Solution :-
Given: A chord AB of length 2cm and radius of the circle is √2cm
To prove: ∠ACB = 45^{0}
Proof: In △AOB,
OA^{2 }+ OB^{2} = (√2)^{2} + (√2)^{2} = 2 + 2 = 4 = AB^{2}
⇒ △AOB is a right triangle right angled at O.
i.e. ∠AOB = 90^{0}
As the angle subtended by an arc at the centre is double the angle subtended by it at remaining part of the circle.
∴ ∠AOB = 2∠ACB
⇒ ∠ACB = 1/2 x 90^{0} = 45^{0}