**Pre-Calculus Help!!! Bearings/Navigation Problem!!!!!?**

You are the navigator for a flight from Mpls./St. Paul to Orlando. When the flight takes off, you are sent to take care of a problem in the passenger compartment. When you return 20 minutes later (20 minutes after the take-off) you realize the pilot has been flying a a bearing of 143 degrees and should be flying at a bearing of 168 degrees. 1. If the plane travels at 300 mph and it is 1300 miles from Mpls./St. Paul to Orlando, at what bearing should the pilot turn to correct the course? - Ok to answer this problem do I just do 168 degrees - 143 degrees = 25 degrees and is that the final answer??? Or is there more since it gives the speed of the plane and the distance??? 2. At what speed should the plane increase to in order to arrive in Orlando on schedule? - I don't even understand how I am supposed to know this because I don't even have the time when the flight took off and when they are supposed to arrive in Orlando. Or is this information technically not needed?? 3. If the plane flies at 30,000 feet, how many ground miles from Orlando should the plane start its descent in order to make a landing at an angle of 3.8 degrees with the ground?? - You use trig for this right???? Like SohCahToa??? then for the 3.8 degrees, which side of the triangle would have that angle??? Please will anyone help me with this?!?!?!?! I'm not looking for full answers just tips and ideas or equations I should use in order to solve this. I would like to try to solve it a little bit by myself but I just have no idea of where I should begin. Please and thank you in advance!!! ^_^

Mathematics - 2 Answers

**Random Answers, Critics, Comments, Opinions :**

1 :

'Stefanie ' I will assume you are now studying vectors let V be the proper direction vector and D1(t) =300 t V..V = <cos 168Â° , sin 148Â° > and t = 1300/300 for correct flight time , ie D(13/3) puts you at Orlando 1st ; W = distance vector for 20 minutes ...W = [20/60] 300 < cos 143Â° , sin 143Â° >...compute ' - W + D(13/3) ' , all it B = < a,b> , | B | = new distance and ( #1)arctan [b / a] = new direction , ( #2) new speed = | B | / [ 13/3 - 1/3] (#3)..plane will bounce hard { landing at 300 mph !!! }.... the angle is between the fight path and the ground...and 30000ft will be side opposite the angle

2 :

1) a. First, draw a picture. Let A be the Mpls/StPl, B be where you currently are, and C be Orlando. This forms your triangle. AC = 1300 miles. b. Figure out how many degrees angle A is from the two bearings you were given, and mark this angle on the diagram. c. Using the distance formula, figure out how far AB is. Be sure and watch the units! d. You now have SAS, so use the Law of Cosines to find side BC. e. Use the Law of Sines to find angle B (be careful, this is an ambiguous case, so you will have two angles to choose from - choose wisely!) f. Add the appropriate bearing from the problem statement to angle B to get the new bearing. 2) You don't need the actual time of takeoff, only the time you have travelled already. a. Use the distance formula (d = rt) to determine how long the trip would take if they had headed the right direction in the first place. b. Take the time from (a) and subtract off the time already travelled to get the time remaining. Watch your units! c. Use the time from (b) and the distance you need to travel to get to Orlando from where you currently are to get the new speed. 3) SohCahToa? Yes. They gave you the height (30,000 feet) and the angle of elevation (3.8 deg). You need to calculate the distance along the ground. Start with a sketch. Draw a right triangle with the plane at A, the airport at B, and the right angle at C. Angle B is 3.8 degrees, and AC = 3000 feet. Sounds like you can probably take it from there. Again, watch your units! Good luck!