Are you saying you knew all along that ∠CAB = ∠BAO, or that you just noticed it in the answer?

I should have, but I didn't. I did

*observe* that ∠CAB = ∠BAO but did not initially appreciate its significance. I calculated ∠ABO and ∠BAO after first determining the distance between the two points with Law of Cosines. Then, I determined ∠ABC as the supplement of ∠ABO and that's where I got stuck. I did not accept your suggestion that all that was needed to solve the two remaining angles was the ratio 12:16. I convinced myself that I could place ∠ACB anywhere along the line BC, but it would not align with AC without taking time into consideration. In other words, I missed the significance of proportionality.

To solve the time element I resorted to my calculator's solve function to find its limit,

**and therein noted ∠CAB = ∠BAO**. The penny still didn't drop, so I had to resort to a thought experiment. I drew two vectors at an arbitrary angle, as 12/

__0__ and 16/ ≈

__30__ and connected their endpoints. I then multiplied both magnitudes by 4 and again drew a line connecting their endpoints. I noted these two lines were parallel, and then the light bulb lit up, and I knew you were right, that the problem could be solved by only knowing the ratios 12:16 and 36:48, but again,

**as long** **as both ships maintained original course and speed**. I also completely solved two other similar problems with different speeds and interior angles with my calculator and again noted the equality ∠CAB = ∠BAO, but again, without resorting to, or being aware of the theorem.

So, a navigator well versed in trig and geometry could answer this and another interesting question almost instantly, or at least, it would appear to be such to observers on the bridge, and get himself out of hack. You said, "At what speed, sir?" The first answer would be, "if we maintain current speed, the course change is 335 true and the ETA is 5 hrs 34 mins. Now if we increase speed to say 20 knots (knots is correct but miles was used in the original statement of the problem,) then the course is 328 true with ETA 3 hrs 20 mins."

sin-1 ( (12 sin 106) / 20) = 35.222°

360 - (180 - (67 + 46.1021 + 35.222)) ≈ 328° true.

ETA is:

[ 43.3 sin (106) ÷ 20 sin (73.8979-35.222) ] = 3.330 = 3 hrs 20 mins.

The ratio would now be 12:20 and changing one number, 16 to 20 would quickly result in the value for ∠CAB, namely 35.222°. The calculator is no longer needed, but could still be used as fail-safe, to verify the math.

I intuited there was a principle involved here, as my first expression of new course to target was 180 - (67 + 2(46.1021)) = 25.2042°, but then I changed that to 180 - (67 + 46.1021 + 46.1021) = 25.2042° to make it more general, in anticipation that a speed change might be requested. You gave this notion a name,

*angle bisector theorem*, which I may have studied many years ago in HS geometry (I minored in math at U of I Chicago but that was class of '71, eons ago), but that was so long ago I'd forgotten it. Thanks for naming it, since that saved me a google search, out of native curiosity.