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.