The diameter of the ball is 20. A plane is drawn through the end of the diameter at an angle

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.