A cone is inscribed in a regular quadrangular pyramid, the side of the base of which is 10 cm, and the flat angle at the top is 60 degrees. find the height of the cone.
Since the pyramid is regular, there is a square at its base, and the side faces are isosceles triangles. Let’s draw the heights МH and MK of the lateral faces of the prism, which are also generators of the inscribed cone.
The diameter of the circle at the base of the cone is equal to the side of the square at the base of the pyramid.
KN = AB = 10 cm.
OH = KM / 2 = 5 cm.
The KMH triangle is isosceles, then MO is the height and bisector of the triangle, and the angle MOH = 60/2 = 300.
In a right-angled triangle MOH tg30 = OH / MO.
MO = OH / tg30 = 5 / (1 / √3) = 5 * √3 cm.
Answer: The height of the cone is 5 * √3 cm.
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