There is a 6 metres wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 metres long, the other 12 metres, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape.
What is the distance from the point of intersection from the ground?
I never thought to use graphing, that's pretty clever. I stumbled around with trig until I found this solution.
Consider the triangle made underneath the two ladders. It has a base of 6 and a height h. The left corner (I anchored the 10 ladder in the left corner) has an angle of ~53.1 degrees (arctan(8/6)). The right corner has an angle of 60 degrees (arccos(6/12)).
Drawing a line straight down from the intersection point to the ground splits this triangle into two. The length of this line is h. Call the distance from the left corner to the point on the ground where this line hits x (with the distance from the right corner being 6-x). Since we have the angles, we get two equations:
tan(53.1 degrees) = h/x and
tan(60 degrees) = h/(6-x)
The first equation gives you x = 0.75h. Putting that into the second equation gives you 1.73(6 - 0.75h) = h. Solve for h and you get 4.52.
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