Draw a diagram.
Consider the triangle
made up by the VOR, the aircraft's 0020 position and the aircraft's 0035
position.
The angle at the VOR =
50º + 40º = 90º.
Since the two sides
VOR-0020 and VOR-0035 are both 40nm it is an isosceles triangle and the other
two angles are 45º each. At 0020 the bearing from the aircraft to the VOR is
310º - 180º = 130º so the aircraft's track is 130º - 45º = 085º. Since the four
answers are 085º, 090º, 080º and 088º it must be 085º - 226 kt.
There is absolutely no
need to work out the groundspeed.
You can work out the
ground speed by considering the VOR-0020-0035 triangle again.
It is a right angle
triangle and we can work out the 0020-0035 distance by trigonometry = 40/sin45º
= 56.6nm.
This could also be done
using Pythagoras.
This distance has been
covered in 15 minutes so the groundspeed is 226 kts.
The problem is that we
still have two possible answers and still have to work out the track angle.
Top tip on a question
like this is to calculate the unique value (in this case track) and not the
duplicated value (in this case groundspeed) because it will save time.