Question

In the figure AB||CD and FP bisected /_ CGH = 150° find the value of /_ EFP . ​

Answer 1

Answer : 75

Give,

<CGH = 150 °

As, AB||CD

Then , <FGD = 150° [ Vertically opposite ]

<EFP = 150°[alternative exterior angle]

Given , FP bisected <EFB

Let ,x be <EFP and <BFP

x+x=150

2x=150

x= 75.