In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that 
(i) PT = QT
(ii) ∠TQR = 15°

Since PQRS is a square
∠PSR = ∠QRS ...... (each 90°)
Now again in equilateral ΔSRT,  
we have ∠TSR = ∠TRS ...... (each 60°)
∠PSR + ∠TSR = ∠QRS + ∠TRS
       ⇒ ∠TSP = ∠TRQ
Now in ΔTSP and ΔTRQ, we have
             TS = TR ...... (sides of equilateral triangle)
         ∠TSP = ∠TRQ ...... (Proved above)
             PS = QR ...... (sides of a square)
          ΔTSP ≅ ΔTRQ
        ∴ PT = QT 
Now in ΔTQR, we have
TR = QR ...... (QR = RS = TR)
∠TQR = ∠QTR and ∠TQR + ∠QTR + ∠TRQ = 180°
             ∠TQR + ∠QTR + ∠TRS + ∠SRQ = 180°
                           2 ∠TQR + 60° + 90° = 180°...........(∵∠TQR = ∠QTR)
                         2 ∠TQR = 180° - 150° = 30°
                                       ∠TQR =  30  = 15°
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