In figure AB=RS , BQ= RP AND ANGLE R = ANGLE
Answers
Answered by
1
Answer:
(i) ∠PRB=∠BAP=35
o
(Angles in the same segment of the circle)
(ii) ∠BPA=90
o
(angle in a semicircle)
∴∠BPQ=90
o
∴∠PBR=∠BQP+∠BPQ=25
o
+90
o
=115
o
(exterior angle in △BPQ)
(iii) ∠ABP=90
o
−∠BAP=90
o
−35
o
=55
o
∴∠ABR=∠PBR=∠ABP=115
o
−55
o
=60
o
∴∠APR=∠ABR=60
o
(Angle in the same segment of circle)
Hence, ∠BPR=90
o
−∠APR=90
o
−60
o
=30
o
Similar questions