1. **Problem statement:** A boat travels at 4.2 m/s relative to the water, which flows at 2.8 m/s horizontally (to the right). We want to find the angle $\theta$ the boat must head upstream to reach a point directly across the river. 2. **Concepts and formula:** The boat's velocity relative to the ground is the vector sum of its velocity relative to the water and the river's velocity. To reach directly across, the horizontal components must cancel out, so the boat's velocity component in the river's direction must exactly oppose the river flow. 3. **Set up the velocity components:** - Let $\theta$ be the angle between the boat's velocity vector and the vertical (upstream direction). - The boat's velocity relative to water has components: - Horizontal (river direction): $4.2 \sin \theta$ - Vertical (across river): $4.2 \cos \theta$ 4. **Condition for reaching directly across:** The horizontal component of the boat's velocity must cancel the river velocity: $$4.2 \sin \theta = 2.8$$ 5. **Solve for $\theta$:** $$\sin \theta = \frac{2.8}{4.2} = \frac{2.8}{4.2} = 0.6667$$ 6. **Calculate $\theta$:** $$\theta = \sin^{-1}(0.6667) \approx 41.81^\circ$$ 7. **Answer:** The boat must head at approximately $42^\circ$ to reach directly across the river. **Correct choice:** C. 42°