In a triangle \(\mathrm{ABC}, \mathrm{m} \angle \mathrm{A}, \mathrm{m} \angle \mathrm{B}, \mathrm{m} \angle \mathrm{C}\) are in A.P. and lengths of two larger sides are 10 units, 9 units respectively, then the length (in units) of the third side is

\(\angle \mathrm{A}, \angle \mathrm{B}, \angle \mathrm{C}\) are in A.P.
\(\begin{aligned}
& \Rightarrow 2 \mathrm{~B}=\mathrm{A}+\mathrm{C} \\
& \Rightarrow 3 \mathrm{~B}=\mathrm{A}+\mathrm{B}+\mathrm{C} \\
& \Rightarrow 3 \mathrm{~B}=180^{\circ} \\
& \Rightarrow \mathrm{B}=60^{\circ} \\
& \cos \mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}
\end{aligned}\)
\(\Rightarrow \cos 60^{\circ}=\frac{\mathrm{c}^2+10^2-9^2}{2 \mathrm{c}(10)}\)
\(\ldots[\) Let \(a=10, b=9]\)
\(\begin{aligned} & \Rightarrow \frac{1}{2}=\frac{\mathrm{c}^2+100-81}{20 \mathrm{c}} \\ & \Rightarrow 10 \mathrm{c}=\mathrm{c}^2+19 \\ & \Rightarrow \mathrm{c}^2-10 \mathrm{c}+19=0 \\ & \Rightarrow \mathrm{c}=\frac{10 \pm \sqrt{100-76}}{2} \\ & =\frac{10 \pm 2 \sqrt{6}}{2} \\ & =5 \pm \sqrt{6}\end{aligned}\)