In a triangle ΔPQR, ∠R = 62°. The perpendicular bisector of PQ at S meets QR at T. If ∠TPR = 38°, what is the measure of ∠PQR?
40°
Because \(T\) lies on the perpendicular bisector of \(PQ\), it is equidistant from the two endpoints, so \(TP=TQ\) and triangle \(TPQ\) is isosceles.
Equal sides face equal base angles, hence \(\angle TPQ=\angle TQP\). Since \(T\) lies on segment \(QR\), the angle \(\angle TQP\) is the same as \(\angle PQR\). Write \(\angle PQR=y\).
Angle \(\angle PTR\) is an exterior angle of triangle \(TPQ\) at \(T\), so it equals the sum of the two remote interior angles: \(\angle PTR=\angle TPQ+\angle TQP=2y\).
In triangle \(PTR\), the angle at \(R\) is the given \(\angle R=52°\), and the angle at \(P\) is \(\angle TPR=38°\).
Apply the angle-sum of triangle \(PTR\): \(\angle PTR+\angle TPR+\angle TRP=180°\), i.e. \(2y+38°+52°=180°\).
Solve: \(2y=90°\), so \(y=45°\). The key tools are the equidistance property of a perpendicular bisector and the exterior-angle theorem.
This construction yields \(\angle PQR=45°\), while the option marked correct in the official answer key is 40°.
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