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How big is the ladder



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I would be interested in how you calculated your answer.

I came up with something slightly different

I assumed that the critical position is at 45 degrees.

At that point the ladder would extend:

  • 7 foot behind the corner on the far side of the 7 foot hallway; and
  • 5 foot beyond the corner on the far side of the 5 foot hallway

Each of these points would be 12 feet from the [outside] corner where the halls meet

Therefore the ladder could be no larger than the distance between those 2 points

D = Sqrt ((12*12)+(12*12)) = Sqrt (144*2) = Sqrt (288) ~ 16.97



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dgreening, I get the same answer as k-man

The critical angle should be 45 degrees only if the two hallways are the same size. In this case, it's about 48.21 degrees.


In the picture below, the total length of the ladder is L = 7/sin(theta) + 5/cos(theta).

This minimizes when dL/dtheta = 0 

7 csc(theta) cot(theta) + 5 sec(theta) tan(theta) = dL/dtheta = 0

5 sin^3(theta) = 7 cos^3(theta)

tan^3(theta) = 7/5

theta = [ atan(7/5) ] ^ (1/3) = 48.21 degrees

This gives L = 16.8914... feet


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