Q: Trigonometry Bearings: Find distance between 2 points based on separate bearings Just finished a trig exam with the following problem and most everyone in the class arrived at a very different answer than me. The Professor and most of the class determined that to solve for side AB you would simply use the law of sines: 

Problem:
A land developer wants to find the distance across a small lake in the middle of his proposed development. The bearing from A to B is N15°W. The developer leaves point A and travels 66 yards perpendicular to AB to point C. The bearing from C to point B is N75°W. Determine the distance, AB, across the small lake. Round distance to nearest yard.
Class Solution:
(using law of sines)
step1: 66 yards / sin15° = AB / sin75°
step2: (66)(sin75°) / sin15° = AB
step3: 246.32 ≈ AB
My Confusion with this is that the angles are not directly stated in the problem, at least from my understanding of trig bearings. It seems like the Professor and the students assumed that (angle B = 15°) and since (angle A = 90°) then (angle C = 75°). 
So I created a little diagram using actual measured angles to figure out what the triangle looked like based on my understanding of bearings (for N15°W = start at North or 90° on a unit circle and move counter clockwise 15°) and from my diagram I determined that (angle C = 30°), (angle A = 90°), therefore (angle B = 60°). I ran through this problem multiple times and even looked at similar problems both in the book and online, and basically want to validate which answer is correct. Thanks for playing!
My solution:
(Using law of sines)
step1: 66 yards / sin60° = AB / sin30°
step2: (66)(sin30°) / sin60° = AB
step3: 38.11 ≈ AB
 A: I’m going to add a couple of points to your diagram for clarity:

Since $\angle{BAC}$ and $\angle{EAD}$ are both right angles, then $\angle{EAB}=\angle{DAC}$. The lines $\overline{AD}$ and $\overline{EC}$ are parallel, so $\angle{EAC}=\angle{DAC}=15°$. Finally, $\overline{EC}\perp\overline{DC}$, so $\angle{ECB}=90°-75°=15°$.
A: Your classmate (who wrote the figures linked in a comment) was correct that
$75 - 15 = 60.$ That's why angle $\angle ABC$, not angle $\angle ACB,$
is equal to $60$ degrees.
It follows that $\angle ACB = 30^\circ.$
One need only draw a scale diagram on graph paper with a ruler and protractor
(something your classmates and your instructor apparently could use some practice with)
to confirm these fact,
or one could simply chase angles around the figure and determine the angles correctly by arithmetic.
The angles in the equation
$$
\frac{66}{\sin 15^\circ} = \frac{AB}{\sin75^\circ}
$$
appear to be derived from the standard trigonometry technique learned by many students,
"Oh look, there's a number in the problem statement, it must plug directly into a formula I memorized by rote, let's put it in the 'nearest' place I can find."
What is a bit shocking is that the instructor fell into the same schoolchild error. It is even more shocking if the figure in the question is (as it appears to be) the actual figure that was presented as an illustration of the problem in the original exam materials.
Your solution is completely correct: the correct angle measurements and the correct application of the law of sines.
