Originally Posted by
rpenmanparker
That looks completely correct. Is it not true that the trig approach I used inherently incorporates the formula you provided? You're right about the lack of need for using the STA, but I wonder if the use of it, just as an approximation doesn't simplify the whole process.
Yes you could solve with angles. I wouldn't because you don't care what the starting or ending angle are, but you will need to solve them, and use them in a subsequent equation. They are intermediate calculations to move from ∆v to ∆h. If you can make a formula that uses only ∆v &∆h, why not solve direct?