Question

u is the incenter of $\triangle ghy$. find each measure.

1. $m\angle ugm$
2. $m\angle phu$
3. hu

Explanation:

Step1: Find $m\angle UGM$

The incenter is the intersection of angle bisectors, so $UG$ bisects $\angle HGY$. Given $\angle UGB = 28^\circ$, then $\angle UGM = 28^\circ$.

Step2: Calculate $\angle PHU$

First, find $\angle GHY$: since $HU$ bisects $\angle GHY$, and $\angle MHU = 21^\circ$, so $\angle GHY = 2\times21^\circ=42^\circ$. $UP\perp HY$, so $\triangle PHU$ is right-angled at $P$.
$\angle PHU = 90^\circ - \angle HUP$. First find $\angle HUP$: $\angle HUP = 90^\circ - \frac{1}{2}\angle GHY$? No, correct: in right $\triangle PHU$, $\angle PHU = 90^\circ - \angle PUH$. Wait, $\angle PUH = 90^\circ - \angle UHP$? No, simpler: $\angle UHP = 21^\circ$, so in right $\triangle PHU$, $\angle PHU = 90^\circ - (90^\circ - 21^\circ)$? No, wrong. Correct: $\angle PHU$ is the same as $\angle UHP = 21^\circ$? No, wait, $UP$ is the inradius, so $\angle UPH=90^\circ$. $\angle PHU$ is the angle at $H$ in $\triangle PHU$, which is equal to $\angle UHY = 21^\circ$? No, no: the incenter's angle bisector splits $\angle GHY$ into two $21^\circ$ angles, so $\angle PHU = 21^\circ$, and $\angle HUP = 90^\circ - 21^\circ = 69^\circ$? No, wait, we need $\angle PHU$: actually, $\angle PHU = 90^\circ - \angle PUH$. Wait, no, let's use triangle angles. The inradius is 5, so $UB=UM=UP=5$. For $\angle PHU$: in $\triangle PHU$, $\angle UPH=90^\circ$, $\angle PUH = 90^\circ - \angle PHU$. But $\angle PHU$ is half of $\angle GHY$, which is $21^\circ$, so $\angle PHU = 21^\circ$? No, that's not right. Wait, no: $\angle GHY$ is split by $HU$ into two $21^\circ$ angles, so $\angle PHU = 21^\circ$, and $\angle HUP = 90^\circ - 21^\circ = 69^\circ$. Wait, no, the question is $m\angle PHU$, which is $21^\circ$? No, wait, no: $\angle PHU$ is the angle at $H$ between $HP$ and $HU$, which is exactly the angle bisected, so yes, $\angle PHU = 21^\circ$. Wait, no, let's calculate the triangle's angles first. $\angle HGY = 2\times28^\circ=56^\circ$, $\angle GHY=2\times21^\circ=42^\circ$, so $\angle GYH=180-56-42=82^\circ$. Then $\angle UYH=41^\circ$. Now, in $\triangle PHU$, $\angle UPH=90^\circ$, $\angle PUH = 180 - 90 - \angle PHU$. But $\angle PHU=21^\circ$, so $\angle PUH=69^\circ$. Wait, no, the question is $m\angle PHU$, which is $21^\circ$? No, I'm confused. Wait, no: $HU$ is the angle bisector, so $\angle GHM = \angle MHP = 21^\circ$, so $\angle PHU = 21^\circ$. Yes, that's correct.

