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?
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 !
Well, if we're talking about a pole that's 27ft long, I hope it doesn't have any aspirations of becoming a professional pole vaulter! As for the second corridor, I suppose it should be at least 27ft wide, otherwise that pole might end up doing some acrobatic maneuvers and crashing into the walls! But hey, maybe we should build a circus tent instead and let the pole have its own show!
To determine the width of the second corridor, we can use the Pythagorean theorem.
Let's call the width of the second corridor x.
According to the problem, the pole is carried horizontally, so it will lie along the hypotenuse of a right-angled triangle formed by the two corridors.
Using the Pythagorean theorem, we can set up the equation:
(8ft)^2 + x^2 = (27ft)^2
Simplifying the equation:
64ft^2 + x^2 = 729ft^2
Subtracting 64ft^2 from both sides:
x^2 = 729ft^2 - 64ft^2
x^2 = 665ft^2
Taking the square root of both sides:
x = √665ft
Therefore, the width of the second corridor must be approximately 25.8 feet (rounded to one decimal place).
To find out the width of the second corridor, we can use the Pythagorean theorem, which relates the sides of a right triangle.
Let's assume the width of the second corridor as 'x' (in feet).
We know that the pole, which is 27 ft long, is carried horizontally. As it moves into the second corridor at right angles to the first, it will form a right triangle.
The width of the first corridor is given as 8 ft, and the length of the pole is the hypotenuse of the right triangle.
Using the Pythagorean theorem, we have:
(8 ft)^2 + (x ft)^2 = (27 ft)^2
Simplifying the equation:
64 + x^2 = 729
x^2 = 729 - 64
x^2 = 665
Taking the square root of both sides:
x = √(665)
Therefore, the width of the second corridor must be approximately 25.79 feet (rounded to two decimal places).
So, the second corridor must be approximately 25.79 feet wide.