Solution

    First, read the directions and draw the triangle that is formed. You depart from Camp Riffle, indicate this by drawing a point to represent Camp Riffle and make a N, S, E, & W for reference. From your point, rotate 350° NW and draw a line, route Wong, to represent 80 nautical miles (nm) and create another point. This point represents Fort Benzene. Once again, create a N, S, E, & W. From point For Benzene, rotate 280° and create a straight line, route Ginger, to represent 60 nm and make another point. This point represents Castle Trig. From Castle Trig create a straight line, route Pythagorean, back to Camp Riffle. You now have your triangle to work from.
    Now fill in what you know. You now that from Camp Riffle to For Benzene is 80 nm and is on a 350° course. You also know that from Fort Benzene to Castle Trig is 60 nm and is on a 280° course. And then, a straight line, which we don't know how long it is, back to Camp Riffle from Castle Trig.
    So, on the triangle you have an extra 10° from Camp Riffle to Fort Benzene to create 360°. Therefore, at Fort Benzene on the N, S axis, the y-axis, in the fourth quadrant is a little 10° angle in the big angle created by Camp Riffle to Fort Benzene to Castle Trig. Also, you know the third quadrant was not taken up at Fort Benzene, so you have a perfect 90° angle. So, so far we have 100° within the big angle. From Fort Benzene we made a 280° line extending through quadrant two. Therefore, you have an extra 80° not used to create the full 360°. This means you have an extra 10° in quadrant two that is not being used, which gives you the third angle you need to create the total of the obtuse angle. The obtuse angle, at point Fort Benzene, equals 110°.
    Now we can use the Law of Cosines to get the length of the third side of the triangle; opposite Fort Benzene's point. The Law of Cosines is c² = a² + b² - (2ab(Cos C)). The " c² " represents the length you wish to find; "a" and "b" represent the two lengths we already have. The "C" represents the angle opposite length "c". All capital letters represent angles and all lowercase letter represent lengths. In the case "C" equals 110°. Now put the formula into action.

c² = 80² + 60² - (2 (80) (60) (Cos 110°) )
c² = 6400 + 4600 - 9600 (Cos 110°)
c² = 10,000 - 9600 (Cos 110°)
c² = 13283.39338	*Don't forget to take the square root!*
c = 115.25
"c" equals your last side length.

Now that all three sides of your triangle are known, find the two missing angles of the triangle at points Camp Riffle and Castle Trig. To do this use the Law of Sines. The Law of Sines is this formula: Sin C/c = Sin B/b = Sin A/a. You only want to use two letters at one time and cross-multiply them. To use the Law of Sines one must have all the parts to one of the A, B, or C's. So, we have all of the C's. You just found out "c" equals 115.25 nm and "C" equals 110°. You also have to say which angle you want to find; the angle opposite Castle Trig or the angle opposite Camp Riffle. Use the angle opposite Camp Riffle first. This is how it would be done.

	*Cross-multiply*

115.25(Sin B) = 60(Sin 110°)	*Now solve for Sin B*



Sin B = .4892109089	*Take Sin inverse of decimal*

Sin B = 29.29°

Now, one can either add the two angles you have, 110° and 29.29°, together and subtract the sum from 180° to find the third angle, but to keep more accuracy, perform the Law of Sines to solve for the last angle.

	*Cross-multiply*


115.25 (Sin A) = 80(Sin110°)	*Now solve for Sin A*





Sin A = .6522812118	*Take Sin inverse of decimal*

Sin A = 40.71°

Now you have all the side lengths: 80 nm, 60 nm, and 15.15 nm.
You also have all the angles of the triangle: 110°, 29.29°, and 40.71°

In conclusion, you now have the distance it was from Castle Trig back to Camp Riffle; 115.25 nm, and on what degree they flew, 40.71 degrees. So, you have solved what needed to be solved. You can also find the total distance of the trip by adding 80 + 60 + 115.25 = 255.25 nm. Do you have enough fuel? If you burn 760 lbs. and hour and the trip is 2 hrs. & 10 minutes, with only 1650 lbs. of fuel, does it work? Here is the suggested way of solving:

2 (760) = 1520 for 2 hrs.
760/6 = 126.67 for 10 min.
1520 + 126.67 = 1646.67 lbs. used in flight

So...I would do the flight; it would be close, but doable.