The sum of two angles of a triangle is 116° and their difference is 24°.

Let $\angle A+\angle B=116^{\circ}$ and $\angle A-\angle B=24^{\circ}$

Then,

$\therefore \angle A+\angle B+\angle A-\angle B=(116+24)^{\circ}$

$\Rightarrow 2 \angle A=140^{\circ}$

$\Rightarrow \angle A=70^{\circ}$

$\therefore \angle B=116^{\circ}-\angle A$

$=(116-70)^{\circ}$

$=46^{\circ}$

Also, in ∆ ABC:

$\angle A+\angle B+\angle C=180^{\circ} \quad[$ Sum of the angles of a triangle $]$

$\Rightarrow 70^{\circ}+46^{\circ}+\angle C=180^{\circ}$

$\Rightarrow \angle C=64^{\circ}$