If an angle of a parallelogram is four-fifths of its adjacent angle,

 Let ABCD be a parallelogram. 

$\therefore \angle A=\angle C$ and $\angle B=\angle D \quad$ (Opposite angles)

Let $\angle A=x^{\circ}$ and $\angle B=\left(\frac{4 x}{5}\right)^{\circ}$

Now, $\angle A+\angle B=180^{\circ}$     (Adjacent angles are supplementary)

$\Rightarrow x+\frac{4 x}{5}=180^{\circ}$

$\Rightarrow \frac{9 x}{5}=180^{\circ}$

$\Rightarrow x=100^{\circ}$

Now, $\angle A=100^{\circ}$ and $\angle B=\left(\frac{4}{5}\right) \times 100^{\circ}=80^{\circ}$

Hence, $\angle A=\angle C=100^{\circ} ; \angle B=\angle D=80^{\circ}$