Question 1181729
Here's the solution, broken down step by step:

**1. Visualize the Setup:** Imagine a sphere perfectly nestled inside a cone, with the sphere's diameter matching the cone's axis. The sphere touches the cone along a great circle.

**2. Key Geometric Relationships:**

*   Let *R* be the radius of the sphere.
*   The cone's height (h) is equal to the sphere's diameter, so h = 2R.
*   The slant height (s) of the cone is the distance from the apex to any point on the edge of the circular base.
*   The radius of the cone's base (r) can be found using the Pythagorean theorem.

**3. (a) Vertical Angle of the Cone:**

*   Consider a right triangle formed by the cone's height (2R), the cone's radius (r), and the cone's slant height (s).  Half of the cone's vertical angle (θ/2) is the angle between the slant height and the axis (height).
*   sin(θ/2) = r / s
*   Since the sphere touches the cone along a great circle, r = R.
*   s = sqrt(r² + h²) = sqrt(R² + (2R)²) = sqrt(5R²) = R√5
*   sin(θ/2) = R / (R√5) = 1/√5
*   θ/2 = arcsin(1/√5)
*   θ = 2 * arcsin(1/√5) ≈ 51.83°

**4. (b) Volume of the Cone:**

*   Volume of a cone = (1/3)πr²h
*   V_cone = (1/3)πR²(2R) = (2/3)πR³

**5. (c) Volume Common to the Two Solids:**

*   The volume common to both is simply the volume of the sphere.
*   V_common = (4/3)πR³

**6. (d) Volume of the Cone Outside the Sphere:**

*   V_outside_sphere = V_cone - V_common
*   V_outside_sphere = (2/3)πR³ - (4/3)πR³ = (-2/3)πR³

**Important Correction:** The volume of the cone *outside* the sphere should be a positive value. We made an error in the calculation. It should be:

V_outside_sphere = V_cone - V_common
V_outside_sphere = (2/3)πR³- (4/3)πR³