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  #1  
Old 01-18-2010, 02:24 PM
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Default Need some math help



if i have a cable 150' away from a fixed line, and two other cables running 200' to the same fixed line making a triange, what distace would be between the 200' cables? (see wonderful paint skills)
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Old 01-18-2010, 02:28 PM
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Is the 150' perpendicular to the fixed line? Or at an angle like you have it drawn? This will determine if you have enough info to solve this or not...
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Old 01-18-2010, 02:28 PM
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264.6 ft rounded up... but I'm probably wrong as hell.
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Old 01-18-2010, 02:30 PM
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Originally Posted by 98AckCL23 View Post
Is the 150' perpendicular to the fixed line? Or at an angle like you have it drawn? This will determine if you have enough info to solve this or not...

it's perpendicular. just horible paint skills.
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Old 01-18-2010, 02:34 PM
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dam i kno this but i fell asleep during class when they tought us this
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Old 01-18-2010, 02:39 PM
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if your 150 foot line is exactly 90 degrees (square) from the line in question, you will have exactly 250 feet from the center to where each end of the 200 foot runs intersect using the 3-4-5 formulation

edit,... this may be a lil off base, been a while
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Old 01-18-2010, 02:42 PM
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132' 3.5"
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Old 01-18-2010, 02:43 PM
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Originally Posted by nismos14 View Post
264.6 ft rounded up... but I'm probably wrong as hell.
I get the same thing: 264.5751311 or 2*SQRT (17500)

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Originally Posted by wettnudel's View Post
if your 150 foot line is exactly 90 degrees (square) from the line in question, you will have exactly 250 feet from the center to where each end of the 200 foot runs intersect using the 3-4-5 formulation
This would be the case if you were given the lengths in question to be the 200' and the other sides were the ones in question. The hypotenuse (200' lengths in this case) are the longest sides of the triangle.
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Old 01-18-2010, 02:44 PM
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Originally Posted by kapone D.O.A View Post
dam i kno this but i fell asleep during class when they tought us this
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Originally Posted by wettnudel's View Post
if your 150 foot line is exactly 90 degrees (square) from the line in question, you will have exactly 250 feet from the center to where each end of the 200 foot runs intersect using the 3-4-5 formulation

edit,... this may be a lil off base, been a while

seems really close to what niz said. i don't need an exsact number, just something within a couple of feet will work.
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Old 01-18-2010, 02:47 PM
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Quote:
Originally Posted by RollerDJ View Post


if i have a cable 150' away from a fixed line, and two other cables running 200' to the same fixed line making a triange, what distace would be between the 200' cables? (see wonderful paint skills)
Assuming the 150' run is perpendicular to the bottom line, it would be a right triangle. That makes this super easy without the need of the law of sine/cosine.

Let's work this step by step so that the people that can't remember high school math can still follow.
Since a^2+b^2=c^2, we can arrange it so that a^2=c^2-b^2. We then arrange it to a=sqrt((c^2)-(b^2)).

Making it like this: sqrt((200^2)-(150^2))=132.29' This is only representative of one set of , so we multiply this by 2 to get the total distance between the outermost points in the base of the triangle.
132.287566*2=264.58'

nismos you were correct.
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