Test bank for trigonometry by ratti

Find the complement and the supplement of the given angle.. Round to the nearest tenth.65 A flagpole casts a shadow 28 feet long.. At the same time, the shadow cast by a 39-inch-tall shr

MULTIPLE CHOICE Choose the one alternative that best completes the statement or answers the question.

Find the complement and the supplement of the given angle If the angle has no complement or supplement state so.

C) No complementSupplement is -23°

A) quadrant IV

B) quadrant I

C) quadrant II

B) quadrant IV

C) quadrant III

D) quadrant II

20) 585°

x

y

x y

21)

A) quadrant II

22)

23) 840°

x

y

x y

A) quadrant IV

B) quadrant I

23)

A) quadrant II

B) quadrant I

C) quadrant III

B) quadrant II

C) quadrant IV

D) quadrant I

x

y

x y

27)

Solve the problem.

36) In the adjoining figure, AOB is a line Find the value of x

Refer to the right triangle ABC with C = 90°.

45) Assuming that a = 24 and b = 7, find c

Find the remaining sides of a 45-45-90 triangle from the given information.

49) The shorter sides each have length 6

49)

50) The shorter sides each have length 2

3.A) 2

55) The length of the hypotenuse is 4.

LN =

BC

MN =

ACLM

B)∠A ≅ ∠L

∠B ≅ ∠M

∠C ≅ ∠NAB

LN =

BC

MN =

ACLMC) ∠A ≅ ∠L

∠B ≅ ∠N

∠C ≅ ∠MAB

LM = BC

MN = ACLN

D) ∠A ≅ ∠L

∠B ≅ ∠M

∠C ≅ ∠NAB

LM = BC

MN = ACLN

PQ = BC

QR = ACPR

B) ∠A ≅ ∠P

∠B ≅ ∠Q

∠C ≅ ∠RAB

PQ = BC

QR = ACPRC)∠A ≅ ∠P

∠B ≅ ∠R

∠C ≅ ∠QAB

PR =

BC

RQ =

ACPQ

D)∠A ≅ ∠P

∠B ≅ ∠Q

∠C ≅ ∠RAB

PR =

BC

RQ =

ACPQ

57)

22

Full file at https://TestbankDirect.eu/Test-Bank-for-Trigonometry-by-Ratti

A)∠L ≅ ∠P

∠M ≅ ∠Q

∠N ≅ ∠RLM

PR =

MN

RQ =

LNPQ

B)∠L ≅ ∠P

∠M ≅ ∠R

∠N ≅ ∠QLM

PR =

MN

RQ =

LNPQC) ∠L ≅ ∠P

∠M ≅ ∠R

∠N ≅ ∠QLM

PQ = MN

QR = LNPR

D) ∠L ≅ ∠P

∠M ≅ ∠Q

∠N ≅ ∠RLM

PQ = MN

QR = LNPR

SR =

PO

RO =

QOSO

B)∠P ≅ ∠S

∠Q ≅ ∠R

∠O ≅ ∠OPQ

SR =

PO

SO =

QORO

59)

A)∠L ≅ ∠L

∠M ≅ ∠P

∠N ≅ ∠QLM

MP =

MN

PQ =

LNNQ

B)∠L ≅ ∠L

∠M ≅ ∠Q

∠N ≅ ∠PLM

LQ =

MN

PQ =

LNLPC) ∠L ≅ ∠L

∠M ≅ ∠Q

∠N ≅ ∠PLM

MQ = MN

PQ = LNNP

D) ∠L ≅ ∠L

∠M ≅ ∠P

∠N ≅ ∠QLM

LP = MN

PQ = LNLQ

60)

Find the length of the unknown segment labeled with the variable x.

61) AB and CD are parallel

Solve the problem Round to the nearest tenth.

65) A flagpole casts a shadow 28 feet long At the same time, the shadow cast by a 39-inch-tall shrub

is 78 inches long Find the height of the flagpole

65)

66) A building casts a shadow 15 meters long At the same time, the shadow cast by a 34-centimeter

tall pole is 59 centimeters long How tall is the building?

66)

The terminal side of θ in standard position contains the given point Find the values of the six trigonometric functions of θ.

