In the given figure, AB is parallel to EF, ED is parallel to CB, and \(\angle BPD = 42^{\circ}\). What is the measure of \(\angle CQF\)?
\(138^{\circ}\)
Since AB is parallel to EF and CD is a transversal cutting both lines, \(\angle BPD\) and \(\angle CQF\) are formed on the same side of the transversal at the two parallel lines.
By the co-interior (same-side interior) angles property, angles on the same side of a transversal between two parallel lines are supplementary: \(\angle BPD + \angle CQF = 180^{\circ}\).
Substitute \(\angle BPD = 42^{\circ}\): \(42^{\circ} + \angle CQF = 180^{\circ}\).
Therefore \(\angle CQF = 180^{\circ} - 42^{\circ} = 138^{\circ}\).
Hence, the measure of \(\angle CQF\) is \(138^{\circ}\).
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