Step3: Find $HU$

$HU$ is the hypotenuse of right $\triangle PHU$, where $UP=5$ (inradius), $\angle PHU=21^\circ$.
$\sin(21^\circ)=\frac{UP}{HU}$, so $HU=\frac{UP}{\sin(21^\circ)}$? No, wait, $\sin(\angle PHU)=\frac{UP}{HU}$, so $HU=\frac{5}{\sin(21^\circ)}$? No, wait, $\angle PHU=21^\circ$, $\angle UPH=90^\circ$, so $\sin(21^\circ)=\frac{UP}{HU}$, so $HU=\frac{5}{\sin(21^\circ)}\approx14.0$. But wait, $HM$ is part of $GH=12$? No, $GH=12$, $HM$ is a segment. Wait, no, $UP=5$, which is the inradius, so $UB=UM=UP=5$. In right $\triangle UMH$, $UM=5$, $\angle UHM=21^\circ$, so $HU=\frac{UM}{\sin(21^\circ)}=\frac{5}{\sin(21^\circ)}\approx14.0$. Alternatively, use cosine: $\cos(21^\circ)=\frac{HM}{HU}$, but we don't know $HM$. Wait, no, $GH=12$, $\angle HGM=28^\circ$, so $GM=GH\cos(28^\circ)=12\cos(28^\circ)\approx10.6$, $HM=GH\sin(28^\circ)=12\sin(28^\circ)\approx5.6$. Then $HU=\sqrt{HM^2 + UM^2}=\sqrt{5.6^2 +5^2}\approx\sqrt{31.36+25}=\sqrt{56.36}\approx7.5$. Wait, that's conflicting. Oh, I see my mistake: $\angle UGM=28^\circ$, which is half of $\angle HGY$, so $\angle HGY=56^\circ$, $\angle GHY=42^\circ$, so $\angle GYH=82^\circ$. The inradius is 5, so the area of $\triangle GHY$ is $\frac{1}{2}(GH+HY+GY)\times5$. But we know $GH=12$, let's find $HY$: using sine law, $\frac{GH}{\sin(82^\circ)}=\frac{HY}{\sin(56^\circ)}$, so $HY=\frac{12\sin(56^\circ)}{\sin(82^\circ)}\approx10.1$, $GY=\frac{12\sin(42^\circ)}{\sin(82^\circ)}\approx8.1$. Then area is $\frac{1}{2}\times12\times HY\times\sin(42^\circ)\approx\frac{1}{2}\times12\times10.1\times0.669\approx40.3$. Also, area is $\frac{1}{2}(12+10.1+8.1)\times5=\frac{1}{2}\times30.2\times5=75.5$, which doesn't match. So my mistake is that $\angle UGM=28^\circ$ is not half of $\angle HGY$, but $\angle UGB=28^\circ$, so $\angle HGY=2\times28^\circ=56^\circ$, correct. $\angle MHU=21^\circ$, so $\angle GHY=2\times21^\circ=42^\circ$, correct. Then $\angle GYH=82^\circ$, correct. The inradius $r=5$, so area $=r\times s$, where $s$ is semiperimeter: $s=\frac{GH+HY+GY}{2}$. Using sine law: $\frac{GH}{\sin\angle GYH}=\frac{HY}{\sin\angle HGY}=\frac{GY}{\sin\angle GHY}=2R$. $2R=\frac{12}{\sin82^\circ}\approx12.1$, so $HY=12.1\sin56^\circ\approx10.0$, $GY=12.1\sin42^\circ\approx8.1$. $s=\frac{12+10+8.1}{2}=15.05$, area $=5\times15.05=75.25$. Also, area $=\frac{1}{2}\times GH\times HY\times\sin42^\circ=\frac{1}{2}\times12\times10\times0.669=40.14$, which doesn't match. So the error is that $GH$ is not 12, but $HM=12$? Oh! I misread the diagram: $H$ to $M$ is 12, not $GH$. That's the mistake. So $HM=12$, $UM=5$ (inradius), right angle at $M$. So $HU$ is the hypotenuse of right $\triangle UMH$, so $HU=\sqrt{HM^2 + UM^2}=\sqrt{12^2 +5^2}=\sqrt{144+25}=\sqrt{169}=13$. That makes sense!
Now, correct Step2: $\angle PHU$: since $HU$ is the angle bisector, $\angle PHU=\angle MHU=21^\circ$, and in right $\triangle PHU$, $\angle UPH=90^\circ$, so $\angle PHU=21^\circ$, and $\angle HUP=90^\circ-21^\circ=69^\circ$, but the question is $m\angle PHU=21^\circ$? No, wait, no: $\angle PHU$ is the angle at $H$ in $\triangle PHU$, which is the same as $\angle UHY$, which is half of $\angle GHY$, so $\angle PHU=21^\circ$. Correct.
Step1: $m\angle UGM$: $UG$ bisects $\angle HGY$, and $\angle UGB=28^\circ$, so $\angle UGM=28^\circ$.

Answer:

1. $m\angle UGM = 28^\circ$
2. $m\angle PHU = 21^\circ$
3. $HU = 13$