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