$CD$, of length $14$ units, is a line fixed on the ground. $AB$ is a thin rod fixed in the air. A ring, which is attached to $AB$, can slide freely on $AB$. An elastic band (remaining taut all the time) passing through the ring is fixed to the ends $C$ and $D$ such that the triangle(s) so formed are always isosceles with the vertical angle $= \theta$, [$\theta$ remains as acute in the whole process.]. At some point $Q$ on $AB$, $∠AQD = 90^0$. Find the change of $\theta$ as the ring moves from $A$ to $Q$ and from $Q$ to $B$.
I found $\cos \theta = ... = 1 - \dfrac {98}{h^2} $