Question

In Figure AB||CD, EF||DQ. Determine the measures of Angle PDQ, Angle AED and Angle DEF.​

Answer 1

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∴∠AED = ∠CDP

→ ∠AED = 34°

Ray EF stands on AB at E.

∴ ∠AEF + ∠FEB = 180°

→ ∠AED + ∠DEF + ∠FEB = 180°

-- 34° + ∠DEF + ∠FEB = 180°

-- 112° + ∠DEF = 180° - 112° = 68°

Now,EF || DQ || and transversal PE intersects DQ at D andEF at E.

∴∠PDQ = ∠DEF

∠PDQ = 68°

Answer 2

Answer:

∵ AB || CD and transversal line intersect E and D respectively.

∴ ∠AED = ∠CDP (corresponding angles)

=> ∠AED = 34°

Ray EF is inclined on line AB at point E

∴ ∠AEF + ∠BEF = 180°

=> (∠AEP + ∠PEF) + ∠BEF = 180° [ ∵ ∠AEF = ∠AEP + ∠PEF]

=> 34° + ∠PEF + 78° = 180°

=> ∠PEF = 180°- 112° = 68°

=> ∠DEF = 68° …(1)

∵ EF || DQ and transversal line DE intersects at E and D respectively.

∵ ∠FED = ∠PDQ = 68° [ ∵ ∠FED = ∠PEF = 68°]

=> ∠PDQ = 68°

So ∠PDQ = ∠DEF = 68°, ∠AED = 34°