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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°
Solution
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.
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