In Fig. 9.40, if AB||CD, EF||BC, BAC = 65° and DHF = 35°, find AGH.

Given,

AB CD and,


EF BC


BAC = 65o and DHF = 35o


BAC = ACD (Alternate angles)


ACD = 65o


DHF = GHC (Vertically opposite angles)


GHC = 35o


In


GCH + GHC + HGC = 180o


65o + 35o + HGC = 180o


HGC = 80o


AGH + HGC = 180o (Linear pair)


AGH + 80o = 180o


AGH = 100o


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