Math problem

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Alienizer 109 Jan 03, 2014 at 00:51

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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Stainless 151 Jan 03, 2014 at 08:04

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);

so ….

C = 10/tan(beta) + 10/tan(alpha);

That should be enough to get you started

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Alienizer 109 Jan 03, 2014 at 19:17

ok, but now we are back to “how to find beta and alpha”?

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Reedbeta 167 Jan 03, 2014 at 20:27

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.

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Alienizer 109 Jan 03, 2014 at 21:47

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?

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Reedbeta 167 Jan 03, 2014 at 22:13

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.

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Alienizer 109 Jan 04, 2014 at 02:03

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.

a=119

b=70

c=30

d=a^2-b^2

f=2*c^2*d*sqrt(d^2/27+c^4)

g=d^3/27+2*c^4*d

h=d/3+(g+f)^(1/3)+(g-f)^(1/3)

j=sqrt(c^2-d+h)

k=sqrt((2*c*d+2*c^3)/j+2*c^2-d-h)

m=(c+j+k)/2

dist = sqrt(b^2-m^2)
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