Answer:
per diagram AB∣∣CD and PQ is transversal
∠DMQ=100
o
[ Given]
∠DMQ=∠CML=100
[ Vertically opposite angles]
∠CML=∠ALP=100
[ Corresponding angles]
∠ALP=∠MLB=100
∠CMQ=180−100=80
[sum of angle on a straight line is 180^o$$]
∠CMQ=∠LMD=80
∠LMD=∠PLB=80
∠PLB=∠ALM=80
∴ ∠DMQ=∠CML=∠ALP=∠MLB=100
and ∠CMQ=∠LMD=∠ALM=∠PLB=80......
here is your answer...
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Answers & Comments
Answer:
per diagram AB∣∣CD and PQ is transversal
∠DMQ=100
o
[ Given]
∠DMQ=∠CML=100
o
[ Vertically opposite angles]
∠CML=∠ALP=100
o
[ Corresponding angles]
∠ALP=∠MLB=100
o
[ Vertically opposite angles]
∠CMQ=180−100=80
o
[sum of angle on a straight line is 180^o$$]
∠CMQ=∠LMD=80
o
[ Vertically opposite angles]
∠LMD=∠PLB=80
o
[ Corresponding angles]
∠PLB=∠ALM=80
o
[ Vertically opposite angles]
∴ ∠DMQ=∠CML=∠ALP=∠MLB=100
o
and ∠CMQ=∠LMD=∠ALM=∠PLB=80......
Answer:
here is your answer...