Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°
In ΔPQR:
∠PQR = ∠PRQ = 35° (Angle opposite to equal sides)
Again, by angle sum property
∠P + ∠Q + ∠R = 180°
∠P + 35° + 35° = 180°
∠P + 70° = 180°
∠P = 180° – 70°
∠P = 110°
Now, in quadrilateral SQTR,
∠QSR + ∠QTR = 180° (Opposite angles of quadrilateral)
110° + ∠QTR = 180°
∠QTR = 70°