On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 195 feet at an angl

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On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 195 feet at an angl

e of 32° to the right of the pitcher’s mound. An outfielder catches the ball and throws it to the pitcher. Approximately how far does the outfielder throw the ball?

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-03-05T15:49:35+03:00

For this scenario, we can visualize that all points form a triangle. The three vertices are at the pitcher's mound, home plate, and the location where the outfielder catches the ball. We know two sides of the triangle and the angle that lies between these two sides.

<span>Using the cosine law, we can find the unknown third side. The formula to apply is:</span>

c^2 = a^2 + b^2 - 2ab cos θ

Where:

a = 60.5 ft

b = 195 ft

θ = 32°

Substituting the provided values results in:

c^2 = (60.5)^2 + (195)^2 - 2(60.5)(195) cos(32)

c^2 = 3660.25 + 38025 - 20009.7

c^2 = 21,675.56

c = 147.23 ft

<span>Thus, the distance the outfielder throws the ball towards home plate is approximately 147.23 ft.</span>