Multiple Choice
Identify the
choice that best completes the statement or answers the question.
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1.
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sin 304°
a. | -0.8103183 | c. | -0.8203183 | b. | -0.8303183 | d. | -0.8213183 |
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2.
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cos 164°
a. | -0.5808308 | c. | -0.9616616 | b. | -0.7616616 | d. | -0.4808308 |
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3.
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tan 55°
a. | 1.4735983 | c. | 1.4738983 | b. | 1.4732983 | d. | 1.4741983 |
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4.
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csc 30°
a. | 1.9949764 | c. | 1.9959764 | b. | 1.9939764 | d. | 1.9969764 |
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5.
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sin ( -34° )
a. | -0.5562956 | c. | -0.5543603 | b. | -0.5640066 | d. | -0.5620834 |
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6.
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a. | 1.245232 | c. | 1.2589754 | b. | 0.7942967 | d. | 0.8030632 |
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7.
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a. | 0.1943803 | c. | 5.144554 | b. | 1 | d. | 0.0174551 |
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8.
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cos 47° cos
133° - sin 47° sin 133°
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9.
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57° +  57°
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10.
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Use a calculator to decide
whether the statement is true or false.
sin ( 150° + 30°) = sin 150° + sin
30°
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11.
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Use a calculator to decide
whether the statement is true or false.
sin ( 180°+ 225°) = sin 180° · cos 225° +
cos 180° · sin 225°
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12.
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Use a calculator to decide
whether the statement is true or false.
sin (2 · 30°) = 2 · sin 30°
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13.
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Use a calculator to decide
whether the statement is true or false.
sin (2 · 150°) = 2 · sin 150° · cos
150°
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14.
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Use a calculator to decide
whether the statement is true or false.
cos (2 · 30°) = 2 · cos 30°
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15.
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Use a calculator to decide
whether the statement is true or false.
cos (2 · 90°) = cos2 90° - sin2
90°
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16.
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Find a value of
è in [0°, 90°] that satisfies the
statement. Leave answer in decimal degrees rounded to seven decimal places, if necessary.
sin
è = 0.81107642
a. | 125.798766° | c. | 234.201234° | b. | 35.7987657° | d. | 54.2012343° |
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17.
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Find a value of
è in [0°, 90°] that satisfies the
statement. Leave answer in decimal degrees rounded to seven decimal places, if necessary.
cos
è = 0.49566425
a. | 299.713561° | c. | 60.2864387° | b. | 29.7135613° | d. | 119.713561° |
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18.
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Find a value of
è in [0°, 90°] that satisfies the
statement. Leave answer in decimal degrees rounded to seven decimal places, if necessary.
tan
è = 0.83799703
a. | 50.0370924° | c. | 219.962908° | b. | 320.037092° | d. | 39.9629076° |
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19.
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Any offset between a stationary
radar gun and a moving target creates a "cosine effect" that reduces the radar mileage
reading by the cosine of the angle between the gun and the vehicle. That is, the radar speed reading
is the product of the actual reading and the cosine of the angle. Find the radar reading to the
nearest hundredth for the auto shown in the figure.
a. | 96.03
mph | c. | 94.51
mph | b. | 97.97 mph | d. | 21.82 mph |
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20.
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The grade resistance F of a car
traveling up or down a hill is modeled by the equation F = W sin è,
where W is the weight of the car and è is the angle of the
hill's grade (è > 0 for uphill travel, è < 0 for downhill travel). What is the grade resistance (to the nearest pound) of
a 3000-lb car traveling uphill on a 3° grade ( )?
a. | 157 lb | c. | -3003 lb | b. | 3003 lb | d. | -157 lb |
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21.
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The grade resistance F of a car
traveling up or down a hill is modeled by the equation F = W sin è,
where W is the weight of the car and è is the angle of the
hill's grade (è > 0 for uphill travel, è < 0 for downhill travel). Find the weight of the car (to the nearest pound) that is
traveling on a  downhill grade and which has a grade
resistance of  lb.
a. | 4000 lb | c. | 4100 lb | b. | 3800 lb | d. | 4300 lb |
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22.
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The grade resistance F of a car
traveling up or down a hill is modeled by the equation F = W sin è,
where W is the weight of the car and è is the angle of the
hill's grade (è > 0 for uphill travel, è < 0 for downhill travel). What is the grade resistance (to the nearest pound) of
a 2100-lb car traveling downhill on a 5° grade (è =
-5°)?
a. | 2105 lb | c. | -2105 lb | b. | -183 lb | d. | 183 lb |
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23.
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The grade resistance F of a car
traveling up or down a hill is modeled by the equation F = W sin è,
where W is the weight of the car and è is the angle of the
hill's grade (è > 0 for uphill travel, è < 0 for downhill travel). A 2025-lb car has just rolled off a sheer vertical
cliff (è = -90°). What is the car's grade
resistance?
a. | 2025 lb | c. | 0 lb | b. | undefined | d. | -2025 lb |
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24.
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If an automobile is traveling
at velocity V (in feet per second) , the safe radius R for a curve with superelevation
á is given by the formula  where f and
g are constants. A road is being constructed for automobiles traveling at 50 miles per hour. If
  and  calculate R. Round to the nearest foot. (Hint: 1 mile = 5280 feet)
a. | R = 706
ft | c. | R = 742
ft | b. | R = 722
ft | d. | R = 1172
ft |
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25.
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A formula used by an engineer
to determine the safe radius of a curve, R, when designing a road is:
 where
á is the superelevation of the road and V is the velocity (in
feet per second) for which the curve is designed. If  ft per sec,
f = 0.1 and  find
R. Round to the nearest foot.
a. | R = 1973
ft | c. | R = 1983
ft | b. | R = 1977
ft | d. | R = 1970
ft |
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26.
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A formula used by an engineer
to determine the safe radius of a curve, R, when designing a particular road is: 
where á is the superelevation of the road and V is the velocity (in
feet per second) for which the curve is designed. If  f = 0.1 , and  find V. Round to the nearest foot per
second.
a. | V = 74 ft per
sec | c. | V = 67 ft per
sec | b. | V = 71 ft per sec | d. | V = 69 ft per sec |
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27.
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The index of refraction for
air, Ia, is 1.0003. The index of refraction for water, Iw, is 1.3. If  and  find W to the nearest tenth.
a. | 21.7° | c. | 23.7° | b. | 20.7° | d. | 22.7° |
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28.
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Snell's Law states that
 Use this law to find the requested
value. If  
and  find  .
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