Answer:
Given, a circle with centre O and tangents QR and QP from point Q and points of contact to the circle R and P, respectively.
OR ⊥ QR (radius is perpendicular to tangent)
=> ∠ORQ = 90°
Also, OP ⊥ PQ (radius is perpendicular to tangent)
=> ∠OPQ = 90°
In quadrilateral ORQP,
∠OPQ + ∠PQR + ∠QRO + ∠ROP = 360° (angle sum property)
=> 90° + 50° + 90° + ∠ROP = 360°
=> ∠ROP = 360° - 230°
=> ∠ROP = 130°
Reflex ∠POR = 360° - 130° = 230° .... (1)
∠ROP = 2∠PSR (angle subtended at the centre of the circle is twice the angle subtended at any other point on the circle)
=> ∠PSR = [tex]\frac{130}{2}[/tex]
=> ∠PSR = 65°
In quadrilateral PORS,
∠POR + ∠ORS + ∠RSP + ∠SPO = 360° (angle sum property)
230° + 20° + 65° + ∠SPO = 360° (from (1))
∠SPO = 360° - 315°
∠SPO = 45°
x + ∠SPO = 90° (∵ OP ⊥ PQ)
X + 45° = 90°
X = 45°
∴ The value of x is 45°.