图中显示了两根长度分别为 2 米和 5 米的细杆 $AB$ 和 $BC$，它们在 $B$ 处连接，使得角 $ABC = 90^\circ$。小环附着在 $A$ 和 $C$ 上，使 $A$ 可以沿垂直杆 $OY$ 移动，而 $C$ 可以沿水平杆 $OX$ 移动。细杆 $BC$ 与 $OX$ 夹角为 $\theta$。 (iii) 已知 $A$ 距离 $O$ 的高度不能超过 3.5 米，计算 $\theta$ 的最大可能值。

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

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