Question

the angle bisectors of \\( \triangle efg \\) are \\( \overline{es} \\), \\( \overline{fs} \\), and \\( \overline{gs} \\). they meet at a single point \\( s \\).
(in other words, \\( s \\) is the incenter of \\( \triangle efg \\).)
suppose \\( ps = 14 \\), \\( gs = 17 \\), \\( m\angle pfq = 94^{\circ} \\), and \\( m\angle res = 22^{\circ} \\).
find the following measures.
note that the figure is not drawn to scale.

Explanation:

Step1: Find \(m\angle REP\)

Since \(ES\) is an angle bisector, \(m\angle REP = 2\times m\angle RES\).
Given \(m\angle RES = 22^{\circ}\), then \(m\angle REP=2\times22^{\circ}\).

Step2: Find \(QS\)

The in - center \(S\) is equidistant from the sides of the triangle. So \(PS = QS\).
Given \(PS = 14\), then \(QS = 14\).

Step3: Find \(m\angle RGS\)

First, find \(m\angle EFG\). Since \(FS\) is an angle bisector, and \(m\angle PFQ = 94^{\circ}\), the sum of angles in a triangle is \(180^{\circ}\). But for the angle at \(F\), we know that the in - center and angle bisectors.
The sum of angles in \(\triangle EFG\): Let's first note that the in - center \(S\) and angle bisectors.
We know that \(m\angle EFG=94^{\circ}\). The sum of angles in \(\triangle EFG\) is \(180^{\circ}\). But we can also use the property of angle bisectors.
Since \(FS\) is an angle bisector of \(\angle EFG\), and we know that the sum of angles in a triangle. But another approach:
The sum of angles in \(\triangle EFG\): Let's not get confused. We know that the in - center \(S\) and the fact that \(GS\) is an angle bisector.
The sum of angles in \(\triangle EFG\): \(m\angle EFG + m\angle FEG+m\angle FGE = 180^{\circ}\). But we can also use the angle - bisector property.
Since \(ES\) is an angle bisector (\(m\angle RES = 22^{\circ}\)), \(m\angle FEG=44^{\circ}\) (from step 1, double of \(m\angle RES\)).
\(m\angle EFG = 94^{\circ}\), so \(m\angle FGE=180^{\circ}-(94^{\circ}+44^{\circ})=42^{\circ}\).
Since \(GS\) is an angle bisector, \(m\angle RGS=\frac{1}{2}m\angle FGE\). So \(m\angle RGS = 21^{\circ}\).

Answer:

\(m\angle REP = 44^{\circ}\), \(QS = 14\), \(m\angle RGS=21^{\circ}\)