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March 25, 2017

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A baseball diamond is a square sides 22.4 m. The pitcher's mound is 16.8 m from home plate on the line joining home plate and second base.. How far is the pitcher's mound from first base? (we are using the LAW OF SINES AND COSINES)

  • Math - ,

    letting home be H, pitcher be P, 2nd base be S, we have triangle HSF where

    HP = 16.8
    HF = SF = 22.4

    we want d = PF

    d^2 = 16.8^2 + 22.4^2 - 2(16.8)(22.4)cosH

    Now, the diagonal from HS to 2nd base is 22.4√2 = 31.7m

    In triangle HSF,

    22.4^2 = 31.7^2 + 22.4^2 - 2(31.7)(22.4)cosH

    cosH = .7076
    Note that H is not quite 45°, since P is not quite at the midpoint of HS.

    So, now we have

    d^2 = 16.8^2 + 22.4^2 - 2(16.8)(22.4)(.7076)
    d = 15.85m

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