1. **State the problem:** Jack is on top of a 65 m high cliff. He observes a man swimming out to sea at an angle of depression of 51° and a boat at an angle of depression of 30°. We need to find: a) Distance $x$ of the man from the base of the cliff. b) Distance $y$ of the boat from the base of the cliff. c) Distance from the man to the boat. 2. **Relevant formula and rules:** The angle of depression from the cliff top corresponds to the angle of elevation from the base looking up. We use the tangent function in right triangles: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ Here, opposite side is the height of the cliff (65 m), adjacent side is the horizontal distance ($x$ or $y$). 3. **Calculate $x$ (distance of the man):** $$\tan(51^\circ) = \frac{65}{x}$$ Rearranged: $$x = \frac{65}{\tan(51^\circ)}$$ Calculate $\tan(51^\circ)$: $$\tan(51^\circ) \approx 1.2349$$ So: $$x = \frac{65}{1.2349}$$ $$x \approx 52.62$$ Rounded to nearest metre: $$x = 53$$ 4. **Calculate $y$ (distance of the boat):** $$\tan(30^\circ) = \frac{65}{y}$$ Rearranged: $$y = \frac{65}{\tan(30^\circ)}$$ Calculate $\tan(30^\circ)$: $$\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.5774$$ So: $$y = \frac{65}{0.5774}$$ $$y \approx 112.58$$ Rounded to nearest metre: $$y = 113$$ 5. **Calculate distance from man to boat:** The man and boat lie on the horizontal line from the base of the cliff, so distance between them is: $$d = y - x = 113 - 53 = 60$$ 6. **Final answers:** a) Distance of the man from the base: $53$ m b) Distance of the boat from the base: $113$ m c) Distance from the man to the boat: $60$ m