Correct option is (b) 28°
Given, ABCD is a kite,
Let the diagonals are meeting at O.
In △ABD and △BCD
AB = BC (Given)
AD = CD (Given)
BD = BD (Common)
Thus, △ABD ≅ △CBD (SSS rule)
Thus, ∠DBA = ∠CBA (by CPCT) .....(1)
Now, In △OAB and △OCB
∠OBA = ∠OBC (From 1)
OB = OB (Common)
AB = BC (Given)
△OAB ≅ △OCB (SAS rule)
Thus, ∠AOB = ∠COB = \(\frac{180°}2\) = 90° (Equal angles by CPCT)
Now, In △OAB
∠OAB + ∠OBA + ∠OAB = 180°
37° + 2y − 3° + 90° = 180°
2y = 56°
y = 28∘