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 $\mathrm{N} 55^{\circ} \mathrm{E}$ ; the call to Station Baker indicates that the ship is located $\mathrm{S} 60^{\circ} \mathrm{E}$ . $\newline$ Use this information to answer the questions below. $\newline$ (a) How far is each station from the ship? $\newline$ The 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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Q. 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

\mathrm{N} 55^{\circ} \mathrm{E}

; the call to Station Baker indicates that the ship is located

\mathrm{S} 60^{\circ} \mathrm{E}

\newline

Use this information to answer the questions below.

\newline

(a) How far is each station from the ship?

\newline

The 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.)

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 $L$ (the distance between the stations), the distance from Station Able to the ship (let's call this $d_A$ ), and the distance from Station Baker to the ship (let's call this $d_B$ ). The angles opposite these sides are given by the SOS calls: $55$ degrees at Station Able and $60$ 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 $180$ degrees.
Angle Calculation: First, let's find the angle at the ship. The sum of angles in a triangle is $180$ degrees. We have two angles: $55$ degrees (from Station Able) and $60$ degrees (from Station Baker). The third angle is $180 - 55 - 60 = 65$ degrees.
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 $(d_A / \sin(60)) = (L / \sin(65)) = (d_B / \sin(55))$ .
Calculate $d_A$ : We can solve for $d_A$ using the Law of Sines. Rearranging the formula, we get $d_A = (L \cdot \sin(60^\circ)) / \sin(65^\circ)$ . Plugging in the values, we have $d_A = (160 \cdot \sin(60^\circ)) / \sin(65^\circ)$ .
Calculate $d_A$ : Now we calculate $d_A$ using a calculator. $\sin(60)$ is approximately $0.8660$ , and $\sin(65)$ is approximately $0.9063$ . So, $d_A \approx (160 \times 0.8660) / 0.9063 \approx 153.86$ miles.
Calculate $d_B$ : Next, we solve for $d_B$ using the Law of Sines. Rearranging the formula, we get $d_B = (L \cdot \sin(55^\circ)) / \sin(65^\circ)$ . Plugging in the values, we have $d_B = (160 \cdot \sin(55^\circ)) / \sin(65^\circ)$ .
Calculate $d_B$ : Next, we solve for $d_B$ using the Law of Sines. Rearranging the formula, we get $d_B = (L \cdot \sin(55)) / \sin(65)$ . Plugging in the values, we have $d_B = (160 \cdot \sin(55)) / \sin(65)$ .Now we calculate $d_B$ using a calculator. $\sin(55)$ is approximately $0.8192$ . So, $d_B \approx (160 \cdot 0.8192) / 0.9063 \approx 144.96$ miles.