Cyclic quadrilateral

The area of a cyclic quadrilateral is given by $$\sqrt{(s-a)(s-b)(s-c)(s-d)}$$, where s is the semi-perimeter.
In this case, s = (14+17+18+23)/2 = 36.
Area = $$\sqrt{(36-14)(36-17)(36-18)(36-23)}$$ = 6$$\sqrt{2717}$$.

$$\angle{QNR} = 130\,^{\circ}$$

=> $$\angle{PNO} = 130\,^{\circ}$$ as they are opposite angles.

The opposite angles of a cyclic quadrilateral add up to 180°.

Hence, $$\angle{PMO}$$ = 180 - $$\angle{PNO}$$ = 50°

In triangle MQO,

$$\angle{OQM}$$ + $$\angle{QMO}$$ + $$\angle{MOQ}$$ =180°

Similarly, In triangle MPR, $$\angle{MRP}$$ + $$\angle{PMR}$$ + $$\angle{MPR}$$ =180°

Adding both equations, we get

$$\angle{Q}$$ + 50° + $$\angle{O}$$ + $$\angle{R}$$ + 50° + $$\angle{P}$$ = 360°

$$\angle{P}$$ + $$\angle{O}$$ = 180°

Hence, $$\angle{Q}$$ + 50° + $$\angle{R}$$ + 50° = 180°

$$=> Q + R = 80\,^{\circ}$$

$$\angle{MRN}$$ = 72$$^{\circ}$$ because angle made by a chord in the major segment is half that of angle made by the chord at the center.

MRON is a cyclic quadrilateral => Sum of opposite angles = 180$$^{\circ}$$

$$\angle{MRN}$$ + $$\angle{MON}$$ = 180$$^{\circ}$$

=> $$\angle{MON}$$ = 180$$^{\circ}$$ - 72$$^{\circ}$$ = 108$$^{\circ}$$