*A board leaning against a fence makes an angle of 30 degrees with the horizontal. If the board is 4 feet long, what is the diameter of the largest pipe which will fit between the board, the fence and the ground?*

We assume the pipe touches the ground, the fence, and the board. Since the board is 4 feet long, the fence is 2 feet high where the board touches it, and the distance from the bottom of the fence to the bottom of the board is `2sqrt(3)` feet, using the 30-60-90 right triangle.

Let the distance from the bottom of the board to the point the board touches the pipe be x feet, and the radius of the pipe be r feet. Then the distance from the bottom of the board to the point the pipe touches the ground is x feet (tangent segments drawn from a point to a circle are congruent), and the distance from where the pipe touches the ground to the base of the fence is r feet. (The radius is perpendicular to the fence and the ground, forming a square of side length r feet)

Thus `x+r=2sqrt(3)` . But we also have a right triangle formed by the bottom of the board, the point the pipe touches the ground, and the center of the circle with an acute angle of `15^circ` . Then `r/x=tan15^circ` or `r=xtan15^circ` .

Substituting we get `x+xtan15^circ = 2sqrt(3)` . Noting that `tan15^circ = 1/(2+sqrt(3))` (Using a half-angle formula) we now have:

`x(1+1/(2+sqrt(3)))=2sqrt(3)`

`x=(2sqrt(3))/(1+1/(2+sqrt(3)))`

`x=(4sqrt(3)+6)/(3+sqrt(3))` by rationalizing

`x=sqrt(3)+1`

Then `x+r=2sqrt(3) => r=2sqrt(3)-(sqrt(3)+1)=sqrt(3)-1`

**Thus the diameter is twice r or `2sqrt(3)-2 ~~1.464` feet.**

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