Originally Posted by RollerDJ
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.
nismos you were correct.
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