11. A boat leaves the harbor and travels 30 miles in the direction of \( N 38^{\circ} \mathrm{W} \). The boat tums \( 90^{\circ} \) and then travels in the direction \( \$ 52^{\circ} \mathrm{W} \) for 12 miles. At that time, how far is the boat from the harbor and what is the bearing of the boat from the harbor entrance?

To solve this problem, we can break down the boat's journey using trigonometry to find the final position relative to the harbor. First, let's represent the initial leg of the journey. The boat travels 30 miles at a bearing of \( N 38^{\circ} W \). We can convert this to a Cartesian coordinate system, where North is the positive Y-axis and West is the negative X-axis. The coordinates can be calculated as follows: - \( x_1 = -30 \sin(38^{\circ}) \) - \( y_1 = 30 \cos(38^{\circ}) \) For the second leg, the boat turns \( 90^{\circ} \) to the right from \( N 38^{\circ} W \) (which gives a new direction of \( N 38^{\circ} E \)), and then travels \( 12 \) miles at a bearing of \( N 52^{\circ} W \). We need to adjust the direction before calculating the coordinates: - After the turn, the bearing of \( 52^{\circ} \) west changes it to \( N 52^{\circ} W = 90^{\circ} - 52^{\circ} = 38^{\circ} \). - New position: \( x_2 = -12 \sin(52^{\circ}) \) and \( y_2 = 12 \cos(52^{\circ}) \). Now we can add the coordinates of both legs: Final coordinates: - \( x_{final} = x_1 + x_2 \) - \( y_{final} = y_1 + y_2 \) To find the distance from the harbor, we can use the Pythagorean theorem: Distance \( D = \sqrt{x_{final}^2 + y_{final}^2} \) Finally, the bearing from the harbor can be calculated using the arctangent of the ratio of the coordinates: Bearing \( = \tan^{-1} \left(\frac{y_{final}}{-x_{final}}\right) \) You can substitute the values and calculate to find both the distance from the harbor and the bearing. Remember to adjust the final bearing based on the quadrant the boat is located in relative to the harbor. For more accurate results, using a calculator is highly recommended!