In a A B C , the lengths of the two larger sides are 10 and 9...

Let A, B and $C$ be the three angles of $\Delta A B C$ and Let $a=10$ and $b=9$ It is given that the angles are in $\mathrm{AP}$. $\therefore 2 \mathrm{ B}=\mathrm{A}+\mathrm{C}$ on adding $\mathrm{B}$ both the sides, we get $$\begin{array}{l}3 \mathrm{ B}=\mathrm{A}+\mathrm{B}+\mathrm{C}\\ \Rightarrow 3 \mathrm{ B}=180^{{}^{\circ}} \Rightarrow \mathrm{B}=60^{{}^{\circ}} \end{array}$$Now, we know \cos \mathrm{B}=\frac{\mathrma^{2}+\mathrmc^{2}-\mathrmb^{2}}{2 \mathrm{ac}} $$\begin{array}{l} \Rightarrow \cos60^{{}^{\circ}}=\frac{10^{2}+\mathrmc^{2}-9^{2}}{2 \times 10 \times \mathrm{c}}\\ \Rightarrow \frac{1}{2}=\frac{100+\mathrmc^{2}-81}{20 \mathrm{c}}\\ \Rightarrow \mathrmc^{2}-10 \mathrm{c}+19=0 \Rightarrow \mathrm{c}=5 \pm \sqrt{6} \end{array}$$$