68) (5, 12)

A) sin θ = 12

13cos θ = 5

13tan θ = 5

12csc θ = 13

12sec θ = 13

5cot θ = 12

5

B) sin θ = 5

13cos θ = 12

13tan θ = 5

12csc θ = 13

5sec θ = 13

12cot θ = 12

5

C) sin θ = 12

13cos θ = 5

13tan θ = 12

5csc θ = 1312sec θ = 13

5cot θ = 512

D) sin θ = 5

13cos θ = 12

13tan θ = 12

5csc θ = 13

5sec θ = 13

12cot θ = 5

25tan θ = - 24

7csc θ = 25

7sec θ = - 25

24cot θ = - 7

24

B) sin θ = - 24

25cos θ = 7

25tan θ = - 24

7csc θ = - 25

24sec θ = 25

7cot θ = - 7

24

C) sin θ = 24

25cos θ = - 7

25tan θ = - 24

7csc θ = 2524sec θ = - 25

7cot θ = - 7

24

D) sin θ = 24

25cos θ = - 7

25tan θ = - 7

24csc θ = 25

24sec θ = - 25

7cot θ = - 24

13tan θ = - 5

12csc θ = - 13

12sec θ = 13

5cot θ = - 12

5

B) sin θ = - 12

13cos θ = 5

13tan θ = - 12

5csc θ = - 13

12sec θ = 13

5cot θ = - 5

12

C) sin θ = - 5

13cos θ = 12

13tan θ = - 12

5csc θ = - 13

5sec θ = 1312cot θ = - 5

12

D) sin θ = 12

13cos θ = - 5

13tan θ = - 12

5csc θ = 13

12sec θ = - 13

5cot θ = - 5

71) (-4, -3)

A) sin θ = - 3

5cos θ = - 4

5tan θ = 3

4csc θ = - 5

3sec θ = - 5

4cot θ = 4

3

B) sin θ = 3

5cos θ = - 4

5tan θ = - 4

3csc θ = 5

3sec θ = - 5

4cot θ = - 3

4

C) sin θ = - 3

5cos θ = - 4

5tan θ = - 3

4csc θ = - 5

3sec θ = - 5

4cot θ = - 4

3

D) sin θ = - 3

5cos θ = 4

5tan θ = - 4

3csc θ = - 5

3sec θ = 5

4cot θ = - 3

6tan θ = 5 11

11csc θ = 6

5sec θ = 6 11

11cot θ = 11

5

B) sin θ = 5

6cos θ = 11

6tan θ = 11

5csc θ = 6

5sec θ = 6 11

11cot θ = 5 11

11

C) sin θ = 11

6cos θ = 5

6tan θ = 11

5csc θ = 6 11

11sec θ = 65cot θ = 5 11

11

D) sin θ = 11

6cos θ = 5

6tan θ = 5 11

11csc θ = 6 11

11sec θ = 6

5cot θ = 11

34tan θ = - 5

3csc θ = - 34

B) sin θ = 5 34

34cos θ = - 3 34

34tan θ = - 5

3csc θ = 34

C) sin θ = 3 34

34cos θ = - 5 34

34tan θ = - 5

3csc θ = 34

D) sin θ = - 5 34

34cos θ = 3 34

34tan θ = - 5

3csc θ = - 34

73)

Use the given information to find the quadrant in which θ lies.

84) tan θ > 0 and sin θ < 0

84)

85) cos θ < 0 and csc θ < 0

85)

86) sin θ > 0 and cos θ < 0

86)

28

Full file at https://TestbankDirect.eu/Test-Bank-for-Trigonometry-by-Ratti

87) cot θ < 0 and cos θ > 0

87)

88) csc θ > 0 and sec θ > 0

88)

89) sec θ < 0 and tan θ < 0

89)

90) tan θ < 0 and sin θ < 0

90)

91) cos θ > 0 and csc θ < 0

91)

92) cot θ > 0 and sin θ < 0

92)

93) sin θ > 0 and cos θ > 0

93)

Find the exact values of the remaining trigonometric functions of θ from the given information.

94) sin θ = - 7

25, θ in quadrant IIIA) sin θ = - 7

25cos θ = - 24

25tan θ = - 24

7csc θ = 25

7sec θ = - 25

24cot θ = - 7

24

B) sin θ = - 7

25cos θ = 24

25tan θ = - 24

7csc θ = - 25

7sec θ = 25

24cot θ = - 7

24

C) sin θ = - 7

25cos θ = - 24

25tan θ = 7

24csc θ = - 25

7sec θ = - 25

24cot θ = 24

7

D) sin θ = - 7

25cos θ = - 24

25tan θ = - 7

24csc θ = - 25

7sec θ = - 25

24cot θ = - 24

7

94)

