60°. Solution: Since ABCD is a parallelogram, AB || CD since opposite angles of a parallelogram are equal. $\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles) In $\triangle\text{ADB},$ $\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property) $\Rightarrow45+\angle\text{BDA}+75=180$ $\Rightarrow\angle\text{BDA}+120=180$ $\Rightarrow\angle\text{BDA}=60^{\circ}$ $\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles) $\Rightarrow\angle\text{CBD}=60^{\circ}$
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