A radio antenna stands on the edge of a cliff, at a height h = 120 m above a calm sea, and points…

Step 1. Path difference (a). The antenna and its image are $2h$ apart, on one vertical line. Parallel rays at elevation $\alpha$ reach them with a path difference $\Delta = 2h\sin\alpha$.

Step 2. Conditions. The wavelength is $\lambda = c/f = 1.50\ \text{m}$. With the phase reversal, maxima need $2h\sin\alpha = \left(m + \tfrac12\right)\lambda$ and minima need $2h\sin\alpha = m\lambda$. On the horizon itself the signal is zero.

Step 3. First maximum and next minimum (b, c). $\sin\alpha_1 = \dfrac{\lambda}{4h} = 0.003125$, so $\alpha_1 = 0.179^\circ$. The next minimum has $\sin\alpha = \dfrac{\lambda}{2h}$, so $\alpha = 0.358^\circ$.

Step 4. Time (d). The second maximum has $\sin\alpha_2 = \dfrac{3\lambda}{4h}$, so $\alpha_2 = 0.537^\circ$. At $0.25^\circ$ per minute, the source takes $\dfrac{0.537^\circ - 0.179^\circ}{0.25^\circ\ \text{per min}} \times 60\ \text{s per min} = 85.9\ \text{s}$.

Step 5. Counting (e). At $5.0^\circ$, $\dfrac{2h\sin\alpha}{\lambda} = 13.94$. The maxima are at $m + \tfrac12 = 0.5, 1.5, \ldots, 13.5$, so there are $14$.

Step 6. Both together (f). At $120\ \text{MHz}$, $\lambda' = 2.50\ \text{m}$. Both are at a maximum where $2h\sin\alpha$ is an odd number of half-wavelengths of each: $5 \times \dfrac{1.50\ \text{m}}{2} = 3 \times \dfrac{2.50\ \text{m}}{2}$. So $\sin\alpha = \dfrac{5\lambda}{4h}$ and $\alpha = 0.895^\circ$.