Question 10 of 20 This test 20 point(g) possible thil question: 1 point() possible Submiltest Two ships leave a port at the same time. The first ship sails on a bearing of \( 32^{\circ} \) at 18 knots (nautical miles per hour) and the second on a bearing of \( 122^{\circ} \) at 26 knots . How far apart are they after 1.5 hours? (Neglect the curvature of the earth.) After 1.5 hours, the ships are approximately \( \square \) nautical miles apart. (Round to the nearest nautical mile as needed.)

To find out how far apart the two ships are after 1.5 hours, we begin by determining the distance each ship travels. The first ship covers \( 18 \text{ knots} \times 1.5 \text{ hours} = 27 \text{ nautical miles} \), while the second ship travels \( 26 \text{ knots} \times 1.5 \text{ hours} = 39 \text{ nautical miles} \). Next, we can break down the movements of the ships into their respective components using trigonometry. For the first ship on a \( 32^{\circ} \) bearing, its components can be calculated as: - \( x_1 = 27 \cos(32^{\circ}) \) - \( y_1 = 27 \sin(32^{\circ}) \) For the second ship on a \( 122^{\circ} \) bearing, the components are: - \( x_2 = 39 \cos(122^{\circ}) \) - \( y_2 = 39 \sin(122^{\circ}) \) Now, calculating these components would yield: - Ship 1: \( x_1 \approx 23.0, y_1 \approx 14.7 \) - Ship 2: \( x_2 \approx -11.9, y_2 \approx 34.1 \) The distance between the two ships can then be determined using the formula: \[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \] Plugging in the values: \[ d \approx \sqrt{(23.0 + 11.9)^2 + (14.7 - 34.1)^2} \approx \sqrt{(34.9)^2 + (-19.4)^2} \] Calculating further: - \( (34.9)^2 \approx 1218.01 \) - \( (-19.4)^2 \approx 376.36 \) Thus, \[ d \approx \sqrt{1218.01 + 376.36} \approx \sqrt{1594.37} \approx 39.9 \text{ nautical miles} \] Rounding to the nearest nautical mile, the ships are approximately **40 nautical miles** apart after 1.5 hours.