Question

In the given figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

Answer 1

In the given figure,,AB || CD,EF || BC ∠BAC = 65° and ∠DHF = 35°

We need to find ∠AGH

Here, GF and CD are straight lines intersecting at point H, so using the property, “vertically opposite angles are equal”, we get,

∠DHF = ∠GHC

∠GHC = 35°

Further, as AB || CD and AC is the transversal

Using the property, “alternate interior angles are equal”

∠BAC = ∠ACD

∠ACD = 65°

Further applying angle sum property of the triangle

In ΔGHC

∠GHC + ∠HCG + CGH = 180°

∠CGH + 35° + 65° = 180°

100 + ∠CGH = 180°

∠CGH = 180°- 100°

∠CGH = 80°

Hence, applying the property, “angles forming a linear pair are supplementary”

As AGC is a straight line

∠CGH + ∠AGH = 180°

∠AGH + 80° = 180°

∠AGH + 180° - 80°

∠AGH = 100°

Therefore, ∠AGH = 100°