A couple of years ago I here gave 2 (cos(c) + cos(b) + cos(a) + 1) acos(-------------------------------------- - 1). (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) Duh, cos(c) + cos(b) + cos(a) + 1 2 acos(----------------------------). a b c 4 cos(-) cos(-) cos(-) 2 2 2 Imaginary result means you violated the triangle inequality. Tsk. --rwg PS, for those interested in trig exercises, this started out as cos(a) - cos(b) cos(c) cos(b) - cos(a) cos(c) acos(----------------------) + acos(----------------------) sin(b) sin(c) sin(a) sin(c) cos(c) - cos(a) cos(b) + acos(----------------------) - %pi sin(a) sin(b) For those with an unhealthy interest in trig exercises, www.ippi.com/rwg/trigbelt.dvi