Question

if p is the incenter of $\triangle jkl$, find each measure.

1. $m\angle jlp$
2. $m\angle jkp$
3. $mp$
4. $kn$
5. $jp$
6. $pl$

Explanation:

Step1: Find $\angle JLK$

The incenter is where angle bisectors meet, so $\angle JLP = \angle KLP = 22^\circ$.

Step2: Find $\angle KJL$

Similarly, $\angle JKP = \angle LJP = 32^\circ$.

Step3: Calculate $\angle JKL$

Sum of angles in $\triangle JKL$ is $180^\circ$.
$\angle JKL = 180^\circ - 2\times32^\circ - 2\times22^\circ = 180^\circ - 64^\circ - 44^\circ = 72^\circ$

Step4: Find $m\angle JLP$

$P$ bisects $\angle JLK$, so $m\angle JLP = 22^\circ$.

Step5: Find $m\angle JKP$

$P$ bisects $\angle JKL$, so $m\angle JKP = \frac{72^\circ}{2} = 36^\circ$.

Step6: Find $MP$

Incenter is equidistant from sides, so $MP = PN = 7$.

Step7: Find $KN$

$KN = KJ - MJ$, $MJ = JO = 11$, $KJ = 10 + 11 = 21$.
$KN = 21 - 11 = 10$.

Step8: Find $JP$

Use Pythagoras in $\triangle JMP$: $JP = \sqrt{MJ^2 + MP^2} = \sqrt{11^2 + 7^2} = \sqrt{121 + 49} = \sqrt{170}$.

Step9: Find $PL$

First find $LN = KL - KN = 17 - 10 = 7$. Use Pythagoras in $\triangle LNP$: $PL = \sqrt{LN^2 + PN^2} = \sqrt{7^2 + 7^2} = \sqrt{98} = 7\sqrt{2}$.

Answer:

1. $22^\circ$
2. $36^\circ$
3. $7$
4. $10$
5. $\sqrt{170}$
6. $7\sqrt{2}$