45º Solution: It is given in the question that, In parallelogram ABCD: $\angle\text{BAD} = 75^\circ, \ \angle\text{CBD} = 60^\circ$ Now, $\angle\text{DAB} = \angle\text{DCB} = 75^\circ$ (Opposite angles) Also, in triangle DBC we know that sum of angles of a triangle is 180º $\angle\text{DBC} + \angle\text{BDC} + \angle\text{DCB} = 180^\circ$ $60^\circ + \angle\text{BDC} + 75^\circ= 180^\circ$ $135^\circ + \angle\text{BDC} = 180^\circ$ $\angle\text{BDC} = 180^\circ – 135^\circ$ $\angle\text{BDC} = 45^\circ$ Hence, 45º is correct.
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