In the figure AB ||CD and EF ||DQ find the value of xyz
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ANSWER
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
∘
Take xyz as Pdq.
BTS army!!!
ANSWER
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
∘
Take xyz as Pdq.
BTS army!!!
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