Question

HP A 19 In fig AB : is Diametere find LAPB
​

Answer 1

Answer:

(i) ∠PRB=∠BAP=35o(Angles in the same segment of the circle)

(ii) ∠BPA=90o(angle in a semicircle)

∴∠BPQ=90o

∴∠PBR=∠BQP+∠BPQ=25o+90o=115o(exterior angle in △BPQ)

(iii) ∠ABP=90o−∠BAP=90o−35o=55o

∴∠ABR=∠PBR=∠ABP=115o−55o=60o

∴∠APR=∠ABR=60o(Angle in the same segment of circle)

Hence, ∠BPR=90o−∠APR=90o−60o=30o