In the diagram above, x₁ = 180^o - 117^o = 63^o(opposite angles of a cyclic quad.)

x₁ = x₂ (base angles of isos. \(\Delta\))

x₁ + x₂ + \(\alpha\) = 180^o (sum of angles of a \(\Delta\)

63^o + 63^o + \(\alpha\) = 180^o

\(\alpha\) = 180^o - (63 + 63)^o

= 54^o