1. **Problem statement:** Alden is on a yacht 15 miles north of an east-west coastline. The hospital is 60 miles east along the coast from the point closest to the yacht. A motorboat travels at 20 mph from the yacht to a point $x$ miles east on the shore, where an ambulance traveling at 90 mph takes Alden to the hospital. We want to find the angle the motorboat should head to minimize total travel time. 2. **Set up variables and equations:** - Let $x$ be the distance east along the shore from the point closest to the yacht where the motorboat lands. - The motorboat travels from $(0,15)$ to $(x,0)$, so distance is $$\sqrt{x^2 + 15^2} = \sqrt{x^2 + 225}.$$ - The ambulance travels from $(x,0)$ to the hospital at $(60,0)$, so distance is $$60 - x.$$ - Motorboat speed $v_m = 20$ mph, ambulance speed $v_a = 90$ mph. 3. **Total travel time $T(x)$:** $$ T(x) = \frac{\sqrt{x^2 + 225}}{20} + \frac{60 - x}{90}. $$ 4. **Minimize $T(x)$:** Take derivative with respect to $x$: $$ T'(x) = \frac{1}{20} \cdot \frac{x}{\sqrt{x^2 + 225}} - \frac{1}{90}. $$ Set $T'(x) = 0$: $$ \frac{x}{20 \sqrt{x^2 + 225}} = \frac{1}{90}. $$ 5. **Solve for $x$:** Multiply both sides by $20 \sqrt{x^2 + 225}$: $$ x = \frac{20}{90} \sqrt{x^2 + 225} = \frac{2}{9} \sqrt{x^2 + 225}. $$ Square both sides: $$ x^2 = \frac{4}{81} (x^2 + 225). $$ Multiply both sides by 81: $$ 81 x^2 = 4 x^2 + 900. $$ Bring terms to one side: $$ 81 x^2 - 4 x^2 = 900 \implies 77 x^2 = 900. $$ Divide: $$ x^2 = \frac{900}{77}. $$ Take positive root: $$ x = \frac{30}{\sqrt{77}}. $$ 6. **Calculate the angle $\theta$ the motorboat should head:** The motorboat heads from $(0,15)$ to $(x,0)$, so the angle below the horizontal is: $$ \theta = \arctan\left(\frac{15}{x}\right) = \arctan\left(\frac{15}{30/\sqrt{77}}\right) = \arctan\left(\frac{15 \sqrt{77}}{30}\right) = \arctan\left(\frac{\sqrt{77}}{2}\right). $$ 7. **Final answer:** The motorboat should head at an angle $$ \boxed{\theta = \arctan\left(\frac{\sqrt{77}}{2}\right)} $$ north of east to minimize total travel time.