a) ZPRB, AR35° O
29. AB is a diameter of the circle APBR as shown
in the figure. APQ and RBQ are straight lines.
Find :
P
(ii) ZPBR
25°
(iii) ZBPR.
.
SO
Q
B.
Po-125
55
R
33+55+2=114
180-
Answers
Answered by
1
Answer:
good question ok bhut badiya
Answered by
3
Answer:
∠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
Step-by-step explanation:
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