A boat attempts to cross a river. The boat's speed with respect to the water is 12.0m/s . The speed of the river current with respect to the river bank is 6.0m/s At what angle should the boat be directed so that it crosses the river to a point directly across from its starting point? a) 45.0^circ b) 26.6^circ C) 30.0^circ d) 53.1^circ e) 60.0^circ

To solve this problem, we need to consider the relative velocity of the boat with respect to the water and the velocity of the river current with respect to the river bank. We want the boat to reach a point directly across the river, which means the component of the boat's velocity perpendicular to the river should cancel out the velocity of the river current.<br /><br />Let's denote the angle at which the boat should be directed as $\theta$. The velocity of the boat parallel to the river is $v_b \cdot \cos(\theta)$, and the velocity of the river current is $v_r = 6.0 \, \text{m/s}$. The velocity of the boat perpendicular to the river is $v_b \cdot \sin(\theta)$.<br /><br />For the boat to reach a point directly across the river, the component of the boat's velocity parallel to the river must equal the velocity of the river current:<br /><br />$v_b \cdot \cos(\theta) = v_r$<br /><br />Solving for $\theta$, we get:<br /><br />$\cos(\theta) = \frac{v_r}{v_b}$<br /><br />Substituting the given values, we have:<br /><br />$\cos(\theta) = \frac{6.0 \, \text{m/s}}{12.0 \, \text{m/s}} = 0.5$<br /><br />Taking the inverse cosine of both sides, we find:<br /><br />$\theta = \cos^{-1}(0.5) = 60.0^{\circ}$<br /><br />Therefore, the correct answer is:<br /><br />e) $60.0^{\circ}$