sejalbandil23 sejalbandil23
  • 29-03-2019
  • Mathematics
contestada

Given: PKHE - inscribed, m∠E=120°
m∠EPK=51°, PF ∥EH
Find: m∠KPF, m∠H.

Given PKHE inscribed mE120 mEPK51 PF EH Find mKPF mH class=

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loumast
loumast loumast
  • 29-03-2019

Answer:

KPF = 9 and H = 129

Step-by-step explanation:

An inscribed quadrilateral in a circle has all diagonal angles add up to 180, so we can use this to find the angles of the quadrilateral.  

H= 180-EPK and K = 180-E so H = 129 and K = 60

Now PF and EH are parallel, so PE is a transversal.   That means FPE = 180 - E = 60.  Now it's pretty easy to solve for KPF = FPE - EPK = 60 - 51 = 9.

Let me know if you don't see how I did any of this and I'll be happy to explain it..

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