The angle at the apex of the axial section of a cone with a height of 1m is equal to 120 degrees, which is equal to the cross-sectional area of the cone drawn through 2 generatrices, the angle between which is 60 degrees.
The axial section of the cone is an isosceles triangle ABC, in which BA = BC as generators of the cone.
The height of the OB of the cone is the height, bisector and median of the triangle ABC, then the segment OA = OC, and the angle ABO = CBO = ABC / 2 = 120/2 = 60.
In the right-angled triangle ABO, we determine the length of the hypotenuse AB.
CosABO = BO / AO.
AO = BО / Cos60 = 1 / (1/2) = 2 m.
The CME section is also an isosceles triangle with lateral sides BK = BM = 2 m and an angle of 60 between them.
Determine the cross-sectional area of the CME.
Sqm = BK * BM * Sin60 / 2 = 2 * 2 * √3 / 2/2 = √3 m2.
Answer: The cross-sectional area is √3 m2.
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