Two adjacent angles of a parallelogram are (3x − 4)° and (3x + 16)°.

Let $A B C D$ be a parallelogram.

$L$ et $\angle A=(3 x-4)^{\circ}$

$\angle B=(3 x+16)^{\circ}$

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

$\Rightarrow 3 x-4+3 x+16=180$

$\Rightarrow 3 x-4+3 x+16=180$

$\Rightarrow 6 x+12=180$

$\Rightarrow 6 x=168$

$\Rightarrow x=\frac{168}{6}$

$\Rightarrow x=28$

$\therefore \angle A=(3 \times 28-4)^{\circ}$

$\quad=(84-4)^{\circ}$

$\quad=80^{\circ}$

$\angle B=((3 \times 28)+16)^{\circ}$

$\quad=(84+16)^{\circ}$

$\quad=100^{\circ}$

The opposite angles of a paralleleogram are equal.

$\therefore \angle C=\angle A=80^{\circ}$

$\angle D=\angle B=100^{\circ}$