Respuesta :

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:

  • Hector throws the ball farther than Tre

Methods used to obtain the correct response

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, b, from home base to the pitcher's mound and the line, c,  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:

https://brainly.com/question/2491835