95) tan θ = - 12

5 , θ in quadrant IIA) sin θ = 5

13cos θ = - 12

13tan θ = - 12

5csc θ = 13

5sec θ = - 13

12cot θ = - 5

12

B) sin θ = 12

13cos θ = - 5

13tan θ = - 12

5csc θ = 13

12sec θ = - 13

5cot θ = - 5

12

C) sin θ = 12

13cos θ = - 5

13tan θ = - 12

5csc θ = 1312sec θ = - 13

5cot θ = - 12

5

D) sin θ = - 12

13cos θ = 5

13tan θ = - 12

5csc θ = - 13

12sec θ = 13

5cot θ = - 5

12

95)

96) cot θ = - 4

3, θ in quadrant IVA) sin θ = - 4

5cos θ = 3

5tan θ = - 3

4csc θ = - 5

4sec θ = 5

3cot θ = - 4

3

B) sin θ = - 3

5cos θ = 4

5tan θ = - 3

4csc θ = - 5

3sec θ = 5

4cot θ = - 4

3

C) sin θ = - 3

5cos θ = 4

5tan θ = - 4

3csc θ = - 5

3sec θ = 54cot θ = - 4

3

D) sin θ = 3

5cos θ = - 4

5tan θ = - 3

4csc θ = 5

3sec θ = - 5

4cot θ = - 4

3

96)

97) csc θ = 8

55, θ in quadrant IIA) sin θ = 3

8cos θ = - 55

8tan θ = - 55

3csc θ = 8 55

55sec θ = - 8 55

55cot θ = - 3 55

55

B) sin θ = 55

8cos θ = - 3

8tan θ = - 3 55

55csc θ = 8 55

55sec θ = - 8

3cot θ = - 55

3

C) sin θ = - 55

8cos θ = 3

8tan θ = - 55

3csc θ = 8 55

55sec θ = 83cot θ = - 3 55

55

D) sin θ = 55

8cos θ = - 3

8tan θ = - 55

3csc θ = 8 55

55sec θ = - 8

3cot θ = - 3 55

98) cos θ = 5

13, sin θ < 0A) sin θ = - 12

13cos θ = 5

13tan θ = - 5

12csc θ = - 13

12sec θ = 13

5cot θ = - 12

5

B) sin θ = - 5

13cos θ = 5

13tan θ = - 12

5csc θ = - 13

5sec θ = 13

12cot θ = - 5

12

C) sin θ = - 12

13cos θ = 5

13tan θ = - 12

5csc θ = - 13

12sec θ = 13

5cot θ = - 5

12

D) sin θ = 12

13cos θ = 5

13tan θ = - 12

5csc θ = 13

12sec θ = - 13

5cot θ = - 5

12

98)

99) sec θ = - 13

5 , tan θ > 0A) sin θ = - 12

13cos θ = 5

13tan θ = - 5

12csc θ = - 13

12sec θ = - 13

5cot θ = - 12

5

B) sin θ = - 12

13cos θ = - 5

13tan θ = - 12

5csc θ = - 13

12sec θ = - 13

5cot θ = - 5

12

C) sin θ = 12

13cos θ = - 5

13tan θ = - 5

12csc θ = 1312sec θ = - 13

5cot θ = - 12

5

D) sin θ = - 12

13cos θ = - 5

13tan θ = 12

5csc θ = - 13

12sec θ = - 13

5cot θ = 5

12

99)

Solve the problem If necessary, round to the nearest tenth.

100) The force acting on a pendulum to bring it to its perpendicular resting point is called the restoring

force The restoring force F, in Newtons, acting on a string pendulum is given by the formula

F = mg sin θwhere m is the mass in kilograms of the pendulum's bob, g ≈ 9.8 meters per second per second isthe acceleration due to gravity, and θ is angle at which the pendulum is displaced from theperpendicular What is the value of the restoring force when m = 0.6 kilogram and θ = 30°?

100)

102) If friction is ignored, the time t (in seconds) required for a block to slide down an inclined plane is

given by the formula

g sin θ cos θwhere a is the length (in feet) of the base and g ≈ 32 feet per second per second is the acceleration ofgravity How long does it take a block to slide down an inclined plane with base a = 11 when

θ = 60°?

102)

103) If friction is ignored, the time t (in seconds) required for a block to slide down an inclined plane is

given by the formula

g sin θ cos θwhere a is the length (in feet) of the base and g ≈ 32 feet per second per second is the acceleration ofgravity How long does it take a block to slide down an inclined plane with base a = 10 when

125) sec 1305°

2 33

Find all values of θ with the given trigonometric function value.

