Let ABCD be the square with home plate at A, 1st base at B. Let the pitcher's mound be at P. Let M be the intersection of the diagonals.
(1) We could use the Law of Cosines: Since AC is a diagonal of the square, and P lies on AC, we have `m/_CAB=45^@` .
Then with AP=16.8,AB=22.8 we have:
So the distance from the pitcher's mound to 1st base is approximately 16.1m
(2) Consider triangle BMP: Since ABCD is a square the diagonals are congruent, they bisect each other, and they are perpendicular.
Again, the distance from the pitcher's mound to 1st base is approximately 16.1m.
Since the diamond is actually a square, the line from home to the pitcher's mound would from a 45 degree angle to the line from home to 1st base. Since we know that the distance from home to the pitcher's mound is 16.8 m. We can calculate the length of a line from the pitcher's mound perpendicular to the base path as follows: `sin 45 = y/16.8` so `y=16.8xsin 45 ~~ 11.88m` . Then we calculate the length from home to the right angle intersection: `cotan 45=z/16.8sin 45` so `cotan 45x16.8sin 45=z` . Since tan 45=1, then cotan 45 =1, so `z=16.8sin 45 ~~ 11.88m` . This then gives us another right triangle, on the other side of line y from intersection to 1st base, to the pitcher's mound. `Y=16.8 sin 45` , and from intersection to first base is `22.8-16.8 sin 45` . so `tan x=16.8 sin 45/22.8-16.8 sin 45 ~~ 1.09` . `arctan 1.09 ~~ 47 degrees` . So the length of the hypontenuse of this triangle (i.e. the distance from Pitcher's mound to first base) can be found using the secant of 47 degrees. `sec 47=x/16.8 sin 45 x~~16.24m` The distance from the Pitcher's mound to First base is approximately 16.24m
I wish I could have drawn a diagram, but I don't have the tools to show the diagram. Hopefully you can follow the steps of my calculations.