The two adjacent sides of a cyclic quadrilate | vedclass

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(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB = 2$,$BC = 5$,$CD = 3$,and $DA = d$. The angle $\angle ABC = 60^o$.Since $ABCD$ is a cyclic

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$2$

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(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB = 2$,$BC = 5$,$CD = 3$,and $DA = d$. The angle $\angle ABC = 60^o$.Since $ABCD$ is a cyclic quadrilateral,the opposite angles are supplementary. Therefore,$\angle ADC = 180^o - 60^o = 120^o$.Using the law of cosines in $	riangle ABC$ for the diagonal $AC$:$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(60^o) = 2^2 + 5^2 - 2(2)(5)(0.5) = 4 + 25 - 10 = 19$.Using the law of cosines in $	riangle ADC$ for the diagonal $AC$:$AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos(120^o) = d^2 + 3^2 - 2(d)(3)(-0.5) = d^2 + 9 + 3d$.Equating the two expressions for $AC^2$:$d^2 + 3d + 9 = 19 \Rightarrow d^2 + 3d - 10 = 0$.Factoring the quadratic equation:$(d + 5)(d - 2) = 0$.Since the side length $d$ must be positive,$d = 2$.

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^o$. If the third side is $3$,the remaining fourth side is

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