The distance the ball is thrown is given by cosine rule and the triangle
formed by a known location, the thrower and the destination.
Correct response:
- <u>Hector throws the ball farther than Tre</u>
<h3>Methods used to obtain the correct response</h3>
The given dimensions are;
Dimension of the base diamond = 90 feet by 90 feet
Distance of pitcher's mound from home base = 60.5 feet
Horizontal distance from pitchers mound to right fielder = 95 feet
The distance between Tre and the 3rd base, d₁, is calculated using cosine rule as follows;
- a² = b² + c² - 2·b·c·cos(A)
Where;
A = The angle formed by the line, <em>b</em>, from home base to the pitcher's mound and the line, <em>c</em>, from home base to 3rd base.
a = Distance from the pitcher's mound to 3rd base = d₁
b = 60.5 feet
c = 90 feet
Therefore;
d₁² = 60.5² + 90² - 2 × 60.5 × 90 × cos(45°) = 11,760.25 - 10,890·cos(45°)
d₁ = √(11,760.25 - 10,890×cos(45°)) ≈ 63.717
- The distance Tre throws the ball, d₁ ≈ 63.717 feet
Similarly, the distance Hector throws the ball, d₂, is calculated as follows;
d₂² = (90·√2 - 60.5)² + 95² - 2 × (90·√2 - 60.5) × 95 × cos(45°)
Which gives;
d₂ = √((90·√2 - 60.5)² + 95² - 2 × (90·√2 - 60.5) × 95 × cos(45°)) ≈ 67.18
- The distance Hector throws the ball, d₂ ≈ 67.18 feet
Therefore;
- The distance Hector throws the ball, d₂ ≈ 67.18 feet, is farther than the distance, d₁ ≈ 63.717 feet, Tre throws the ball
Learn more about the law of cosines here:
brainly.com/question/2491835
