The diameter of the ball is 4 m. Through the end of the diameter, a plane is drawn at an angle of 3 degrees

| February 7, 2021 | education

The diameter of the ball is 4 m. Through the end of the diameter, a plane is drawn at an angle of 3 degrees to it. Find the cross-sectional area of the ball by this plane.

Let’s build the radius of the OB. The AOB triangle is isosceles, since OA = OB = R.

The angle ОАВ, by condition, is equal to 30, then the angle ОВА = ОАВ = 30, the angle АВ = (180 – 30 – 30) = 120.

ОА = ОВ = D / 2 = 4/2 = 2 m.

In an isosceles triangle AOB, according to the cosine theorem: AB ^ 2 = OA ^ 2 + OB ^ 2 – 2 * OA * OB * Cos120 = 4 + 4 – 2 * 2 * 2 * (-1/2) = 8 + 4 = 12.

AB = 2 * √3 m.

AB is the diameter of the section, then its radius is: r = AB / 2 = 2 * √3 / 2 = √3 m.

Then the cross-sectional area is: Ssection = π * r ^ 2 = 3 * π m2.

Answer: The sectional area is 3 * π m2.



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