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:

1. ∠ORS = y°
2. write an expression connecting x and y.

Answer 1

∠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°