Your formula is a "simple" rule of thumb based on trigonometry or use of Pythagorean Logic.
Basically, you need to draw a consistent triangle on the ground or on your piece of paper to do this "estimate".
The problem with WIND is that if you account for wind on one leg of your triangle then you would have to account for wind everywhere and simplicity just left the barn.
Let's say you are doing the old "turn perpendicular to a station and time 10 degrees thing". Your theorectical model is a triangle with a right angle where you made your turn perpendicular. (For those challenged - you are flying West with a VOR directly in front of you and no DME aboard - you turn North - 360 degrees and watch for the VOR needle to change from 270 TO to 260 TO while you time this ten degree change.)
Your triangle is defined by the points A, B and C where A is the VOR, B is the right angle where you made the turn and C is the acute angle back to the station made between heading North and the the 260 TO indication.
OK, so we can do all the math in a "NO WIND" situation where technically ground speed = true airspeed, right? Ah ha!
But if we introduce wind into the formula, how the heck do we do that? Let's say you are 3,000ft and the nice Winds Aloft forecast tells you that today winds are 190/32. OK, well when you turn North, you basically have a 30-31 kt tailwind and you are going much faster over the ground. On your turn West-Southwest back to the station you will have a crosswind with a little bit of headwind - you'll lose about 5-10 knots of groundspeed.
Without wind and in a 120kt airplane, let's say we fly the 10 degrees in 5 minutes and yield by your formula, 60NM or 30 minutes from the station.
With wind, we will fly the 10 degrees (at 150kts) in only 4 minutes. Now since we are aware of the tailwind, we do the math at 150kts and arrive with 60NM, but only 24 minutes to the station. (Note: the mileage came out right but the time is going to be way off.) When we turn back to the station in the wind, we will make about 115kts across the ground - instead of 24 minutes or even 30 minutes, the trip will take 31.3 minutes.
Finally, let's take wind and you are completely unaware of the winds. TAS is 120kts. So you compute 120kts in 4 minutes or 48 miles from the station and 24 minutes to get there. You will be off by 7 and a half minutes using the "simple" estimate.
So, what did we prove. Well, you were exactly right to question TAS vs Ground Speed. Since you are drawing a physical triangle on the ground with your airplane, GS is important to accuracy as long as you are measuring distance.
Now, how are you going to do ground speed when you are in the clouds, somewhat lost and without a DME? Are you going to:
A.) Read the dials and use TAS for a rough.
B.) Think, I have the WA forecast and will "fudge".
C.) I have the WA forecast and will do all the math.
D.) Ask ATC for your position in miles from the XYZ VOR.
For the written test, you will do C. In real life you will do D or A or you will do the old "cross radial" fix thing because you really don't want to be asking ATC for your own position.
PS. a 30 knot wind to 120 kt airplane is 25% of the plane's airspeed at max values. Notice your estimates were off by 25% when doing no-wind calculations.
Fly safe, fly smart.
