The sum of two opposite angles of a parallelogram is 130°.

Solution:

Let $A B C D$ be a parallelogram and let the sum of its opposite angles be $130^{\circ}$.

$\angle A+\angle C=130^{\circ}$

$T$ he opposite angles are equal in a parallelogram.

$\therefore \angle A=\angle C=x^{\circ}$

$\Rightarrow x+x=130$

$\Rightarrow 2 x=130$

$\Rightarrow x=\frac{130}{2}$

$\Rightarrow x=65$

$\therefore \angle A=65^{\circ}$ and $\angle C=65^{\circ}$

$\angle A+\angle B=180^{\circ} \quad\left[s\right.$ ince the sum of adjacent angles of a parallelogram is $\left.180^{\circ}\right]$

$\Rightarrow 65^{\circ}+\angle B=180^{\circ}$

$\Rightarrow \angle B=(180-65)^{\circ}$

$\Rightarrow \angle B=115^{\circ}$'

$\angle D=\angle B=115^{\circ} \quad[o$ pposite angles of parallelogram are equal $]$