In figure XP and XQ are two tangents

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to

(A) 50°    (B) 60°
(C) 70°    (D) 80°.

In the given figure, we have an external point X from where two tangents, XP and XQ, are drawn to the circle.XP = XQ    (The lengths of the tangents drawn from an external point to the circle are equal.)Similarly, we have:AP = ARBQ = BRNow, XP = XA + AP        ...(1)XQ = XB + BQ                ...(2)On putting AP = AR in equation (1) and BQ = BR in equation (2), we get:XP = XA + AR XQ = XB + BRSince XP and XQ are equal, we have:XA + AR = XB + BR

Hence, proved.

In figure, XP and XQ are two tangents to a circle with center O from a point X outside the circle. ARB is tangent to circle at R. Prove that XA + AR = XB+ BR.

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