Homework Question 7, 9.7.27 HW Score: \( 83.33 \% \), 8.33 of 10 points Points: 0 of 1 Savo Two ships leave a harbor at the same time. One ship travels on a bearing \( \mathrm{S} 12^{\circ} \mathrm{W} \) at 17 miles per hour. The other ship travels on a bearing \( \mathrm{N} 75^{\circ} \mathrm{E} \) at 10 miles per hour. How tar apart will the ships be after 2 hours? The distance is approximately \( \square \) miles. (Round to the nearest tenth as needed.)
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The Deep Dive
To find the distance between the two ships after 2 hours, we first determine their positions after that time. The first ship travels \( 17 \text{ miles/hour} \times 2 \text{ hours} = 34 \text{ miles} \) on a bearing \( S 12^\circ W \), and the second ship travels \( 10 \text{ miles/hour} \times 2 \text{ hours} = 20 \text{ miles} \) on a bearing \( N 75^\circ E \). Next, let's visualize their positions as points on a coordinate plane. For the first ship, the direction implies that it's traveling southwest. We'll convert the bearing into standard angles, where \( S 12^\circ W \) corresponds to \( 180^\circ + 12^\circ = 192^\circ \), and for the second ship, \( N 75^\circ E \) translates to \( 75^\circ \). Using trigonometry, we can find the coordinates: - For the first ship: - \( x_1 = 34 \cos(192^\circ) \) - \( y_1 = 34 \sin(192^\circ) \) - For the second ship: - \( x_2 = 20 \cos(75^\circ) \) - \( y_2 = 20 \sin(75^\circ) \) Calculating these gives us: - \( x_1 \approx -34 \cdot 0.974 = -33.116 \) (approx), \( y_1 \approx -34 \cdot 0.207 = -7.038 \) (approx) - \( x_2 \approx 20 \cdot 0.258 = 5.16 \) (approx), \( y_2 \approx 20 \cdot 0.966 = 19.32 \) (approx) Now, to find the distance between the two ships, we can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting: \[ d = \sqrt{(5.16 - (-33.116))^2 + (19.32 - (-7.038))^2} \] Calculating inside the square root gives: \[ = \sqrt{(38.276)^2 + (26.358)^2} \approx \sqrt{1463.072 + 694.819} \approx \sqrt{2157.891} \approx 46.5. \] Thus, after rounding to the nearest tenth, the distance between the two ships after 2 hours is approximately 46.5 miles.
