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