Posted by T on Thursday, May 6, 2010 at 7:18pm.
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)

Algebra 2 with trigonometry  Reiny, Thursday, May 6, 2010 at 7:22pm
I see it as a simple Pythagorean problem
x^2 + 60.5^2 = 90^2
....
....
x = 66.6 feet

Correction  Algebra 2 with trigonometry  Reiny, Thursday, May 6, 2010 at 7:26pm
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
Answer This Question
Related Questions
 math  A baseball diamond is actually a square with 90 feet between the bases. ...
 Math (trigonometry)  A baseball diamond is actually a square with each side ...
 Math  A baseball diamond is a square sides 22.4 m. The pitcher's mound is 16.8 ...
 Math  A baseball diamond is a square with sides 22.4m. The pitcher's mound is ...
 Algebra  HELP! I can't figure the formula out for this problem. a pitcher threw...
 Algebra  a pitcher threw a baseball clocked at 105 miles per hour. The pitcherâ€™...
 geometry  I asked my teacher for a hint and he said that the pitcher's mound is...
 Trigonometry  A baseball diamond is a square with sides 27.4m. The pitchers ...
 physics  a pitcher threw a baseball clocked at 105 miles per hour. The pitcherâ€™...
 Calculus  A baseball diamond is a 90 foot square with the pitcher's mound at ...
More Related Questions