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Impossible Math Question?-?
An alley with high wall on both sides has a 40 foot ladder leaning from the left wall to the right wall and a 30 foot ladder leaning from the right wall to the left. They cross 10 feet above the ground. How wide is the alley?
3 Answers
- 1 decade agoFavorite Answer
The answer is 26 ft
It looks like a really fancy similar triangles problem. But I cheated and used a drafting program to figure it out.
Source(s): Microstation - TychaBraheLv 71 decade ago
Here's a start. Sorry, but I've run out of time.
Let's call the width of the alley x.
The angle formed by the 40 foot ladder at the base is F (for Forty).
The angle formed by the 30 foot ladder at the base is T (for Thirty).
cos(F) = x/40
cos(T) = x/30
F = arccos(x/40)
T = arccos(x/30)
The ladders meet above a point that is y feet across the alley at a height of 10 feet.
tan(F) = 10/y
tan(T) = 10/(x - y)
y = 10/tan(F)
x - y = 10/tan(T)
x = 10/tan(F) + 10/tan(T)
x = (10tan(T) + 10tan(F)) / tan(T)tan(F)
- 1 decade ago
Idk , I'm gonna take a wild guess:
60!
I don't think this is impossible though - watch, some smart person is gonna prove all of us wrong that this is an impossible question...
Source(s): Meh.
