In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = 72∘, find ∠PTR.
∠OQR = ∠ORQ (OR=OQ (radius))
But ∠ROP = ∠OQR + ∠ORQ (Exterior angle theorem)
72∘ = 2 ∠OQR
∠OQR = 36∘
∠PTQ = 180∘ - ∠PQT - ∠QPT
∠PTQ = 180∘ – 36∘ - 90∘
∠PTQ = 54∘
Therefore, ∠PTR = 54∘
In figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If ∠POR=130∘ and S is a point on the circle, find ∠1+∠2
In the given figure, PT is a tangent of a circle, with centre O, at point R. If diameter SQ is produced, it meets with PT at point P with ∠SPR=x and ∠QSR=y,then find the value of x+2y.