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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Important Questions from Triangles, Congruence and Similarity
Angle between the internal bisectors of two angles ∠B and ∠C of a ΔABC is 132°, then the value of ∠A is
In ΔPQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
In Δ ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:
Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm. then find the measure of AB (in cm).
The base and altitude of an isosceles triangle are 10 cm and 12 cm respectively. Then the length of each equal side is:
