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Question
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P. Given ∠SPR = x° and ∠QRP = y°;
Prove that:
- ∠ORS = y°
- write an expression connecting x and y.
Solution
∠QRP = ∠OSR = y ...(Angles in the alternate segment)
But OS = OR ...(Radii of the same circle)
∴ ∠ORS = ∠OSR = y°
∴ OQ = OR ...(Radii of the same circle)
∴ ∠OQR = ∠ORQ = 90° – y° ...(i) (Since OR ⊥ PT)
But in ΔPQR,
Ext. ∠OQR = x° + y° ...(i)
From (i) and (ii)
x° + y° = 90° – y°
`=>` x° + 2y° = 90°
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