In the given figure, ‘ O’ is the centre of circle and AB = BC and ∠AOB = 90°, then ∠a is.
a) 35°
b) 55°
c) 45°
d) 65°
Sol:
∆AOB ∆COB
AB = BC
AO = CO
OB = common
⇒ ∠BOC = 90° as ∠AOB = 90°
⇒ AC = diameter ⇒ ∠ABC = 90°
∠ABO = 45°= ∠a
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