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How wide is the alley?

 Mathematics Tutor: None Selected Time limit: 1 Day

Say you have a building on each side of a level alley. You place a 10 foot ladder at the base of one building and lean it against the other building. Then you place a 14 foot ladder at the base of the other building and lean it across to the opposite building. The ladders cross each other at a vertical distance of 5 feet from the surface of the alley. Calculate the width of the alley.

Apr 27th, 2015

Let x1 = the angle the 10 ft ladder makes with the ground

Let x2 = the angle the 14 ft ladder makes with the ground

Then cosx1 = width of the alley / 14

cosx2 = width of the alley / 10

therefore,

14cosx1 = 10cosx2

Where the ladders cross is 5 ft off the ground. Draw a line directly from that point to the ground.

Let y = where the line divides the alley to the left

then, w - y = where the line divides the line to the right

therefore,

tanx1 = 5 / y

y = 5 / tanx1

tanx2 = 5 / w - y

wtanx2 - ytanx2 = 5

ytanx2 = wtanx2 - 5

y = wtanx2 / tanx2 - 5 / tanx2

y = w - 5 / tanx2

therefore,

5/ tanx1 = w - 5 / tanx2

w = 5 / tanx1 + 5 / tanx2

I'm going to need more time to figure this one out, if you'll let me.

Apr 27th, 2015

OK Vincent, take more time to figure it out.

Apr 27th, 2015

Ok, thanks.

Apr 27th, 2015

Fortunately, it turns out the relationship from the beginning

14cosx1 = 10cosx2  is easily solved as

14cos(60) = 10cos(45)

therefore, x1 = 60 degrees

and x2 = 45 degrees

therefore,

w = 5 / tan(60) + 5 / tan(45)

= 5 / 1.732 + 5 / 1

= 2.887 + 5 = 7.887 ft

Apr 27th, 2015

How did you determine that the angles were 45 degrees and 60 degrees?

Apr 27th, 2015

If you draw a diagram of the two ladders leaning up against building, you'll see that you can establish two relationships right away.

Let x1 = the angle the 10 ft ladder makes with the ground

Let x2 = the angle the 14 ft ladder makes with the ground

I hope you can see that, since I can't draw it here. If you're having trouble seeing that then I can draw it and attach it here.

Then,

cosx1 = width of the alley / 14

cosx2 = width of the alley / 10

Therefore,

width of the alley = 14cosx1

and, width of the alley = 10cosx2

that means,

14cosx1 = 10cosx2

By trial and error I found very quickly that x1 = 60 and x2 = 45,  because

14cos(60) = 10cos(45)   check it out

once you determine those two angles the rest is easier to solve.

Apr 27th, 2015

Is there a way of calculating those two angles with out using trial and error?  Using the trial and error method, I find that the two angles do not come out to be exactly 60 and 45 degrees.

Apr 27th, 2015

But they are within 7/1000. That's close enough. Really, this problem is much harder than it seems using other methods. You wind up getting 4th degree polynomials. It's a lot more work.

Apr 27th, 2015

I still do not see where the angles would be 60 and 45 degrees. Suppose the ladders (diagonals) were 21.5 feet and 14 feet long, how would you determine the angles?

May 14th, 2015

The angles depend on the ratio of the ladders.  If you change the lengths, you change the ratios.  Then the angle calculation changes too.  Then it would be 21.5cosx1 = 14cosx2.  You get a different set of angles.

May 14th, 2015

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Apr 27th, 2015
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Apr 27th, 2015
Dec 5th, 2016
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