Well Phil has become a lonesome person with no friends after having invested so much time on his record store. BUT then his great friend Scott from freshman years comes back home to San Francisco and they quickly reconnect through spontaneous adventures through great 'ole San Fran.
a) Law of Sines
Problem: Scott and Phil agreed to meet at the record store so Phil can show Scott what he's been up to in life. They realize they're 15 feet apart. Scott heads to the store at an angle of N25E while Phil is heading off at an angle of N55W. Considering that Phil is a slowpoke, how far is he from the record store?
b) Law of Cosines
Problem: Phil and Scott decide to go on a boat ride to relax and just ride the waves out after such a long day. Phil gets to the harbor before Scott however and told him to amount the SS Kirch at 5pm. At 5pm, Scott gets aboard the SS Cheesbucket. Phil's boat, the SS Kirch, is traveling at 80 mph and is at a bearing of 220* while Scott's boat is traveling at 115 mph at a bearing of 335*. It takes four hours for both "men" to realize they aren't aboard the same boat (they're huge ships and cell reception isn't exactly great out in the waters.) How far apart are the boats when Phil and Scott finally realize they're aboard different ships?
Now Scott's boat, the SS Cheesebucket is traveling at 115mph for 4 hours then it has traveled a distance of 460 miles. Phil's boat, SS Kirch travels at 80 mph for 4 hours so really has traveled 320 miles.
Our triangle is SAS, making sure our included angle is between two sides. Now that we have this information we can plug it into our equation of Law of Cosines, a^2 = b^2 + c^2 - 2bcCosA, which we plug straight into our calculator and then square root to find that the distance between Scott and Phil is a tragic 662.13 miles apart...someone better start swimming now.