Question

Suppose a certain baseball diamond is a square 60 feet on a side. The pitching rubber is located 40.5 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 first base is about feet.
(Round to two decimal places as needed.)
(b) The distance from the pitching rubber to 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.)

Answer 1

Answer: 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

Explanation: