Question

In the given figure, AB//CD and DE//FG. Determine angle PDE , angle AFD and angle DFG.

Answer 1

/_PDC =/_ AFD = 55 (corresponding angles)

AFB is a straight line so

/_AFD +/_DFG + /_BFG =180

55  + /_DFG  + 70 =180

/_DFG + 125 =180

/_DFG =180-125

/_DFG =55

And /_ DFG = /_PDE =55 (corresponding angles)

Answer 2

$\huge\underline\mathfrak\red{Answer:}$

Since AB$\parallel$CD and transversal DF intersects them at F and D respectively.

•°•$\angle$AFD = $\angle$CDF (corresponding angles)

$\implies$$\angle$AFD = 55°.......(1)

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Now, ray FG stands on AB at F, we have

$\angle$AFG + $\angle$BFG = 180°

($\angle$AFP)+($\angle$PFG) + $\angle$BFG = 180°

55° + $\angle$PFG + 70° = 180°

$\rightarrow$$\angle$PFG + 125° = 180°

$\rightarrow$$\angle$PFG = 180° - 125°

$\rightarrow$$\angle$PFG = 55°

$\rightarrow$$\angle$DFG = 55°

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Now, FG$\parallel$DE and transversal DF intersects them at F and D respectively.

•°•$\angle$DFG = $\angle$PDE

$\implies$ 55° = $\angle$PDE .