The question asks >distance from A<.

TE=5x60/50=6deg

The question asks you to turn 12deg left at B. First 6deg will bring you parallel to the track, then PLUS another 6deg will bring you to the C. The second 6deg is Closing Angle.

You now have a triangle where both TE and CA are 6deg each. That means the distance from B to C is the same as distance from A to B.

The answer: distance from A to C is 50nm + 50nm = 100nm.

Time is irrelevant, not asked, but you can tell it will be another 18mins from B to C.

Hope above make sense.

Can I ask… this will only work if you have that on track fix? Like you needed to divert around weather?

[quote="MissSoph" post=13862]Can I ask… this will only work if you have that on track fix? Like you needed to divert around weather?

The principle is the same.
1. calculate drift (what winds have done to you). at this point in time you already know that you are off the track.
2. when you apply drift to your heading, it will bring you to the track [b][u]parallel[/u][/b] to original.

[attachment=2073]20211204_183533.png[/attachment]

3. to intercept orginal track you need to correct more, otherwise you will end up abeam (parallel) to your destination

For weather avoidance in real life, read paragraph "Other applications" on page 112 Cpl Navigation book (ed. 2015). Basically use same angles (from your DG) and same times (using your watch).

Hope this helps