See Fig 1 and Fig 2.
Given that the right pyramid has a square base, to find the angle between faces #TAB and TBC#, you can draw the altitude #AP# and #CP# of the two faces #TAB and TBC#. #P# lies on the common edge #TB#, as shown in Fig 2, then the angle between the two faces is #angleAPC#, as shown in Fig 4. Then #angleAPC# can be found using the law of cosines.
1) Consider #DeltaTBC#, see Fig 3.
#TM=sqrt(TB^2-BM^2)=sqrt(6^2-2^2)=sqrt32=4sqrt2#
Area of #DeltaTBC=1/2*BC*TM=1/2*TB*CP#
#=> CP=(BC*TM)/(TB)=(4*4sqrt2)/6=(8sqrt2)/3#
As #DeltaTAB and DeltaTBC# are congruent,
#AP=CP=(8sqrt2)/3#
2) Consider #DeltaABC#, see Fig 2.
#AC=sqrt(AB^2+BC^2)=sqrt(4^2+4^2)=4sqrt2#
3) Consider #DeltaAPC#, see Fig 4.
Let #angleAPC=alpha#
By the law of cosines,
#AC^2=AP^2+CP^2-2*AP*CP*cosalpha#
#=> cosalpha=(AP^2+CP^2-AC^2)/(2*AP*CP#
#=(((8sqrt2)/3)^2+((8sqrt2)/3)^2-(4sqrt2)^2)/(2*(8sqrt2)/3*(8sqrt2)/3#
#=((128+128-288)/9)/((2*128)/9)#
#=(-32)/256=-1/8#
Hence, #cosalpha=-1/8#
