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.**