Q10 of 31 Page 234

, touch ⨀(O, 15) at A and B. If m∠AOB = 80, then m∠OPB =

Given PA and PB touch circle (O, 15) at A and B and m ∠AOB = 80°.



Here, ΔPOA and ΔPOB are congruent right angled triangle.


⇒ ∠BOP = 1/2 ∠AOB


= 1/2 × 80°


= 40°


In right angled ΔOBP,


We know that sum of angles in a triangle is 180°.


⇒ ∠BOP + ∠B + ∠OPB = 180°


⇒ 40 + 90 + ∠OPB = 180°


⇒ 130 + ∠OPB = 180°


∴ ∠OPB = 50°

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