Convert everything to True and note that the aircraft follows a Track of 002ºT.

The calculated Heading and Groundspeed are 006ºTand 415 kts respectively.

At 1000 (Position A), the calculated heading = 006ºT.

To calculate True Bearing to PY:

HDG + RB = QUJ
006ºT + 311º = 317ºT

Remainder to True North = 360º - 317º = 043º

However, the aircraft's flight path will be along the track of 002ºT.

Therefore, the angle from the aircraft's track to the bearing to PY = 043º + 002º = 045º Left of track

At 1003 (Position B), the calculated heading is still 006ºT.

To calculate True Bearing to PY:

HDG + RB = QUJ
006ºT + 266º = 272ºT

Remainder to True North = 360º - 272º = 088º

However, the aircraft's flight path will be along the track of 002ºT.

Therefore, the angle from the aircraft's track to the bearing to PY = 088º + 002º = 090º Left of track

We now have a triangle A - B - PY with internal angles at A (045º) and B (090º), thus leaving the last angle at PY = 180º - 045º - 090º = 045º.

A triangle with 2 equal internal angles (A = 045º and PY = 045º) is an Isosceles Triangle, which will have two equal sides adjacent to the two equal angles. i.e Side A - B = Side PY - B.

Side A - B = 3 minutes @ 415 kts = 21 NM

As Side A - B = Side PY - B, the latter also equals 21 NM.