# In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

EGB = QHP (Alternate Exterior Angles) = 35O

QPH = 90O

So, in triangle QHP we have,

QPH + QHP + PQH = 180O

90O + 35O + PQH = 180O

PQH = 180O – 90O – 35O

= 55O

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