In a \(\triangle ABC\), the lengths of the two larger sides are 10 and 9 units respectively. If the angles are in A.P., then the length of the third side can be
Step 1: If angles are in A.P., the middle angle is \(60^\circ\).
Step 2: By cosine rule:
\[
a^2=10^2+9^2-2(10)(9)\cos60^\circ
\]
\[
a^2=181-90=91
\Rightarrow a=\sqrt{91}=5\pm\sqrt6
\]