At 0020 UTC an aircraft is crossing the 300° radial at 50 NM of a VOR/DME station. At 0040 UTC the radial is 050° and DME distance is 50 NM. Magnetic variation is zero. The true track and ground speed are:
ans is 212.132 kts and 85 deg T
I understood the 85 deg part but Shabez how did you get 212.132 kts..
This is the diagram I came up with
Now since Triangle ABD is congruent to Triangle ADC,
BD = DC
cos 35 = BD/50 => BD = cos35*x 50 = 40.95 nm
=> DC =40.95
=> BC which is the distance travelled by the A/C in 20 minutes is 40.95 + 40.95 = 81.9 nm
Therefore speed is 81.9/0:20 = 245.74
I think you mistook the Angle BAC as 90 deg when it is actually 110 (60 + 50)