I´d say "d" is the distance from the eye to the wall. Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles. B+alfa=45° tan^-1(0.2/d)=B tan^-1(1.3/d)=alfa THEN: tan^-1(0.2/d)+tan^-1(1.3/d)=45° Now you have 3 ecs and 3 variables. alfa,B and "d"
