Angel M.
asked • 06/07/21Pre Calculus Final Please Help
Suppose a certain baseball diamond is a square 85 feet on a side. The pitching rubber is located
57.5 feet from home plate on a line joining home plate and second base.
(a) How far is it from the pitching rubber to first base?
(b) How far is it from the pitching rubber to second base?
(c) If a pitcher faces home plate, through what angle does he need to turn to face first base?
(a) The distance from the pitching rubber to the first base is about
_____ feet.
(Round to two decimal places as needed.)
(b) The distance from the pitching rubber to the second base is about
_______feet.
(Round to two decimal places as needed.)
(c) He needs to turn about ______° to his left.
(Round to one decimal place as needed.)
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1 Expert Answer
Raymond B. answered • 06/07/21
Tutor
5
(2)
Math, microeconomics or criminal justice
41.7 feet from pitcher to 1st
13.2 feet from pitcher to 2nd
58 degrees to turn from home to 1st
distance from pitcher to the line from home to 1st is 57.5(sin45) = 57.5(.707) = 40.66
construct a right triangle with legs 40.66 and 50-40.66 = 9.34. then sum the legs' squares, and take the square root to get 41.72 feet
distance from home to 2nd = hypotenuse of a right triangle with sides 50 and 50, and angle of 45 degrees
50^2 + 50^2 = 2(2500) = 5000. sqr5000= 70.71, subtract 57.5. 70.71-57.5= 13.21 feet from pitcher to 2nd
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Mark M.
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