Let

$\angle OLM = \angle OLN =
a^{\circ},$

$\angle OML = \angle OMN = b^{\circ}$
and

$\angle LOM = c^{\circ}$

Angles in a triangle add up to $180^{\circ}$, so from $\triangle LMN$, $$2a^{\circ}+2b^{\circ}+68^{\circ} = 180^{\circ}$$ which gives $$ 2(a^{\circ}+b^{\circ})=112^{\circ}$$ In other words $$a^{\circ}+b^{\circ}=56^{\circ}$$

Also, from $\triangle LOM$, $$a^{\circ}+b^{\circ}+c^{\circ}=180^{\circ}$$ and so

$$ \eqalign{

c^{\circ}&= 180^{\circ} - (a^{\circ}+b^{\circ})\cr &= 180^{\circ}-56^{\circ}\cr &=124^{\circ}}$$

*This problem is taken from the UKMT Mathematical Challenges.*