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On a baseball field, the pitchers mound is 60.5 feet from home plate. During practice, a batter hits a ball 186 feet at an angle of 40 degrees to the right of the pitchers mound. An outfielder catches the ball and throws it to the pitcher. Approximately how far does the outfielder throw the ball?!

Respuesta :

Use cosine law to solve for x

[tex]x^2=60.5^2+186^2-2\times60.5\times186\times \cos{40^o} \\x^2=38,256.25-17,240.6 \\x^2=21,015.65 \\x= \sqrt{21,015.65} \\x=145[/tex]

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Answer:

The outfielder threw the ball 145 feet away.

Step-by-step explanation:

We are given that,

Distance of mound from home plate = 60.5 feet

Distance of outfielder from home plate = 186 feet

Angle made by the batter = 40°

So, using the cosine law, we have,

[tex]c^{2}=a^{2}+b^{2}-2ab\cos \theta[/tex]

i.e. [tex]c^{2}=(60.5)^{2}+(186)^{2}-2\times 60.5\times 186\times \cos 40[/tex]

i.e. [tex]c^{2}=38256.25 -22506\times 0.766[/tex]

i.e. [tex]c^{2}=38256.25 -17239.596 [/tex]

i.e. [tex]c^{2}=21016.654 [/tex]

i.e. c = ±144.97 ≈ ±145

Since, the distance cannot be negative.

Thus, the outfielder threw the ball 145 feet away.

Ver imagen wagonbelleville

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