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Question
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find : ∠PBR
Solution
∠BPA = 90°
(Angle in a semicircle is a right angle)
∴ ∠BPQ = 90°
∴ ∠PBR = ∠BQP + ∠BPQ
= 25° + 90°
= 115°
(Exterior angle of a ∆ is equal to the sum of pair of interior opposite angles)
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