The diagram shows two rods $AB$ and $BC$ of length 2 m and 5 m respectively, joined together at $B$ such that angle $ABC = 90^\circ$. Small rings are attached to $A$ and $C$ so that $A$ can move along the vertical pole $OY$ and $C$ can move along the horizontal pole $OX$. The rod $BC$ makes an angle $\theta$ with $OX$. (iii) Given that $A$ can be no higher than 3.5 m above $O$, calculate the greatest possible value of $\theta$.

1. Find the vertical distance from $B$ to the horizontal axis $OX$.

Let $D$ be the foot of the perpendicular from $B$ to $OX$. In the right-angled triangle $BDC$, the angle $BCD$ is $\theta$.
$BD = BC \sin\theta = 5\sin\theta$

2. Find the vertical distance from $B$ to $A$ and hence find $OA$.

Let $E$ be the foot of the perpendicular from $A$ to the vertical line $BD$. Since $\angle ABC = 90^\circ$ and the angle of $BC$ with the horizontal is $\theta$, the angle $AB$ makes with the vertical is $\theta$.
In the right-angled triangle $AEB$, the vertical distance $BE$ is given by:
$BE = AB \cos\theta = 2\cos\theta$
Since $A$ is vertically below $B$, the height of $A$ above $O$ is:
$OA = BD - BE = 5\sin\theta - 2\cos\theta$

1. Find the value of $R$

$R = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$

2. Find the value of $\alpha$

$\tan\alpha = \frac{2}{5}$
$\alpha = \tan^{-1}\left(\frac{2}{5}\right) \approx 21.801^\circ \approx 21.8^\circ$ (to 1 d.p.)

3. Express $OA$ in the required form

$OA = \sqrt{29}\sin(\theta - 21.8^\circ)$