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(B) Given: Radius $r = \sqrt{2} \ cm$ and chord length $AB = 2 \ cm$. In $	riangle APB$,$AP = BP = r = \sqrt{2} \ cm$ and $AB = 2 \ cm$. Check if $

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$45$

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(B) Given: Radius $r = \sqrt{2} \ cm$ and chord length $AB = 2 \ cm$. In $	riangle APB$,$AP = BP = r = \sqrt{2} \ cm$ and $AB = 2 \ cm$. Check if $	riangle APB$ is a right-angled triangle: $AP^2 + BP^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$. Since $AB^2 = 2^2 = 4$,we have $AP^2 + BP^2 = AB^2$. By the converse of the Pythagoras theorem,$	riangle APB$ is a right-angled triangle with $\angle APB = 90^{\circ}$. The angle subtended by the chord $AB$ at the centre is $\angle APB = 90^{\circ}$. The angle subtended by the same chord at any point $C$ on the major arc is half the angle subtended at the centre. Therefore,$\angle ACB = \frac{1}{2} \angle APB = \frac{1}{2} 	imes 90^{\circ} = 45^{\circ}$.

In a circle with centre $P$,the length of a radius is $\sqrt{2} \ cm$. It is divided into two segments by chord $AB$ of length $2 \ cm$. If $C$ is a point on the major arc $AB$,then find the value of $\angle ACB$. (in $^{\circ}$)

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