The diameter of the ball is 20. A plane is drawn through the end of the diameter at an angle of 30 degrees to it. Find the cross-sectional area of the ball by this plane.
Consider a right-angled triangle ABC, in which the angle C = 90, since it is based on the diameter of the circle, the angle ABC = 30, and the hypotenuse AB is equal to the diameter of the ball AB = 20 cm.
Then Cosavs = CB / AB.
CB = AB * Cos30 = 20 * (√3 / 2) = 10 * √3 cm.
The section radius is O1B = CB / 2 = 5 * √3 cm.
Then the cross-sectional area will be equal to:
S = n * R ^ 2 = n * (5 * √3) ^ 2 = n * 75 cm2.
Answer: The length of the section line is n * 75 cm2.
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