Trigonometry (11th Edition) Clone

Let A be the location of home plate. Let B be the location of the pitcher's position. Let C be the location of first base. The points ABC form a triangle. The angle $A = 45^{\circ}$, the side $b = 60.0~ft$, and the side $c = 46.0~ft$. We can use the law of cosines to find $a$, which is the distance from the pitcher's position to first base: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(60.0~ft)^2+(46.0~ft)^2-(2)(60.0~ft)(46.0~ft)~cos~45^{\circ}}$ $a = \sqrt{1812.77~ft^2}$ $a = 42.6~ft$ The distance from the pitcher's position to first base is 42.6 feet. By symmetry, the distance from the pitcher's position to third base is 42.6 feet. The total distance from home plate to second base is $\sqrt{2}\times 60.0~ft$. The distance from the pitcher's position to second base is $\sqrt{2}\times 60.0~ft-46.0~ft = 38.9~ft$