In the cone, through its apex at an angle phi to the base plane, a plane is drawn that cuts off the arc alpha from the base circle. The height of the cone is h. find the volume of the cone.
The section of the BKM is an isosceles triangle since BK = BM as generators of the cone. Let’s build the height and median of BO1. In a right-angled triangle OO1B, the angle OBO1 = β, then
ctgβ = OO1 / OB = OO1 / h.
OO1 = ctgβ * h see.
Let’s build the radii OK and OM. The OKM triangle is isosceles, since OM = OK, then OO1 is its height, median and bisector.
Then in the right-angled triangle OO1M the angle O1OM = α / 2, and Cos (α / 2) = OO1 / OM = OO1 / R = (ctgβ * h) / R.
R = (h * ctgβ) / Cos (α / 2).
Then Sop = π * R ^ 2 = π * ((h * ctgβ) / Cos (α / 2)) ^ 2 = h2 * ctgβ / Cos (α / 2) 2.
V = Sosn * h / 3 = h ^ 3 * (ctgβ / Cos (α / 2) ^ 2/3 = h3 * ctg2β / 3 * Cos2 (α / 2).
Answer: The volume of the cone is h ^ 3 * ctg2β / 3 * Cos2 (α / 2).
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