Question

A baseball diamond is a square 90 feet on a side. The pitcher's mound is 60.5 feet from home. How far does the pitcher have to run to cover first? (we are using the LAW OF SINES AND COSINES)
I see it as a simple Pythagorean problem

x^2 + 60.5^2 = 90^2
....
....
x = 66.6 feet
OOPS, I assumed wrongly that the pitcher was halfway between home and second.

the way you state it results in a a Law of Cosine problem

x^2 = 90^ + 60.5^2 - 2(90)(60.54059.857)cos 45
= 4059.857
x = 63.72 feet

Related Questions

I asked my teacher for a hint and he said that the pitcher's mound is not the radius. Am I supposed... A baseball diamond is a 90 foot square with the pitcher's mound at the square's center. A runner ra... A baseball diamond is actually a square with each side approximately 27.4 m. The pitcher's mound is... A baseball diamond is actually a square with 90 feet between the bases. The pitcher's mound is 60.5...