I know the length of the lines A and B, and that they cross at 10’ but can’t find a way to calculate C. Anyone knows how?
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Ok this is just thinking aloud… but
d = A sin (alpha); // d is the third side of the triangle formed by A and C, alpha is the angle between A and C
e = B sin (beta); // e is the third side of the triangle formed by B and C, beta is the angle between B and C
c1 = 10 / tan(alpha); // c1 is x coordinate of the intersection point
C - c1 = 10 / tan(beta);
C = 10/tan(beta) + 10/tan(alpha);
That should be enough to get you started
ok, but now we are back to “how to find beta and alpha”?
You have also C = A*cos(alpha) and C = B*cos(beta). Together with Stainless’ equation C = 10/tan(beta) + 10/tan(alpha), that’s three equations in three unknowns, so should be solvable, in principle.
C = A*cos(alpha)
C = B*cos(beta)
C = 10/tan(beta) + 10/tan(alpha)
That’s 3 separate equations, and Stainless’ equation uses 2 unknown and 1 known, unsolvable. Yours is the same, 1 known and 1 unknown, still unsolvable. Each of the 3 equations can’t really be incorporated together can they?. Would the quadratic equation be useful here?
There’s three equations in my post, and the three unknowns are alpha, beta, and C. You know A and B. Of course C is the only one you care about, but my point is there are three equations and three unknowns, so it should be solvable. It is not a linear equation or a quadratic one, so it would have to be solved by substitution. You could use trig identities to eliminate the trig functions, expressing cos in terms of tan or vice versa. I tried putting it into Maxima to solve but it did not come up with a result. That might only be because Maxima isn’t smart enough to do it.
I found a solution some place on the net, the problem is exactly the same as the “Ladders in the alley” problem. Here is the way to solve it. I tried to understand the way it’s done, but I can’t. The author is unknown and left no explanation at all (maybe it was Einstein LOL). Do you think you are capable enough to understand it?
a = longest ladder.
b = shortest ladder.
c = ladders crossing height.
Problem = solve distance between ladders at the base.
dist = sqrt(b^2-m^2)