In the given figure, lines PQ and RS intersect at point T, such that
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°. In ∆PRT
∠PRT + ∠RPT + ∠PTR = 180°
[Angle sum property of a triangle]
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 180° – 135° = 45°
Also, ∠PTR = ∠STQ
[Vertical opposite angles]
∴ ∠STQ = 45°
Now, in ∆STQ,
∠STQ + ∠TSQ + ∠SQT = 180° [Angle sum property of a triangle]
⇒ 45° + 75° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 180° – 120° = 60°
Hence, ∠SQT = 60° Ans.
