Q16 of 62 Page 193

If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of 40° then ∠POA =


In ΔAOP and ΔBOP,


OP = OP {Common}


PB = PA {Tangents from the same external point are equal}


OB =OA {Radii}


⇒ ΔAOP ΔBOP


⇒ ∠ BPO = ∠ APO {Corresponding parts of congruent triangles}


Given: ∠ BPA = 40°


⇒ ∠ APO=20°


Also, ∠ OAP = 90° {a tangent at any point on a circle is perpendicular to the radius through the point of contact}


So, By angle sum property


∠ POA = 180° – 90° – 20°


⇒ ∠ POA = 70°

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