Q3 of 141 Page 420

In the given figure, is the center of the circle. If and find the value of

∠BPC + ∠APB = 180°[Because APC is a straight line]


⇒ ∠BPC + 110° = 180°


⇒ ∠BPC = 70°


In triangle BPC,


∠BPC + ∠PBC + ∠PCB = 180°[Sum of angles of triangle]


⇒ 70°+ 25° + ∠PCB = 180°


⇒ ∠PCB = 85°


∴ ∠ADB = ∠PCB = 85°[Angles in the same segment of a circle]


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