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(C) Let the angles of the triangle be $A, B, C$ in $A.P.$ such that $A \le B \le C$. Since $A + B + C = 180^{\circ}$ and $2B = A + C$,we have $3B = 18

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$5 - \sqrt{6}$ or $5 + \sqrt{6}$

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(C) Let the angles of the triangle be $A, B, C$ in $A.P.$ such that $A \le B \le C$. Since $A + B + C = 180^{\circ}$ and $2B = A + C$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.Given the sides are $10 \, cm$ and $9 \, cm$,and $B = 60^{\circ}$ is the middle angle,the side opposite to $B$ (let it be $b$) must be the middle side. Since $10$ and $9$ are the two larger sides,$b$ must be $9 \, cm$ and the largest side $a = 10 \, cm$.Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting the values: $\cos 60^{\circ} = \frac{10^2 + c^2 - 9^2}{2(10)(c)} \Rightarrow \frac{1}{2} = \frac{100 + c^2 - 81}{20c}$.$10c = 19 + c^2 \Rightarrow c^2 - 10c + 19 = 0$.Solving for $c$ using the quadratic formula: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2} = 5 \pm \sqrt{6}$.Since both values are positive and satisfy the triangle inequality,the third side can be $5 - \sqrt{6}$ or $5 + \sqrt{6}$.

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

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