For a triangle with
#color(white)("XXXX")#side #a# opposite angle #A#
#color(white)("XXXX")#side #b# opposite angle #B#
#color(white)("XXXX")#side #c# opposite angle #C#
The Law of Cosines says:
#color(white)("XXXX")##c^2 = a^2+b^2-2abcos(C)#
or
#color(white)("XXXX")##cos(C) = (a^2+b^2-c^2)/(2ab)#
#color(white)("XXXX")##color(white)("XXXX")#(and similarly for #A# and #B#)
So
#color(white)("XXXX")##cos(C) = (2.5^2+10.2^2-9^2)/(2(2.5)(10.2))#
#color(white)("XXXX")##color(white)("XXXX")##=0.574314#
#C = "arccos"(cos(C))#
#color(white)("XXXX")##C = "arccos"(0.574314)#
#color(white)("XXXX")##color(white)("XXXX")##=0.959031# radians
#color(white)("XXXX")##color(white)("XXXX")#(using a calculator)
Similar calculations can be made for angles #A# and #B#