137) cos θ = 1

2A) 60° + n · 360°; 120° + n · 360°, n any integerB) 60° + n · 360°; 300° + n · 360°, n any integerC) 210° + n · 360°; 330° + n · 360°, n any integerD) 150° + n · 360°; 210° + n · 360°, n any integer

137)

138) tan θ = 1

A) 135° + n · 360°; 225° + n · 360°, n any integerB) 45° + n · 360°; 225° + n · 360°, n any integerC) 45° + n · 360°; 315° + n · 360°, n any integerD) 225° + n · 360°; 315° + n · 360°, n any integer

138)

139) cos θ = - 3

2A) 210° + n · 360°; 330° + n · 360°, n any integerB) 60° + n · 360°; 120° + n · 360°, n any integerC) 60° + n · 360°; 300° + n · 360°, n any integerD) 150° + n · 360°; 210° + n · 360°, n any integer

139)

140) sin θ = - 1

2A) 150° + n · 360°; 210° + n · 360°, n any integerB) 210° + n · 360°; 330° + n · 360°, n any integerC) 60° + n · 360°; 300° + n · 360°, n any integerD) 60° + n · 360°; 120° + n · 360°, n any integer

140)

141) sec θ = - 2

A) 45° + n · 360°; 225° + n · 360°, n any integerB) 135° + n · 360°; 225° + n · 360°, n any integerC) 225° + n · 360°; 315° + n · 360°, n any integerD) 45° + n · 360°; 315° + n · 360°, n any integer

141)

142) sin θ = - 2

2A) 45° + n · 360°; 315° + n · 360°, n any integerB) 135° + n · 360°; 225° + n · 360°, n any integerC) 225° + n · 360°; 315° + n · 360°, n any integer

142)

153)

154) cos β = 1

10, find sec βA)- 1

155)

36

Full file at https://TestbankDirect.eu/Test-Bank-for-Trigonometry-by-Ratti

157) tan β = 3

4, find cot βA) 7

159)

160) sin α = 12

13, cos α = - 5

13; find cot αA) 5

161)

162) cos θ = 6

7, tan θ = 13

6 ; find sin θA) 7

162)

Use the reciprocal and quotient identities to find the indicated function value.

165) sec θ = - 9

8, tan θ = 17

8 ; find sin θA)- 17

166)

Use Pythagorean identities to find the indicated function value.

167) sin θ = - 3

5, 270° < θ < 360°; find cos θA) 2

167)

168) cos θ = - 6

7, 180° < θ < 270°; find sin θA)- 85

Use the basic trigonometric identities and the given information to find the exact values of the remaining trigonometric functions.

173) cos θ = 7

25 and 270° < θ < 360°

A) sin θ = - 7

25cos θ = 24

25tan θ = - 24

7csc θ = - 25

7sec θ = 25

24cot θ = - 7

24

B) sin θ = - 24

25cos θ = 7

25tan θ = - 24

7csc θ = - 25

24sec θ = 25

7cot θ = - 7

24

C) sin θ = - 24

25cos θ = 7

25tan θ = - 7

24csc θ = - 25

24sec θ = 25

7cot θ = - 24

7

D) sin θ = 24

25cos θ = - 7

25tan θ = - 24

7csc θ = 25

24sec θ = - 25

7cot θ = - 7

13tan θ = 133

6csc θ = - 13

6sec θ = - 13 133

133cot θ = 6 133

133

B) sin θ = - 6

13cos θ = - 133

13tan θ = 6 133

133csc θ = - 13

6sec θ = - 13 133

133cot θ = 133

6C) sin θ = - 6

13cos θ = 133

13tan θ = - 6 133

133

D) sin θ = - 133

13cos θ = - 6

13tan θ = 6 133

133

174)

