Question

On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 214 feet at an angle of 36° 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?

Answer 1

168.84 ft.

Explanation:

Please find the attachment.

We have been given that on a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 214 feet at an angle of 36° to the right of the pitcher’s mound.

We can see from our attachment that point home plate, pitcher's mound and outfielder form a triangle and we need to figure out distance between pitcher's mound and outfielder (c).

As we have been given two sides and included angle of our triangle, so we will use law of cosines to solve our given problem.

`c^2=a^2+b^2-2ab\cdot \text{cos}(C)`, where a, b and c are opposite sides to angles A, B and C respectively.

Upon substituting our given values in above formula we will get,

`c^2=214^2+60.5^2-2(214)(60.5)\cdot \text{cos}(36^(\circ))`

`c^2=45796+3660.25-25894\cdot 0.809016994375`

`c^2=49456.25-20948.68605`

`c^2=28507.56395`

Let us take square root of both sides of our equation.

`c=√(28507.56395)`

`c=168.84183116159`

`c\approx 168.84`

Therefore, the outfielder thrown the ball approximately 168.84 ft.

Answer 2

See the attached picture for the answer: