Question

In the given figure, PT touches a circle with centre O at R. Diameter SQ when produced to meet the tangent PT at P. If ∠SPR = x° and ∠QRP = y°; Show that x° + 2y° = 90°

Answer 1

PRT is tangent at R and QR is a chord.
∠QRP = ∠QSR     ...(Angle is an alternate segment)
∠QRP = y°
and ∠QSR = 90°   ...( QS is diameter and angle in a semicircle is right angle)

Now, in Δ PRS,
∠SPR + ∠PRS + ∠RSP = 180°
x° + y° + 90° + y° = 180°
x° + 2y° = 180° - 90°
x° + 2y° = 90°
Hence proved.