PreCal
posted by Anonymous on .
tan1=43\9.3

totally unclear
is there an argument after tan1 ??
did you mean tan^1 Ø or something like that?
is the right side = 43/9.3 ?
I will assume you meant:
tan^1 Ø = 43/9.3
= 4.6236...
Ø = 77.796°