To solve this problem, we need to determine the angle at which the swimmer should swim to cross the river along the shortest path.$\\$Given:$\\$• Speed of the swimmer in still water $(v_s) = 20$ m/s$\\$• Speed of the river $(v_r) = 10$ m/s$$ (flowing due east)$\\$The shortest path across the river is the one where the swimmer's resultant velocity is perpendicular to the river banks.$\\$Let's denote:$\\$• $\theta$ as the angle the swimmer should make with the north direction (upstream).$\\$To achieve the shortest path, the swimmer's velocity component perpendicular to the river flow should cancel out the river's velocity.$\\$The swimmer's velocity can be decomposed into two components:$\\$• $v_{s\_north} = v_s \cos(\theta)$ (northward component)$\\$• $v_{s\_east} = v_s \sin(\theta)$ (eastward component)$\\$For the swimmer to move directly across the river, the eastward component of his velocity should cancel the river's velocity:$\\ v_{s\_east} = v_r \\$Substitute the known values:$\\ v_s \sin(\theta) = v_r \\ 20 \sin(\theta) = 10 \\$Solve for $\sin(\theta): \\ \sin(\theta) = \frac{10}{20} \\ \sin(\theta) = \frac{1}{2} \\$The angle $\theta$ for which $\sin(\theta) = \frac{1}{2}$ is $30^\circ. \\$Since the swimmer needs to swim west of north to counteract the eastward flow of the river, the angle is $30^\circ$ west.$\\$Therefore, the correct option is Option 1: $30^\circ$ west.