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Answer:
AOB=90
o
Join OC OC⊥AB
(Tangent at any pt of circle is far to radius)
∠ACO=∠BCO=90
o
In ΔAOP & ΔAOC
OP=OC
AP=AC (length of drawn from external point to circle are equal)
⇒ΔAOP≅ΔAOC
In ΔBOC & ΔBOQ
with same argument
ΔBOC≅ΔBOQ
∠BOC=∠BOQ (CPCT)
for line PQ
∠AOP+∠AOC+∠BOC+∠BOQ=180
2∠AOC=2∠BOC=180
⇒∠AOC+∠BOC=90
⇒∠AOB=90
hope it helps dear
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