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'Incentre Angle' printed from https://nrich.maths.org/
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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.
You can find more short problems, arranged by curriculum topic, in our short problems collection.