Question 6:
In the given figure, the side QR of ∆PQR is produced to a point S. If the bisectors of
∠PQR and ∠PRS meet at point T, then
Answer:
In ∆QTR, ∴TRS is an exterior angle.
∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS − ∠TQR (1)
For ∆PQR, ∠PRS is an external angle.
∠QPR + ∠PQR = ∠PRS
∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)
∠QPR = 2(∠TRS − ∠TQR)
∠QPR = 2∠QTR [By using equation (1)]
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