#### 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:

(i) ∠ORS = y°

(ii) Write an expression connecting x and y.

#### Solution

`∠`QRP = `∠`OSR = y (angles in alternate segment)

But OS = OR (Radii of the same circle)

∴`∠`ORS = `∠`OSR = y

∴ OQ = OR (radii of same circle)

∴ OQR = `∠` ORQ= 90° - y …………(i) (Since OR ⊥ PT )

But in Δ PQR ,

Ext `∠` OQR = x + y ……(i)

From (i) and (ii)

x + y = 90°- y

⇒ x + 2 y = 90°

Is there an error in this question or solution?

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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°; (Ii) Write an Expression Connecting X and Y. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.

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