Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°.

Solution:

$O$ ppostie angles of a parallelogram are congurent.

$\therefore(3 x-2)^{\circ}=(50-x)^{\circ}$

$3 x^{\circ}-2^{\circ}=50^{\circ}-x^{\circ}$

$3 x^{\circ}+x^{\circ}=50^{\circ}+2^{\circ}$

$4 x^{\circ}=52^{\circ}$

$x^{\circ}=13^{\circ}$

Putting the value of $x$ in one angle :

$3 x^{\circ}-2^{\circ}=39^{\circ}-2^{\circ}$

$=37^{\circ}$

$O$ pposite angles are congurent:

$\therefore 50-x^{\circ}$

$=37^{\circ}$

Let the remaining two angles be $y$ and $z$.

Angles $y$ and $z$ are congurent because they are also opposite angles.

$\therefore y=z$

The sum of adjacent angles of a paralle $\log$ ram is equal to $180^{\circ}$.

$\therefore 37^{\circ}+y=180^{\circ}$

$y=180^{\circ}-37^{\circ}$

$y=143^{\circ}$

So, the anlges measure are :

$37^{\circ}, 37^{\circ}, 143^{\circ}$ and $143^{\circ}$