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
Best Answer
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 $\angle{ECB}=90°-75°=15°$.