A spherical balloon subtends an angle α = ^ -1 ((7)/(9)) at the eye...

Let the radius of the balloon be $r$ and the distance from the observer's eye to the center of the balloon be $d$.

The balloon subtends an angle $\alpha$ at the observer's eye. From the geometry of tangents to a sphere, the right triangle formed by the observer's eye, the center of the balloon, and the point of tangency gives:
$\sin\left(\dfrac{\alpha}{2}\right) = \dfrac{r}{d}$

We are given $\cos\alpha = \dfrac{7}{9}$. Using the half-angle identity $\cos\alpha = 1 - 2\sin^2\left(\dfrac{\alpha}{2}\right)$:
$\dfrac{7}{9} = 1 - 2\sin^2\left(\dfrac{\alpha}{2}\right)$
$2\sin^2\left(\dfrac{\alpha}{2}\right) = \dfrac{2}{9} \implies \sin\left(\dfrac{\alpha}{2}\right) = \dfrac{1}{3}$

Thus, $d = \dfrac{r}{\sin(\alpha/2)} = 3r$.

Let $h_c$ be the height of the center of the balloon above the observer's eye level. Since the angle of elevation of the center is $30^\circ$:
$h_c = d \sin 30^\circ = (3r)\left(\dfrac{1}{2}\right) = 1.5r$

The height of the topmost point of the balloon from the observer's eye level is the height of the center plus the radius of the balloon:
$\text{Height of topmost point} = h_c + r = 1.5r + r = 2.5r$

Given that this height is $30$ m:
$2.5r = 30 \implies r = \dfrac{30}{2.5} = 12$ m.

Answer: $12$