Total weight = 3400
Winch pull = 20° × 3400 ÷ 60 = 1134
Ground factor = (pavement) 3400 × 4% = 136
Winch load = 1134 + 136 = 1270

That looks better, but please note that the angle input is not proportionate to the actual forces. The usual test with a formula is to exaggerate a variable in both directions and see if the results make sense.
Here, a vertical pull would mean multiplying by 90, but 90 ÷ 60 isn't the vertical pull, it's much more.
I have a feeling that the variable you want is a trig function of the ramp angle, rather than the angle ÷ 60.
Example: with a 20° ramp, the variable should be .333 (20 ÷ 60). The sine of 20° is .342, pretty close (2.6% error).
Use 10°, should get .1666667 (10° ÷ 60). The sine is .174 (4.2% off)
Really steep ramps, like 45° and up, are where the angle/60 method begins to fail: 45 ÷ 60 = .75.
The sine of 45° is .707, a lower figure.
Using 60°: the above method says the load is 1.00 (same as vertical), but of course it isn't, it's probably the sine, .866.
Go to 90°: the sine of 90° is 1.0, so the calculation exactly equals a vertical pull (whereas the above method predicts 1.5).

If the winch is too small but has a long cable, just use a pulley (or more than 1). Running the line to a pulley at the load, then back to the winch almost doubles capacity (some minor loss through friction and cable bending). The cable length must be doubled, but of course if splicing it the added cable can safely be as small as 50% of the original rated load.