175) cot θ = - 33

4 and 90° < θ < 180°

A) sin θ = - 33

7cos θ = 4

7tan θ = - 4 33

33csc θ = - 7 33

33sec θ = 7

4cot θ = - 33

4

B) sin θ = 33

7cos θ = - 4

7tan θ = - 4 33

33csc θ = 7 33

33sec θ = - 7

4cot θ = - 33

4

C) sin θ = - 4

7cos θ = 33

7tan θ = - 4 33

33csc θ = - 7

4sec θ = 7 33

33cot θ = - 33

4

D) sin θ = 4

7cos θ = - 33

7tan θ = - 4 33

33csc θ = 74sec θ = - 7 33

33cot θ = - 33

13cos θ = - 5

13tan θ = - 5

12csc θ = - 13

12sec θ = 13

5cot θ = - 12

5

B) sin θ = 12

13cos θ = - 5

13tan θ = - 5

12csc θ = 13

12sec θ = - 13

5cot θ = - 12

5

C) sin θ = 12

13cos θ = - 5

13tan θ = - 12

5csc θ = 1312sec θ = - 13

5cot θ = - 5

12

D) sin θ = 12

13cos θ = - 5

13tan θ = - 12

5csc θ = - 13

12sec θ = 13

5cot θ = - 5

177) cos θ = 23

12 and sin θ = - 11

12A) sin θ = - 11

12cos θ = 23

12tan θ = - 23

11csc θ = - 12

11sec θ = 12 23

23cot θ = - 11 23

23

B) sin θ = - 11

12cos θ = 23

12tan θ = - 11 23

23csc θ = - 12 23

23sec θ = 1211cot θ = - 23

11C) sin θ = - 11

12cos θ = 23

12tan θ = - 11 23

23csc θ = 1211sec θ = - 12 23

23cot θ = - 23

11

D) sin θ = - 11

12cos θ = 23

12tan θ = - 11 23

23csc θ = - 12

11sec θ = 12 23

23cot θ = - 23

13cos θ = - 12

13tan θ = 12

5csc θ = - 13

B) sin θ = - 5

13cos θ = - 13

12tan θ = 12

5csc θ = - 12

C) sin θ = - 12

13cos θ = - 5

13tan θ = 12

5csc θ = - 13

D) sin θ = - 13

12cos θ = - 5

13tan θ = 12

5csc θ = - 12

178)

179) csc θ = - 5

2 and cot θ = - 21

2A) sin θ = - 5 21

21cos θ = 2

5tan θ = - 2 21

21csc θ = - 5

2sec θ = 21

5cot θ = - 21

2

B) sin θ = - 2

5cos θ = 21

5tan θ = - 2 21

21csc θ = - 5

2sec θ = 5 21

21cot θ = - 21

2

C) sin θ = - 2

5cos θ = 5 21

21tan θ = - 2 21

21csc θ = - 5

2sec θ = 21

5cot θ = - 21

2

D) sin θ = - 21

5cos θ = 2

5tan θ = - 2 21

21csc θ = - 5

2sec θ = 5 21

21cot θ = - 21

2

179)

180) sin θ = - 5

13 and tan θ > 0A) sin θ = - 5

13cos θ = - 12

13tan θ = - 5

12csc θ = - 13

5sec θ = - 13

12cot θ = - 12

5

B) sin θ = - 5

13cos θ = 12

13tan θ = - 12

5csc θ = - 13

5sec θ = 13

12cot θ = - 5

12

C) sin θ = - 5

13cos θ = - 12

13tan θ = 5

12csc θ = - 13

5sec θ = - 13

12cot θ = 12

5

D) sin θ = - 5

13cos θ = - 12

13tan θ = - 12

5csc θ = 13

5sec θ = - 13

12cot θ = - 5

12

180)

181) tan θ = - 13

6 and sec θ < 0A) sin θ = 13

7cos θ = - 6

7tan θ = - 6 13

13csc θ = 7 13

13sec θ = - 7

6cot θ = - 13

6

B) sin θ = - 13

7cos θ = 6

7tan θ = - 13

6csc θ = 7 13

13sec θ = 7

6cot θ = - 6 13

13

C) sin θ = 6

7cos θ = - 13

7tan θ = - 13

6csc θ = 7 13

13sec θ = - 7 13

13cot θ = - 6 13

13

D) sin θ = 13

7cos θ = - 6

7tan θ = - 13

6csc θ = 7 13

13sec θ = - 7

6cot θ = - 6 13

182) sec θ = 10

3 and cot θ < 0A) sin θ = - 3

10cos θ = 91

10tan θ = - 3 91

91csc θ = - 10 91

91sec θ = 10

3cot θ = - 91

3

B) sin θ = 91

10cos θ = - 3

10tan θ = - 91

3csc θ = - 10 91

91sec θ = 10

3cot θ = - 3 91

91C) sin θ = - 91

10cos θ = 3

10tan θ = - 3 91

91csc θ = - 10 91

91sec θ = 10

3cot θ = - 91

3

D) sin θ = - 91

10cos θ = 3

10tan θ = - 91

3csc θ = - 10 91

91sec θ = 10

3cot θ = - 3 91

91

182)

Use the basic identities to simplify the expression.

183) (1 - cos θ)(1 + cos θ) + cos2 θ

191) cot2 θ + 3 cot θ - 4

cot θ + 4A) (cot θ - 4)(cot θ - 1)

Solve the problem.

193) A ladder x feet long leans against a house and makes an angle θ with the horizontal The base of

the ladder is 10 feet from the house Then x = 10

cos θ Use a reciprocal identity to rewrite thisformula

10sin θ

193)

194) From a distance of x feet to a 110-meter-high radio tower, the angle of elevation is θ degrees Then

x = 110tan θ Use a reciprocal identity to rewrite this formula.