Sphere Inscribed In Frustum 2.

Cut the frustum and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the two bases.

Ignore the numbers in the diagram above.

Let $EC = w$ , $GB = 3w$ , and the radius of the circle $r = OE = OG = OF$ .

$\triangle{EOC} \cong \triangle{COF} \implies CF = w$ and $\triangle{FOB} \cong \triangle{BOG} \implies FB = 3w$ . From $C$ draw a perpendicular to base $AB$ at point $I$ , then $\triangle{CIB}$ is a right triangle, where $CI = 2r, IB = 2w$ and $CB = 4w \implies 4w^2 + 4r^2 = 16w^2 \implies r^{2} = 3 w^2 \implies r = \sqrt{3} w \implies \tan(\angle{CBI}) = \dfrac{2\sqrt{3} w}{2 w} = \sqrt{3} \implies \theta = 60^{\circ}.$ .