Question

If in figure AB||CD, angel PCD=130° and angel PBA=140°, then find angel BPC.​

Answer 1

Draw HQ parallel to AB and CD.

Then, angleB + angleP=180°(co-interior)

140°+angle P=180°

angleBPH=180-140

BPH=40°

similarly ,130+CPQ=180(co-interior)

CPQ=180-130

CPQ=50°

Now,

CPQ+BPH+BPC=180°(linear pair)

50° + 40°+BPC=180°

BPC=180°-90°

BPC=90°

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Answer 2

Constructions :- Draw a line EF parallel to AB.

=> EF|| CD|| AB

BP is a transversal.

angle ABP + angle BPE = 180° [ Sum of angles on same side of transversal is 180°]

140° + angle BPE = 180°

angle BPE = 180° - 140°

angle BPE = 40°

CP is a transversal.

angle PCD + angle CPF = 180° [ Sum of angles on same side of transversal is 180°]

130° + angle CPF = 180°

angle CPF = 180° - 130°

angle CPF = 50°

angle CPF + angle BPE + angle BPC = 180° [ Sum of angles on a straight line is 180°]

=> 50° + 40° + angle BPC = 180°

=> 90° + angle BPC = 180°

=> angle BPC = 180° - 90°

=> angle BPC = 90°

HOPE IT HELPS YOU....