Speaker height...should known better

Good idea on the tilt. How would figure out how much tilt (in degrees) to accomplish a 2 to 3 in difference at the listening position? 

Not to long ago, I had a pair of Vandersteen 2ci and required certain amount of tilt for them only specifically for the Vandys. Can I apply the same concept to the Triangles?

Thanks in advance 

You could always put a ribbon tweeter on top.parts express 50$each .you can adjust it to your seating position. This is why I like line arrays cuz no matter where you are ,sit or stand there always there. Enjoy the hunt and the experience and the music

It seems like he’s sitting in nearfield. He can tilt it, turn it upside down, do a 360 with it,  whatever, doesn’t take away from the fact it is a terrible design.....letting that very mediocre woofer get all the way down to 70, 60 hz and all the way up to 3600hz is just a bad idea....a lot got compromised w.r.t achievable fidelity. He may salvage something by bass managing, preventing that woofer from moving too much, rolling it off around 120hz to a subwoofer pair (needs to be a pair to help prevent localization with a high sub crossover of 120-ish hz)...still a terrible li’l box.

On the other hand, a Andrew Jones speaker (suggested as a replacement earlier) with a point source concentric driver, 4 inch mid covering 260 to 1800hz before hand off to the tweeter....A woofer taking over 260hz and under kept that mid driver very stress free.... Especially for near field, it seems like a very good idea, doesn’t it? point source n all (vd not a concern)....

I am sure this goofy triangle made all the rounds with the youtube reviewer promo club (the new apparently exciting kid on the block) and made a bunch of guys hop on it. 

And you got your degree in acoustical engineering where? 

I appreciate all the help. I did some mathematical calculation to figure out the angle tilt figured out I need a 3 deg angle at a distance of 5 ft. This correlates to a 3 inch increase at the listening position. Will see how this works out. 

Um maybe this one? A2 + B2 = C2 (pythagorean theorem). You would need the √C to find the hypotenuse. Provided you’re looking at a perfect isosceles triangle otherwise it would require a few more calculations.