SOLUTION: An 8 foot long pole with a 5 foot long string attached to its top is blown over so that it forms a 30 degree angle with the level ground. The string is pulled taut by its end to th

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Question 412371: An 8 foot long pole with a 5 foot long string attached to its top is blown over so that it forms a 30 degree angle with the level ground. The string is pulled taut by its end to the ground and moved in a circle. What is the volume of the figure enclosed by the string's path and the ground?
Answer by Theo(13342) About Me  (Show Source):
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See the picture below for reference:

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When you lean the pole to make a 30 degree angle with the ground, you are forming a triangle called ABC with a vertical drop from B to D which is the altitude of that triangle.

Since sine (30) = opposite / hypotenuse, you get:

sine (30) = BD / 8.

Multiply both sides of this equation by 8 and you get BD = 8 * sine (30) = 8 * .5 = 4

The altitude of your triangle is 4.

The sine of angle BCD is equal to opposite over hypotenuse is equal to 4/5.

You get sine of angle BCD = 4/5 = .8

You get angle BCD = arcsine of (4/5) = 53.13020235 degrees.

Cosine of angle BCD = adjacent / hypotenuse = DC / 5.

Multiply both sides of this equation by 5 to get:

DC = 5 * cosine (BDC) = 5 * cosine (53.13020235) = 3

You get DC = 3 which is the radius of the circle formed with D at the center.

The area of this circle is equal to pi * r^2 = pi * 3^2 = 9 * pi.

The solid figure formed is a right circular cone.

The figure in the diagram that shows this is DBCFEG. DB is the height of this figure. EGCF is the circular base of this figure. The sides of this figure are represented by the lines BC and BE.

The volume of this figure is equal to 1/3 * the height * the area of the base.

This winds up being 1/3 * 4 * 9 * pi which winds up being 12 * pi.