Given: △pst, m∠s=90°, m∈pt,
pm≅mt, mk⊥pt, m∠spk
m∠kpm = 5/2
find: m∠p, m∠t, m∠skp, and m∠mkt

m∠P = 70°, m∠T = 20°, m∠SKP = 40° , and m∠MKT = 70°

Step-by-step explanation:

* Lets explain how to solve the problem

- In Δ PST

∵ m∠S = 90°

∴ m∠T + m∠P = 90° ⇒ interior angles of a triangle

∵ m∠SPK/m∠KPT = 5/2

- The ratio between the two angles are 5 : 2 , multiply the parts of the

  ratio by x, where x is a real number

∴ m∠SPK = 5x

∴ m∠KPT = 2x

∵ m∠SPK + m∠KPT = m∠P

∴ m∠P = 5x + 2x = 7x

- In ΔPKT

∵ KM ⊥ PT

∵ MP = Mt

∴ KM is perpendicular bisector of PT

∴ ΔPKT is an isosceles triangle with KP = KT

∵ KP = KT

∴ m∠KPT = m∠T

∵ m∠KPT = 2x

∴ m∠T = 2x

∵ m∠T + m∠P = 90°

∵ m∠P = 7x

∵ m∠T = 2x

∴ 2x + 7x = 90 ⇒ solve for x

∴ 9x = 90 ⇒ divide both sides by 9

∴ x = 10

∵ m∠P = 7x

∴ m∠P = 7(10) = 70°

∴ m∠P = 70°

∵ m∠T = 2x

∴ m∠T = 2(10) = 20°

∴ m∠T = 20°

- In ΔSKP

∵ m∠S = 90°

∵ m∠SPK = 5x = 5(10) = 50°

∴ m∠SKP = 180° - (90° + 50°) = 180° - 140° = 40° ⇒ interior angles of a Δ

∴ m∠SKP = 40°

- In Δ KMT

∵ m∠KMT = 90°

∵ m∠T = 20°

∴ m∠MKT = 180° - (90° + 20°) = 180° - 110° = 70° ⇒ interior angles of a Δ

∴ m∠MKT = 70°

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step-by-step explanation: