The height of the cone is 6 cm, the radius of the base is 2√3 dm. Find the area of the section drawn through the two generatrices of the cone if the angle between them is 60 degrees.
From the right-angled triangle BOS, according to the Pythagorean theorem, we determine the length of the hypotenuse CB, which is the generatrix of the cone.
CB ^ 2 = CO ^ 2 + BO ^ 2 = 0.6 ^ 2 + (2 * √3) ^ 2 = 0.36 + 12 = 12.36.
CB = √12.36 dm.
Section ABC is an equilateral triangle, since AC = AC, as generators of the cone, and the angle C = 60.
Determine the area of an equilateral triangle.
S = BC ^ 2 * √3 / 4 = (√12.36) ^ 2 * √3 / 4 = 3.09 * √3 dm2.
Answer: The cross-sectional area is 3.09 * √3 dm2.
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