At 0020 UTC an aircraft is crossing the 310 radial at 40 DME.
At 0035 UTC the a/c crosses radial 040 at 40 DME.
Variation Zero, Find True track & Groundspeed
A) 085 - 226 kts B) 090 - 232 kts C) 080 - 226 kts D) 088 – 232 kts
At 0020 a/c is at B and at 0035 reaches A.
Angular difference between radial 310 and 040 = (360-310) + 040 = 90 deg.
In triangle AOB
OB = OA = 40 nm.
angle AOB = 90 deg.
Therefore it is a 45-45-90 triangle.
angle OBA = OAB = 45 deg.
Variation = 0 so true track = magnetic track, QDM = QUJ
From B if you want to go to the station O you will fly heading (QDM) 310-180 = 130.
Track to find (track BA ) = QUJ(QDM) - angle OBA = 130 - 45 = 085.
OB^2 + OA^2 = BA^2
40*40 + 40*40 = BA^2
dist BA = 56.5 nm
time = 0035 - 0020 = 15 min
ground speed = 56.5 nm / 15 min = 226 kts.
Ans A) 085 - 226 kts