Make a diagram showing the two corridors and the ladder just fitting into the corner
You should have two similar right-angled triangles.
Label the hypotenuse of the one triangel L1 and the hypotenuse of the other L2
then L1 + L2 = 27
label each of the bottom left angles Ø
let the width of the second hall-way be x ft
then sinØ = x/L2 and cosØ = 8/L2 from the 2 triangles.
L2 + x/sinØ and L1 = 8/cosØ
x/sinØ + 8/cosØ = 27
xcosØ + 8sinØ = 27sinØcosØ
x = (27sinØcosØ - 8sinØ)/cosØ = 27sinØ - 8tan‚
we want the minimum length of x for the ladder to fit around the corner, (obviously if the value of x is large, there would be no problem)
dx/dØ = 27cosØ - 8sec^2 Ø = 0 for a min of x
27cosØ = 8/cos^2 Ø
cos^3 Ø = 8/27
cosØ = 2/3
Ø = 48.19°
subbing back into the x = .... equation, I got
x = 11.2 ft
Nice Question !
a pole 27ft long is carried horizontally along a corridor 8ft wide and into a second corridor at right angles to the first. how wide must the secon corridor be?
1 answer