In the diagram, MN//PO, < PMN = 112^o, < PNO = 129^oo and < MPN = y^o. Find the value of y

51^o

54^o

56^o

68^o

Correct answer is B

In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180^o(sum of interior angles of a \(\bigtriangleup\) )

i.e. NPO + 129 + 37 = 180

< NOP = 180 - (129 + 37) = 14^o

< MNP = < NOP = 14^o (alt. < s)

In \(\bigtriangleup\) MPN

< PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))

i.e. 112 + 14 + y = 180^o

y = 180 - (112 + 14) = 180 - 126 = 54^o

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In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y

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In the diagram, MN//PO, < PMN = 112^o, < PNO = 129^oo and < MPN = y^o. Find the value of y