θ = 18.43494882° → θ ≈ 18.4°
tan θ =
T
10 miles
U
Bearing = N 18.4° E
30 miles
θ
S
23)
∠ ACB = 15° + 75° = 90°
125
125
125
sin15° =
→ 0.2588190451 =
→c=
→ c = 482.9629131 → c ≈ 483 miles
c
c
0.2588190451
C
15°
75°
125 miles
15°
15°
A
24)
c
75°
B
h
h
, tan 41.2° =
→ x ⋅ tan 57.5° = h, ( x + 168 ) ⋅ tan 41.2° = h →
x
x + 168
x ⋅ tan 57.5° = x ⋅ tan 41.2° + 168 ⋅ tan 41.2 → x ⋅ tan 57.5° − x ⋅ tan 41.2° = 168 ⋅ tan 41.2 →
168 ⋅ tan 41.2
x ⋅ ( tan 57.5° − tan 41.2° ) = 168 ⋅ tan 41.2 → x =
= 211.8437316
tan 57.5° − tan 41.2°
h = x ⋅ tan 57.5° → h = 211.8437316 ⋅ tan 57.5° → h = 332.52805 → h ≈ 333 m
NOTE – diagram not drawn to scale
tan 57.5° =
41.2°
168 m
57.5°
x
h
25)
∠ ACB = 25° + 65° = 90°, ∠ CAB = 61° − 25° = 36°
x
x
tan 36° =
→ 0.726542528 =
→ 0.726542528 ⋅ 38.5 = x → 27.97188733 = x → x ≈ 28.0 miles
38.5
38.5
C
115°
65°
25°
x
38.5 miles
B
25°
61°
36°
A