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

Given : AB∣∣CD;EF∣∣DQ

To find : ∠PDQ

∵AB∣∣CD and PE is transversal

∠PDC and ∠PEA are corresponding angles.

∴PEA=34∘

∠PEA+∠PEF+∠FEB=180∘ (Linear pair)

34∘+∠PEF+∠FEB=180∘

∠PEF=180∘−34∘−78∘

∠PEF=68∘

∵EF∣∣DQ

∠PDQ=∠PEF=68∘

∴∠PDQ=68∘