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Coast Guard Station Able is located L =160 miles due south of Station Baker. A ship at sea sends an SOS call that is received by each station. The call to Station Able indicates that the ship is located N55^(@)E; the call to Station Baker indicates that the ship is located S60^(@)E. Use this information to answer the questions below. (a) How far is each station from the ship? The distance from Station Able to the ship is ◻ miles. (Do not round until the final answer. Then round to two decimal places as needed.)

Coast Guard Station Able is located L =160 =160 miles due south of Station Baker. A ship at sea sends an SOS call that is received by each station. The call to Station Able indicates that the ship is located N55∘E \mathrm{N} 55^{\circ} \mathrm{E} ; the call to Station Baker indicates that the ship is located S60∘E \mathrm{S} 60^{\circ} \mathrm{E} .\newlineUse this information to answer the questions below.\newline(a) How far is each station from the ship?\newlineThe distance from Station Able to the ship is □ \square miles.\newline(Do not round until the final answer. Then round to two decimal places as needed.)

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Full solution

Q. Coast Guard Station Able is located L =160 =160 miles due south of Station Baker. A ship at sea sends an SOS call that is received by each station. The call to Station Able indicates that the ship is located N55∘E \mathrm{N} 55^{\circ} \mathrm{E} ; the call to Station Baker indicates that the ship is located S60∘E \mathrm{S} 60^{\circ} \mathrm{E} .\newlineUse this information to answer the questions below.\newline(a) How far is each station from the ship?\newlineThe distance from Station Able to the ship is □ \square miles.\newline(Do not round until the final answer. Then round to two decimal places as needed.)
  1. Triangle Formation: To solve this problem, we can use the Law of Sines in a triangle formed by the two stations and the ship. The triangle has sides of length LL (the distance between the stations), the distance from Station Able to the ship (let's call this dAd_A), and the distance from Station Baker to the ship (let's call this dBd_B). The angles opposite these sides are given by the SOS calls: 5555 degrees at Station Able and 6060 degrees at Station Baker. The angle at the ship is the remaining angle in the triangle, which we can find by subtracting the other two angles from 180180 degrees.
  2. Angle Calculation: First, let's find the angle at the ship. The sum of angles in a triangle is 180180 degrees. We have two angles: 5555 degrees (from Station Able) and 6060 degrees (from Station Baker). The third angle is 180−55−60=65180 - 55 - 60 = 65 degrees.
  3. Law of Sines Setup: Now we can set up the Law of Sines. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. In formula terms, this is written as (dA/sin⁡(60))=(L/sin⁡(65))=(dB/sin⁡(55))(d_A / \sin(60)) = (L / \sin(65)) = (d_B / \sin(55)).
  4. Calculate dAd_A: We can solve for dAd_A using the Law of Sines. Rearranging the formula, we get dA=(L⋅sin⁡(60∘))/sin⁡(65∘)d_A = (L \cdot \sin(60^\circ)) / \sin(65^\circ). Plugging in the values, we have dA=(160⋅sin⁡(60∘))/sin⁡(65∘)d_A = (160 \cdot \sin(60^\circ)) / \sin(65^\circ).
  5. Calculate dAd_A: Now we calculate dAd_A using a calculator. sin⁡(60)\sin(60) is approximately 0.86600.8660, and sin⁡(65)\sin(65) is approximately 0.90630.9063. So, dA≈(160×0.8660)/0.9063≈153.86d_A \approx (160 \times 0.8660) / 0.9063 \approx 153.86 miles.
  6. Calculate dBd_B: Next, we solve for dBd_B using the Law of Sines. Rearranging the formula, we get dB=(L⋅sin⁡(55∘))/sin⁡(65∘)d_B = (L \cdot \sin(55^\circ)) / \sin(65^\circ). Plugging in the values, we have dB=(160⋅sin⁡(55∘))/sin⁡(65∘)d_B = (160 \cdot \sin(55^\circ)) / \sin(65^\circ).
  7. Calculate dBd_B: Next, we solve for dBd_B using the Law of Sines. Rearranging the formula, we get dB=(L⋅sin⁡(55))/sin⁡(65)d_B = (L \cdot \sin(55)) / \sin(65). Plugging in the values, we have dB=(160⋅sin⁡(55))/sin⁡(65)d_B = (160 \cdot \sin(55)) / \sin(65).Now we calculate dBd_B using a calculator. sin⁡(55)\sin(55) is approximately 0.81920.8192. So, dB≈(160⋅0.8192)/0.9063≈144.96d_B \approx (160 \cdot 0.8192) / 0.9063 \approx 144.96 miles.

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