43°

(X refers to $\angle$ AXY and Y to $\angle$ AYX, all numeric values unspecified are angles)

Since AXY is a triangle (yds), 43+X+Y=180, X+Y=137.

$\angle$ BAY = Y $\rightarrow$ $\angle$ BAX = Y-43

$\angle$ CAX = X $\rightarrow$ $\angle$ CAY = X-43

Therefore $\angle$ BAC

= $\angle$ BAX + $\angle$ XAY + $\angle$ YAC

=X-43 + 43 + Y-43

=X+Y-43

=137-43

=94

As BY=BA so in ABY <BAY=<AYB let BAY=43+x so in ABY <BAY+<AYB+<B=180 or <BAY+<BAY+<B=180 or 2<BAY+<B=180 or 2(43+x)+<B=180 or 86+2x+<B=180 or <B=94-2x..........(1) Now as CX=CA,so in AXC <AXC=<XAC let <XAC=43+y then in AXC <AXC+<XAC+ <C=180 or 2<XAC+<C=180 or 2(43+y)+<C=180 or 86+2y+<C=180 or <C=94-2y..............................................(2) Now in the whole ABC <A+<B+<C=180 or x+43+y+94-2x+94-2y=180 (putting the value of (1) and (2) or 231-x-y=180 or 231-180=x+y or x+y=51 so <BAC=<A=x+43+y=x+y+43=51+43=94

Anyone has solution? Needed…~