Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 322: 54

Answer

The distance from the pitcher's position to first base is 42.6 feet. The distance from the pitcher's position to second base is 38.9 feet. The distance from the pitcher's position to third base is 42.6 feet.

Work Step by Step

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 $\sqrt{2}\times 60.0~ft-46.0~ft = 38.9~ft$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.