In the adjoining figure,AB||QP,BC||RQ and CA||PR, angle AEQ=110° and angle RPQ=50°,find angle A, angle B, angle C, angle Q, angle R
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Draw a line KH passing through the point F which is parallel to both AB and CD
We know that KF∥CD and FG is a transversal
from the figure we know that ∠KFG and ∠FGD are alternate angles
So we get
∠KFG=∠FGD=r
o
...(i)
We also know that AE∥KF and EF is a transversal
from the figure we know that ∠AEF and ∠KFE are alternate angles
So we get
∠AEF+∠KFE=180
o
By substituting the values we get
p
o
+∠KFE=180
o
∠KFE=180
o
−p
o
....(ii)
By adding both the equations (i) and (ii) we get
∠KFG+∠KFE=180
o
−p
o
+r
o
from the figures ∠KFG+∠KFE can be written as ∠EFG
∠EFG=180
o
−p
o
+r
o
We know that ∠EFG=q
o
q
o
=180
o
−p
o
+r
o
It can be written as
p+q−r=180
o
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