SOLUTION: A batter in a baseball game drops a bunt down the first-base line. It rolls 42ft at an angle of 36 degrees with the base path. The pitcher's mound is 60.5ft from home plate: How fa

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Question 164709: A batter in a baseball game drops a bunt down the first-base line. It rolls 42ft at an angle of 36 degrees with the base path. The pitcher's mound is 60.5ft from home plate: How far must the pitcher travel to pick up the ball? Answer to the nearest foot. Note a baseball field is a square.
I drew the field and the triangle within the square, but i dont think my picture is correct. i came up with 36.23 ft; that xoesnt seem correct to me. Please help
Found 2 solutions by edjones, gonzo:
Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
We have 2 sides of a triangle, 60.5 and 42. The angle between them is 45-36=9 deg. SAS triangle.
a^2=b^2+c^2-2bc*cosA
a^2=60.5^2+42^2-(2*60.5*42*cos(9)
=3660.25+1764-(5082*.987688)
=5424.25-5019.43
=404.818
a=sqrt(404.818)
=20.1201
Answer 20 ft.
.
Ed

Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
hard to do without pictures but i'll try.
let AB be the distance between home plate and the pitcher's mound.
let AC be the distance that the ball traveled.
AB = 60.5
AC = 42.0
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ball traveled at angle of 36 degrees from base path.
pitcher's mound is 45 degrees from base path.
AB is therefore 45 degrees from base path.
AC is 36 degrees from base path.
AC is therefore 9 degrees from AB (45-9).
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if you draw the line AB vertically on your paper.
then draw the line AC slanting 9 degrees to the right of that line.
then draw BC to complete the triangle.
next draw an altitude from point C to AB. this will be a horizontal line that will intersect with AB at point D which is between points A and B somewhere closer to B (roughly 40 feet from A and 20 feet from B as you will see).
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you now have 2 right triangles.
they are triangles ACD and BCD.
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triangle ACD is composed of the following:
line AC which is 42 feet which is the distance and direction the ball traveled.
line AD which is about 40 feet up the path towards the pitcher's mound.
line CD which is the altitude from point C perpendicular to AB and intersecting with AB at point D.
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triangle BCD is composed of the following:
line BD which is about 20 feet and is the distance from the pitcher's mound to point D.
line BC which is the distance from the pitcher's mound to where the ball stopped traveling (what we want to find).
line CD which is the same altitude from point C perpendicular to AB and intersecting with AB at point D (described above).
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if you can draw the picture correctly, you should be able to follow the rest easy enough.
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if you can't draw the picture, then send me an email and i'll send you the picture (gonzo@gmx.us)
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first you need to find the distance AD and the distance CD.
cosine angle CAD = AD / AC.
sine angle CAD = CD / AC.
we know angle CAD is 9 degrees.
we know AC is 42 feet.
from the formulas we can solve for AD and CD.
AD = AC * cosine angle CAD.
CD = AC * sine angle CAD.
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AD = 42 * cosine 9 degrees.
CD = 42 * sine 9 degrees.
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AD = 41.4829.... feet.
CD = 6.5702... feet.
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only approximate values are shown. full values are stored in the computer.
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AD is the distance from home plate to point D that is where CD intersects with AB and is perpendicular to AB.
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now you want to know the distance from that point to the pitcher's mound.
that is 60.5 feet - 41.48291031 feet = 19.0170897 feet.
that distance is the length of the line BD.
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now you turn your attention to your second triangle which is BCD where BC is the distance from the pitcher's mound to the ball and BD is the distance from the pitcher's mound to point D.
you already know the length of CD because it was calculated earlier.
you already know the length of BD because you just calculated it.
you still want to find the length of BC (pitcher's mound to ball).
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CD = 6.0572... feet (calculated earlier).
BD = 19.017... feet (just calculated above).
tangent angle DBC = CD / BD = 6.5702... feet / 19.017... feet.
tangent angle DBC = .34549...
angle DBC = 19.059... degrees.
sine of angle DBC = CD / BC
solving for BC, equation becomes
BC = CD / sine DBC
angle DBC = 19.059... degrees
sine of angle DBC = .32655...
CD = 6.5702 ...
therefore,
BC = 6.5702... / .32655... = 20.12008... feet.
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answer is BC = 20.12008... feet.
that's the distance from the pitcher's mound to where the ball stopped.
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