An aircraft is over flying a VOR at 30.000 ft, at a groundspeed of 300 kt. The maximum time during which no usable signals will be received (in minutes and seconds) is:
As per ICAO, VOR shall provide signal up to a minimum elevation angle of 40 deg.
Point B is directly overhead VOR at height 30,000 ft.
So angle AOB = 90 - 40 = 50 deg
AB x 2 = total distance in which there will no usable signal.
AB = OB x tan 50 = 30,000 x tan 50 ft.
AB x 2 = 60,000 x tan 50 ft = (60,000 x tan 50) / 6080 nm = 11.76 nm
groundspeed = 300 kts
Time = 11.76 / 300 = 2 min 21 sec