Question

In the figure, AB is a diameter. APQ and RBQ are straight lines. Find angle BPR.​

Answer 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