Solution

$$ABCD$$ is a parallelogram.
$$\angle B+\angle D=130^{\circ}$$
$$\angle B=\angle D$$ (opposite angles of a parallelogram are equal)
$$\angle B=\angle B=130^{\circ}$$
$$2\angle B=130^{\circ}$$
$$\angle B=\dfrac{130^{\circ}}{2}$$
$$\angle B=65^{\circ}$$
Now $$\angle B+\angle C=180^{\circ}$$ (adjacent angles of a parallelogram are supplementary )
$$\angle C=180^{\circ}-\angle B=180^{\circ}-65^{\circ}$$
$$\angle C=115^{\circ}$$
$$\therefore \angle A=\angle C=115^{\circ}$$ (opp. angles of parallelogram are equal)
$$\therefore \angle B=\angle D=65^{\circ}$$ and $$\angle A=\angle C=115^{\circ}$$