For any heights and distances problem where the angles of elevation from two points at distances \(a\) and \(b\) from the base of a tower are complementary, the height of the tower is always given by:
\[ h = \sqrt{ab} \]
Here, \(a = 7\) and \(b = 16\), so \(h = \sqrt{7 \times 16} = 4\sqrt{7}\text{ m}\) instantly.
Step 1: Recall a neat property of this classic complementary-angle setup. Whenever a tower's angle of elevation from two points on the same side of its base are complementary, the height of the tower turns out to be the geometric mean of the two ground distances, that is $h^2 = CD \times BD$. This comes directly from the fact that $\tan\theta$ and $\tan(90^\circ-\theta) = \cot\theta$ multiply to give 1. Step 2: Identify the distances. Here $CD = 7$ m and $BD = CD + CB = 7 + 9 = 16$ m. Step 3: Apply the geometric mean relation. \[ h^2 = CD \times BD = 7 \times 16 = 112 \] Step 4: Take the square root. \[ h = \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \text{ m} \]
\[ \boxed{4\sqrt{7} \text{ m}} \]
