In the given figure, if AB||CD||EF, PQ||RS, angle RQD=25° and angle CQP=60°,then find angle QRS and angle RQP.
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(i) AB || CD || EF (Given)
PQ || RS (Given)
∠ RQD = 25° (Given)
∠ CQP = 60° (Given)
According to the figure -
∠ DQR = ∠ QRA = 25° ( Alternate interior angles)
∠ PQC = ∠ BRS = 60° ( Alternate exterior angles)
∠ QRA + ∠ ARS = ∠QRS
= ∠QRA + (180° – ∠BRS) [ Linear pair axioms]
= 25° + 180° - 60°
= 205° - 60°
= 145°
Therefore, the value of ∠QRS is 145°.
(ii) ∠PQC + ∠PQD = 180° ( Linear pair axioms )
60° + ∠PQD = 180°
∠PQD = 180° - 60°
∠PQD = 120°
∠PQD + ∠RQD = ∠RQP
∠RQP = 120° + 25°
∠RQP = 145°
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