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(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB=5$,$AD=2$,$BC=3$,and $\angle DAB = 60^{\circ}$.In $	riangle ABD$,by the Law of Cosines:$BD^

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(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB=5$,$AD=2$,$BC=3$,and $\angle DAB = 60^{\circ}$.In $	riangle ABD$,by the Law of Cosines:$BD^{2} = AB^{2} + AD^{2} - 2(AB)(AD)\cos(60^{\circ})$$BD^{2} = 5^{2} + 2^{2} - 2(5)(2)(\frac{1}{2})$$BD^{2} = 25 + 4 - 10 = 19$.Since $ABCD$ is a cyclic quadrilateral,the opposite angles sum to $180^{\circ}$. Thus,$\angle BCD = 180^{\circ} - 60^{\circ} = 120^{\circ}$.In $	riangle BCD$,let $CD = x$. By the Law of Cosines:$BD^{2} = BC^{2} + CD^{2} - 2(BC)(CD)\cos(120^{\circ})$$19 = 3^{2} + x^{2} - 2(3)(x)(-\frac{1}{2})$$19 = 9 + x^{2} + 3x$$x^{2} + 3x - 10 = 0$$(x+5)(x-2) = 0$Since $x$ must be positive,$x = 2$.

If two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the third side is $3$,then the remaining fourth side is :-

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