the angle bisectors of △efg are $\overline{es}$, $\overline{fs}$, and $\overline{gs}$. they meet at a single point s. (in other words, s is the incenter of △efg.) suppose $ps = 12$, $es = 17$, $m \angle per = 98^\circ$, and $m \angle rgs = 14^\circ$. find the following measures. note that the figure is not drawn to scale. image of triangle efg with incenter s, right angles at p, q, r find $m \angle rgq$, $m \angle qfs$, $qs$.

Question

Accepted Answer

# Explanation:
## Step1: Find $ m\angle RGQ $
Since $ S $ is the incenter, $ GS $ bisects $ \angle RGQ $. Given $ m\angle RGS = 14^\circ $, then $ m\angle RGQ = 2\times m\angle RGS = 2\times14^\circ = 28^\circ $? Wait, no, wait. Wait, $ \angle RGQ $ is a right angle? Wait, no, looking at the diagram, $ Q $ and $ R $ are right angles (since $ S $ is incenter, the distances from $ S $ to sides are equal, and $ SQ \perp FG $, $ SR \perp EG $, $ SP \perp EF $). Wait, no, $ \angle RGQ $: $ R $ is on $ EG $, $ Q $ is on $ FG $, and $ GQ $ and $ GR $ are sides? Wait, no, $ GS $ is an angle bisector, so $ \angle RGS = \angle QGS = 14^\circ $, and $ \angle RGQ $ is a right angle? Wait, no, the diagram shows $ SQ \perp FG $ (since $ \angle SQF = 90^\circ $) and $ SR \perp EG $ ( $ \angle SRE = 90^\circ $)? Wait, no, the problem says $ PS = 12 $, $ ES = 17 $, and $ S $ is incenter, so the distances from $ S $ to the sides (the inradius) are equal. So $ PS = QS = RS $, because $ PS \perp EF $, $ QS \perp FG $, $ RS \perp EG $. So $ QS = PS = 12 $.

Wait, let's re-express:

1. $ m\angle RGQ $: Wait, $ \angle RGQ $ is at $ G $, between $ RG $ and $ QG $. But $ RG $ is perpendicular to $ GS $? No, $ SR \perp EG $ (so $ \angle SRG = 90^\circ $) and $ SQ \perp FG $ ( $ \angle SQG = 90^\circ $)? Wait, no, the diagram has $ P $ on $ EF $, $ R $ on $ EG $, $ Q $ on $ FG $, with $ SP \perp EF $, $ SR \perp EG $, $ SQ \perp FG $. So $ SP $, $ SQ $, $ SR $ are the inradii, so they are equal. So $ SQ = PS = 12 $.

For $ m\angle RGQ $: Wait, $ \angle RGS = 14^\circ $, and $ GS $ bisects $ \angle EGF $, so $ \angle EGF = 2\times14^\circ = 28^\circ $? No, wait, $ \angle RGS $ is half of $ \angle EGF $, so $ \angle EGF = 2\times14^\circ = 28^\circ $, but $ \angle RGQ $: Wait, maybe $ \angle RGQ $ is a right angle? No, that can't be. Wait, maybe I misread. Wait, the problem says "Find the following measures: $ m\angle RGQ $, $ m\angle QFS $, $ QS $".

Wait, let's tackle each:

- $ QS $: Since $ S $ is the incenter, the distances from $ S $ to the sides are equal. $ PS $ is the distance from $ S $ to $ EF $ (since $ SP \perp EF $), $ QS $ is the distance from $ S $ to $ FG $ ( $ SQ \perp FG $), so $ QS = PS = 12 $.

- $ m\angle RGQ $: Wait, $ \angle RGQ $ is at $ G $, between $ RG $ and $ QG $. But $ RG $ is perpendicular to $ SR $ ( $ \angle SRG = 90^\circ $) and $ QG $ is perpendicular to $ SQ $ ( $ \angle SQG = 90^\circ $)? No, $ SR \perp EG $ (so $ \angle SRG = 90^\circ $) and $ SQ \perp FG $ ( $ \angle SQG = 90^\circ $), so $ \angle RGQ $ is the angle between $ EG $ and $ FG $ at $ G $, but $ GS $ bisects $ \angle EGF $, so $ \angle EGF = 2\times m\angle RGS = 2\times14^\circ = 28^\circ $? Wait, no, $ \angle RGS = 14^\circ $, so $ \angle QGS = 14^\circ $, so $ \angle RGQ = \angle RGS + \angle QGS = 28^\circ $? No, that's not right. Wait, maybe $ \angle RGQ $ is a right angle? No, the diagram shows $ SQ \perp FG $ and $ SR \perp EG $, so $ \angle SRG = 90^\circ $ and $ \angle SQG = 90^\circ $, so $ \angle RGQ $ is the angle between $ EG $ and $ FG $, which is $ \angle EGF $, and since $ GS $ bisects it, $ \angle EGF = 2\times14^\circ = 28^\circ $, but that seems small. Wait, maybe I made a mistake.

Wait, $ m\angle PER = 98^\circ $, which is $ \angle PER $ at $ E $, between $ PE $ and $ RE $. $ ES $ bisects $ \angle PE R $, so $ \angle PES = \angle RES = \frac{98^\circ}{2} = 49^\circ $. Then, in triangle $ EFG $, the sum of angles is $ 180^\circ $. Let's denote $ \angle E = 98^\circ $ (wait, $ \angle PER = 98^\circ $, which is $ \angle at E $, so $ \angle E = 98^\circ $? Wait, $ \angle PER $ is a straight angle? No, $ P $ is on $ EF $, $ R $ is on $ EG $, so $ \angle PER $ is $ \angle at E $, between $ PE $ (which is part of $ EF $) and $ RE $ (part of $ EG $), so $ \angle E = 98^\circ $. Then $ ES $ bisects $ \angle E $, so $ \angle PES = \angle RES = 49^\circ $.

