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


Question

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

Options

A) 51o

B) 54o

C) 56o

D) 68o

The correct answer is B.

Explanation:

In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180\(^{\circ}\)(sum of interior angles of a \(\bigtriangleup\) )
i.e. NPO + 129 + 37 = 180
< NOP = 180 - (129 + 37) = 14\(^{\circ}\)
< MNP = < NOP = 14\(^{\circ}\) (alt. < s)
In \(\bigtriangleup\) MPN
< PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))
i.e. 112 + 14 + y = 180\(^{\circ}\)
y = 180 - (112 + 14) = 180 - 126 = 54\(^{\circ}\)

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Dicussion (1)

  • In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180\(^{\circ}\)(sum of interior angles of a \(\bigtriangleup\) )
    i.e. NPO + 129 + 37 = 180
    < NOP = 180 - (129 + 37) = 14\(^{\circ}\)
    < MNP = < NOP = 14\(^{\circ}\) (alt. < s)
    In \(\bigtriangleup\) MPN
    < PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))
    i.e. 112 + 14 + y = 180\(^{\circ}\)
    y = 180 - (112 + 14) = 180 - 126 = 54\(^{\circ}\)

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