As angles #A# and#B# are acute, we have #0 < A < pi/2# as well as #0 < B < pi/2# i.e. both lie in #Q1#
Now regarding #A+B#, observe that as #A# and #B# both lie in #Q1#, #A+B#. if both #A# and #B# are low enough, #A+B# may be be in #Q1# but if they are larger #A+B# may lie in #Q2#.
Note that if #A+B# is in #Q1#, #cos(A+B)# is positive, but if #A+B# is in #Q2#, #cos(A+B)# would be negative. Note that #sin(A+B)#, however would be positive in bot quadrants. Hence, let us workout #cos(A+B)#.
As #sinA=1/2# #cosA=sqrt(1-(1/2)^2)=sqrt(1-1/4)=sqrt(3/4)=sqrt3/2#
and as #cosB=1/3#, #sinB=sqrt(1-(1/3)^2)=sqrt(1-1/9)=sqrt(8/9)=sqrt8/3#
and #cos(A+B)=cosAcosB-sinAsinB#
= #sqrt3/2*1/3-1/2*sqrt8/3=(sqrt3-sqrt8)/6#
As #sqrt8>sqrt3#, #cos(A+B)# is negative
and hence #A+B# is i #Q2# and #pi/2 < A+B < pi#
i.e. #A+Bin(pi/2,pi)#
