In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O,

Solution:

We know that if two lines intersect, then the vertically opposite angles are equal.

$\therefore \angle B O D=\angle A O C=90^{\circ}$

Hence, $t=90^{\circ}$

Also,

$\angle D O F=\angle C O E=50^{\circ}$

Hence, $z=50^{\circ}$

Since, AOB is a straight line, we have:

$\angle A O C+\angle C O E+\angle B O E=180^{\circ}$

$\Rightarrow 90+50+y=180^{\circ}$

$\Rightarrow 140+y=180^{\circ}$

$\Rightarrow y=40^{\circ}$

Also,

$\angle B O E=\angle A O F=40^{\circ}$

Hence, $x=40^{\circ}$

$\therefore x=40^{\circ}, y=40^{\circ}, z=50^{\circ}$ and $t=90^{\circ}$