Now, $ \angle RGS = 14^\circ $, so $ \angle EGF = 2\times14^\circ = 28^\circ $ (since $ GS $ bisects $ \angle EGF $). Then, in triangle $ EFG $, angles sum to $ 180^\circ $, so $ \angle F = 180^\circ - \angle E - \angle EGF = 180^\circ - 98^\circ - 28^\circ = 54^\circ $. Then $ FS $ bisects $ \angle F $, so $ \angle QFS = \frac{54^\circ}{2} = 27^\circ $.

Now, $ QS $: Since $ S $ is incenter, $ PS = QS = RS $ (the inradius), and $ PS = 12 $, so $ QS = 12 $.

Now, $ m\angle RGQ $: Wait, $ \angle RGQ $ is a right angle? Wait, $ SQ \perp FG $ (so $ \angle SQG = 90^\circ $) and $ SR \perp EG $ ( $ \angle SRG = 90^\circ $), so $ \angle RGQ $ is the angle between $ EG $ and $ FG $, which is $ \angle EGF = 28^\circ $? But that seems odd. Wait, maybe $ \angle RGQ $ is $ 90^\circ $? No, the diagram shows $ SQ $ and $ SR $ as perpendicular to $ FG $ and $ EG $, so $ \angle RGQ $ is $ 90^\circ $? Wait, no, $ RG $ is on $ EG $, $ QG $ is on $ FG $, so $ \angle RGQ $ is $ \angle EGF $, which we found as $ 28^\circ $, but that contradicts. Wait, maybe $ \angle PER = 98^\circ $ is not $ \angle E $, but $ \angle at E $ between $ PE $ and $ RE $, so $ \angle PER = 98^\circ $, so $ \angle PES + \angle RES = 98^\circ $, and since $ ES $ bisects it, each is $ 49^\circ $. Then, $ \angle at E $ in triangle $ EFG $ is $ 98^\circ $, $ \angle at G $ is $ 2\times14^\circ = 28^\circ $, so $ \angle at F $ is $ 180 - 98 - 28 = 54^\circ $, so $ FS $ bisects $ \angle F $, so $ \angle QFS = 54/2 = 27^\circ $.

Now, $ QS = PS = 12 $, since $ PS $ and $ QS $ are both inradii (distances from $ S $ to sides $ EF $ and $ FG $).

For $ m\angle RGQ $: Wait, maybe $ \angle RGQ $ is $ 90^\circ $? No, the diagram shows $ SQ \perp FG $ (so $ \angle SQG = 90^\circ $) and $ SR \perp EG $ ( $ \angle SRG = 90^\circ $), so $ \angle RGQ $ is the angle between $ EG $ and $ FG $, which is $ \angle EGF = 28^\circ $, but that seems small. Wait, maybe I misread $ \angle RGS $. The problem says $ m\angle RGS = 14^\circ $, so $ \angle RGS = 14^\circ $, so $ \angle QGS = 14^\circ $, so $ \angle RGQ = \angle RGS + \angle QGS = 28^\circ $. Yes, that makes sense because $ GS $ bisects $ \angle EGF $, so $ \angle EGF = 2\times14^\circ = 28^\circ $, so $ m\angle RGQ = 28^\circ $.

So summarizing:

- $ m\angle RGQ = 28^\circ $ (because $ GS $ bisects $ \angle EGF $, so $ \angle EGF = 2 \times 14^\circ = 28^\circ $)
- $ m\angle QFS = 27^\circ $ (because $ \angle F = 180^\circ - 98^\circ - 28^\circ = 54^\circ $, and $ FS $ bisects $ \angle F $, so $ 54^\circ / 2 = 27^\circ $)
- $ QS = 12 $ (because $ QS $ is equal to $ PS $, the inradius, so $ QS = PS = 12 $)

## Step1: Find $ m\angle RGQ $
$ GS $ is an angle bisector, so $ \angle RGS = \angle QGS = 14^\circ $. Thus, $ m\angle RGQ = \angle RGS + \angle QGS = 14^\circ + 14^\circ = 28^\circ $.

## Step2: Find $ m\angle QFS $
First, find $ \angle F $ in $ \triangle EFG $. The sum of angles in a triangle is $ 180^\circ $. We know $ \angle E = 98^\circ $ (from $ m\angle PER = 98^\circ $, as $ ES $ bisects $ \angle E $) and $ \angle G = 28^\circ $ (from Step1). So $ \angle F = 180^\circ - 98^\circ - 28^\circ = 54^\circ $. $ FS $ bisects $ \angle F $, so $ m\angle QFS = \frac{54^\circ}{2} = 27^\circ $.

## Step3: Find $ QS $
$ S $ is the incenter, so the distances from $ S $ to the sides (inradii) are equal. $ PS \perp EF $ and $ QS \perp FG $, so $ QS = PS = 12 $.

# Answer:
$ m\angle RGQ = \boxed{28}^\circ $
$ m\angle QFS = \boxed{27}^\circ $
$ QS = \boxed{12} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Find \( m\angle RGQ \)

Answer: