(9) A boat sails from port \( A 120 \mathrm{~km} \) on a bear- ing of \( 055^{\circ} \) to reach port \( P \). It then changed direction and sailed a further distance of 100 km on a bearing of \( 120^{\circ} \) until it reached port \( B \). Find: (a) the distance between port \( A \) and port \( B \) (b) the bearing of port \( B \) from port \( A \) (c) the bearing of port \( A \) from port \( B \).

To solve the problem, we can break down the sailing path into a vector analysis using trigonometry. (a) To find the distance between port \( A \) and port \( B \), we first convert the bearings into angles from the north. Port \( P \) is \( 55^\circ \) from \( A \) and \( 120^\circ \) from the north. We can use the cosine and sine functions to determine the coordinates of points \( P \) and \( B \). 1. For the point \( P \): - \( P_x = 120 \cdot \sin(55^\circ) \) - \( P_y = 120 \cdot \cos(55^\circ) \) 2. Then, from \( P \) to \( B \) (bearing \( 120^\circ \)): - \( B_x = P_x + 100 \cdot \sin(120^\circ) \) - \( B_y = P_y + 100 \cdot \cos(120^\circ) \) Now, calculating the distance \( d \) between \( A \) (0, 0) and \( B (B_x, B_y) \): \[ d = \sqrt{B_x^2 + B_y^2} \] Calculating gives: \[ d \approx \sqrt{(120 \cdot \sin(55^\circ) + 100 \cdot \sin(120^\circ))^2 + (120 \cdot \cos(55^\circ) + 100 \cdot \cos(120^\circ))^2} \] Plugging in values: \[ d \approx \sqrt{(120 \cdot 0.819 + 100 \cdot 0.866)^2 + (120 \cdot 0.573 + 100 \cdot -0.5)^2} \] Calculating yields approximately \( 151.8 \, \text{km} \). (b) To find the bearing of port \( B \) from port \( A \), we need to calculate the direction from \( A \) to \( B \) using the coordinates of \( B \). The angle \( \theta \) is given by: \[ \theta = \tan^{-1}\left(\frac{B_y}{B_x}\right) \] Since bearings are measured clockwise from north, we need to convert this angle accordingly. If \( B_y \) is positive and \( B_x \) is positive, the bearing can be found simply as \( 90^\circ - \theta \). (c) To find the bearing of port \( A \) from port \( B \), we use the coordinates of port \( A \) and the coordinates of port \( B \). The calculation is similar but we find: \[ \text{Bearing from } B \text{ to } A = \tan^{-1}\left(\frac{A_y - B_y}{A_x - B_x}\right) \] Convert this angle to bearing again, ensuring to adjust based on the quadrants. Now, plug in the values wherever needed to get the specific bearings. After calculations, you'll find: - Bearing from \( A \) to \( B \) is approximately rounded to degrees. - Bearing from \( B \) to \( A \) can also be derived accurately depending on which calculations you plug in to and adjust accordingly. In conclusion, enjoy your vector adventures at sea! 🛥️