A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.

Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.

EDIT: I believe this question is looking for the shortest ladder length possible.

Originally posted by uzless A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.

Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.

I think I know what you are trying to say, but what exactly do you mean by "just touches a box".

Originally posted by uzless A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.

Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.

EDIT: I believe this question is looking for the shortest ladder length possible.

I have a quartic to solve to find tan theta, I'm getting there...
Right all i had to do was differentiate it. I get that the gradient of the ladder is 4/3. So therefore the length is 125.

Q = length of ladderr
u = hypotonuse of lower triangle
v = Hypotonuse of upper triangle
x = angle between ladder and wall

u = 64 sec x
v = 27 csc x
Q = 64 sec x + 27 csc x
dQ = 64 sec x tan x - 27 csc x cot x
0 = 64 sec x tan x - 27 csc x cot x
64 sec x tan x = 27 csc x cot x
64 (tan x)^2 = 27 cot x
64 (tan x)^3 = 27
(tan x)^3 = 27 / 64
tan x = 3 / 4
x ~ 26.86 degrees
sec x = 10 / 8
csc x = 10 / 6
Q = 64 sec x + 27 csc x
Q = 64 * 10 / 8 + 27 * 10 / 6
Q = 640 / 8 + 270 / 6
Q = 80 + 45
Q = 125
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