2. Coast Guard Station A is located 120 miles due west of Station B. A ship at sea sends an SOS call that is received by each station. The call to Station A indicates that the location of the ship is N 40 degrees E of Station A and the call to Station B indicated that the ship is N 30 degrees W of Station B. How far is each station from the ship?

3. To start solving this problem first you need a picture or diagram visibly showing what is going on so you can see what you have to start your solution. First the problem stated that Station A is 120 miles from station B. This is shown in start of my diagram below. B stands for station B while A stands for station A. The line connecting them stands for the 120 mile distance between them. B is to the left on my diagram because the problem stated that B is due west of A. 120

4. Next the problem stated that the call to station A was N 40 degrees E of Station B and the call to station B was N 30 degrees W of Station B. In my diagram below I made the point C be the point of the call and drew lines that represented a N 40 degrees E from Station A and a N 30 degrees W from station B. 120 40 30

5. Now you can see that there are two angles known on the diagram which means we can figure out the third angle. You can do this my adding the two known angles up and subtracting that from 18o. (You subtract from 180 because all three angles of a triangle should always add up to 180 no matter what.) 40+30=70 180-70=110 Then add the new angle to the diagram. 120 4030 110

6. Since the diagram has everything we know we can start solving for the two missing sides. When we solve these two sides we will solve the problem because the two sides a and b represent the distance from the call to station A and the distance from the call to station B respectively. 120 4030 110

7. First we will find the side a or the distance form station A to the call. We can find this by using the Law of Sines. We can use the Law of Sines because we know two angles and a side and one of the angles (angle C) matches up with the side (side AB). You set up the problem as side c over the sine of theta C equals the side a over the sin of theta C or or in our case = ca sinCsinA = 120 sin110 sin30 a

8. Then you do the same for the other side and set it up the same way so you end up with and the one the we made 120 sin110 = b sin40 120 sin110 = a sin30 Now we need solve for a and b which will give the us the distance from the call each one are. To do this you divide 120 by the sin110. Then you times that by the sin40 or on the other one the sin30. 120 sin110 = a sin30 x sin30 x 63.9 = a 120 = b sin110 sin40 x sin40sin40 x 82.1 = b

9. 120 miles 30 110 40 Then you now know that a=63.9 and that b=82.1. This means that the distance from station A to the call was 63.9 miles, and the distance from station B to the call was 82.1 miles. This is the answer to the problem. 63.9 miles 82.1 miles