Question 36

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waecmaths question: 

In the diagram, $MN\parallel PO,$ $\angle PMN={{112}^{{}^\circ }},\angle PNO={{129}^{{}^\circ }}\angle NOP={{37}^{{}^\circ }}$ and$\angle MPN=y$ find the value of y

Option A: 

51o

Option B: 

54o

Option C: 

56o

Option D: 

66o

waecmaths solution: 

\[\begin{align}  & \angle OPN={{180}^{\circ }}-({{129}^{\circ }}+{{37}^{\circ }})={{14}^{\circ }}\text{  }\!\!\{\!\!\text{ }\angle \text{s in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle PNM=\angle OPN={{14}^{\circ }}\text{    }\!\!\{\!\!\text{ alternate angles }\!\!\}\!\!\text{ } \\ & \angle MPN={{180}^{\circ }}-(\angle PNM+\angle PMN)\text{   }\!\!\{\!\!\text{ }\angle \text{s in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & y=\angle MPN={{180}^{\circ }}-({{112}^{\circ }}+{{14}^{\circ }})={{54}^{\circ }} \\\end{align}\]

maths year: