A pentagon has interior angles measuring 120°, 100°, 108°. The remaining two angles are such that one is three times the other. Find the measure of the two remaining angles.

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.
For a pentagon, $n=5$.

$$\text{Sum of angles} = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$$

The given angles are $120^\circ$, $100^\circ$, and $108^\circ$.
Let one of the remaining angles be $x$.
The other remaining angle is three times the first, so it is $3x$.
The sum of all five angles must be $540^\circ$.

$$120^\circ + 100^\circ + 108^\circ + x + 3x = 540^\circ$$

Combine the known angles and the terms with $x$:

$$328^\circ + 4x = 540^\circ$$

Subtract $328^\circ$ from both sides:

$$4x = 540^\circ - 328^\circ$$ $$4x = 212^\circ$$

Divide by 4:

$$x = \frac{212^\circ}{4}$$ $$x = 53^\circ$$

One angle is $x$:

$$x = 53^\circ$$

The other angle is $3x$:

$$3x = 3 \times 53^\circ = 159^\circ$$

The remaining two angles are $\text{53}^\circ \text{ and } 159^\circ$.