If p is the incenter of triangle jkl, find each measure

  By using the property of incenter, measures of the sides and the angles will be,

JK = 21.2 units

KL = 26.63 units

JL = 28.33 units

m∠K = 36°

 By the property of incenter,

PM = PN = PO = 7 units

∠MJP = ∠OJP = 32°

∠NLP = ∠OLP = 22°

 By the triangle sum theorem,

m∠JKL + m∠KLJ + m∠LJK = 180°

2(m∠NKP) + 2(m∠NLP) + 2(m∠MJP) = 180°

2(m∠NKP) + 2(22°) + 2(32°) = 180°

2(m∠NKP) = 180° - 108°

m∠NKP = 36°

From ΔJMP,

tan(32°) =

tan(32°) =

MJ =

MJ = 11.20

From ΔPOL,

tan(22°) =

tan(22°) =

OL = 17.33

From ΔKNP,

tan(36°) =

tan(36°) =

KN = 9.63

Therefore, JK = MJ + MK = 21.2 units

KL = LN + KN = 26.63 units

JL = JO + OL = 28.33 units

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PO = 7

PM = 7

MJ = 11

∠PJO = 32°

∠KJL = 64°

PL ≈ 18.385

OL = 17

∠PLO = 22°

∠NLO = 44°

∠JKL = 72°

∠MKP = 36°

∠NKP = 36°

∠PKN = 36°

KN = 10

PL ≈ 13.04

PK ≈ 12.21

JL = 28

JK = 21

LK = 27

Step-by-step explanation:

The given parameters are;

The point representing the incenter of the triangle = P

Therefore PO = PM = PN = 7

tan(32°) = PM/JM = 7/JM

∴JM = 7/(tan(32°)) ≈ 11.2

∠PJO = tan⁻¹(7/11)≈ 32.47°

∠PJO = ∠PJM = 32° similar triangles

∠KJL = ∠KJP + ∠PJO = 32 + 32 = 64°

∠KJL ≈ 64°

PL = √(7² + 17²) ≈ 18.385

OL = NL = 17 similar triangles

∠PLO = sin⁻¹(7/18.385) ≈ 22.380°

∠PLO = ∠PLN = 22°

∠NLO = ∠PNL + ∠OLP ≈ 22° + 22° ≈ 44°

∠NLO ≈ 44.380°

∠JKL = 180 - (∠KJL + ∠NLO)

∠JKL = 180° - (64° + 44°) ≈ 72°

∠JKL  ≈ 72°

∠MKP = ∠NKP = 72°/2 = 36°

∠MKP = 36°

∠NKP = 36°

∠PKN = ∠JKL - ∠MKP = 72° - 36° ≈ 36°

∠PKN  ≈ 36°

KN = KM = 10

MJ = OJ = 11

PL = √(7² + 11²) ≈ 13.04

PK = √(7² + 10²) ≈ 12.21

JL = JO + OL = 11 + 17 = 28

JK = JM + MK = 11 + 10 = 21

LK = LN + NK = 17 + 10 = 27

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