What is the angle made by the line joining the centre of the sphere and any point on the rim of the circular opening with a vertical line passing through the centre?

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

From the previous calculation, the radius of the circular opening (r) = 10\(\sqrt{3}\) cm

From the previous calculation, the perpendicular distance from the center of the sphere to the plane of the circular opening (d) = 10 cm

Calculations:

The perpendicular distance from the center of the sphere to the plane of the circular opening (d) as the adjacent side to the angle θ (this is the vertical line segment from the center to the center of the opening).

The inner radius of the circular opening (r) as the opposite side to the angle θ.

We can use trigonometric ratios:

cos(θ) = Adjacent / Hypotenuse = d / R

cos(θ) = d / R = 10 cm / 20 cm = 1/2

The angle whose cosine is 1/2 is 60°.

⇒ θ = 60° = π/3

∴ The angle made by the line joining the center of the sphere and any point on the rim of the circular opening with a vertical line passing through the center is 60°.