Where Are The Alternate Angles?

Extend $NO$ to meet line $l$ at $P$ . Extend $MO$ to meet line $m$ at Q $\angle MQN=35^{\circ}$ and $\angle NOQ=180^{\circ}-105^{\circ}=75^{\circ}$ .

Therefore acute angle between $m$ and $NO$ is $180^{\circ}-(35^{\circ}+75^{\circ})=\boxed{70^{\circ}}$ .

First extend $\angle MON$ and as well as the components of $\angle lMO$

Next, we can observe the opposite angle of $\angle MON$ , which according to the Vertical Angles properties, opposite angles are equal, thus $\delta$ = $\angle MON$

We also apply the supplementary angle property, as the two parallel lines are are cut by the same segment, thus splitting the straight angles (180º) in two supplementary angles, which have the same measurements in line $l$ and $m$ . Knowing that, we can trace the other half in line $m$ [The two green angles].

We can also trace the straight angle forming in the segment that extends from $\overline{ON}$

Using simple supplementary angle properties, we find the supplementary angle of $\delta$ (which would also be the supplementary angle of $\angle MON$ if seen in the extended segment of $\overline{MO}$ ).

And lastly , observing the formed triangle $\triangle NOP$ , we use the interior angle sum property (sum of internal angles of a triangle equals 180º) to solve easily for the desired angle.

$35^{\circ} +75^{\circ} = 110^{\circ}$

$180^{\circ}-110^{\circ} = \boxed{70^{\circ}}$

Sum of angles of a pentagon=540°

540-[(180-35)+105+90+90] = 110.

180-110= 70

70

@Sharky Kesa does this seems right to you ?

simple solution: draw a line parallel to lines l and m that passes through point O (call it line o). The acute angle between MO and o must be same as between MO and m, i.e. 35 degrees, so angle MON minus 35 degrees must be the same as the acute angle between m and NO - i.e. 105-35 = 70

make a line n through O parallel to the other lines with end points P on the left and Q on the right. then angle MOP = 35 and angle PON = 70 and since line n is parallel to line m our angle in question is also 70.

Imagine a third parallel line called PQ cutting through the centre O. The angle MOQ would therefore equal to 145 as the angles are supplementary. 145 + 105 =250, 360- 250 = 110, therefore NOQ would equal to 110 as there is 360 degrees around a point. Again as the angles are supplementary, 180 -110 = 70, therefore the angle mNO equals to 70.