In figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

∠QRT = ∠RQS + ∠QSR
| ∵ The exterior angle is equal to the sum of the two interior opposite angles. ⇒    65°
= 28° + ∠QSR
⇒ ∠QSR = 65° - 28° = 37°
∵ PQ ≠ SP
∴ ∠QPS = 90°
∵ PQ || SR
∴ ∠QPS + ∠PSR = 180°
| ∵ The sum of consecutive interior angles on the same side of the transversal is 180°
⇒ 90° + ∠PSR = 180°
⇒    ∠PSR = 180° - 90° = 90°
⇒ ∠PSQ + ∠QSR = 90°
⇒    y + 37° = 90°
⇒    y = 90° - 37° = 53°
In ∆PQS,
∠PQS + ∠QSP + ∠QPS = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ x + y + 90° = 180°
⇒ x + 53° + 90° = 180°
⇒    x + 143° = 180°
⇒ x = 180° - 143° = 37°.