The claw of a claw machine is designed such that $AB = BC = CD = DE = EF = k\text{ cm}$ with angle $ABC = \text{angle } FED = 90^\circ$. The two sides of the claw, $ABC$ and $FED$, hinge at $C$ and $D$ respectively at angle $\theta^\circ$ with $\theta > 45$. The claw closes when $A$ and $F$ touches one another.

1. Determine the horizontal distance from C to B

Let the horizontal line through $C$ and $D$ be the reference line. The horizontal distance from $C$ to $B$ is $k \cos \theta$.

2. Determine the horizontal distance from B to A

Since $\angle ABC = 90^\circ$, the angle $AB$ makes with the horizontal is $90^\circ - \theta$. The horizontal distance from $B$ to $A$ (moving rightwards) is $k \cos(90^\circ - \theta) = k \sin \theta$.

3. Calculate the total horizontal distance between A and F

The horizontal position of $A$ relative to $C$ is $-k \cos \theta + k \sin \theta$. By symmetry, the horizontal position of $F$ relative to $D$ is $k \cos \theta - k \sin \theta$. The distance $CD = k$. Thus, the distance $AF = k - 2(k \sin \theta - k \cos \theta) = k - 2k(\sin \theta - \cos \theta)$.

1. Expand $R\sin(\theta - \alpha)$ and equate coefficients

$R\sin(\theta - \alpha) = R(\sin\theta\cos\alpha - \cos\theta\sin\alpha)$
Comparing with $\sin\theta - \cos\theta$:
$R\cos\alpha = 1$
$R\sin\alpha = 1$

2. Find the value of $R$

$R = \sqrt{1^2 + 1^2} = \sqrt{2}$

3. Find the value of $\alpha$ and write the final expression

$\tan\alpha = \frac{1}{1} = 1$
Since $0^\circ < \alpha < 90^\circ$, $\alpha = 45^\circ$
$\therefore \sin\theta - \cos\theta = \sqrt{2}\sin(\theta - 45^\circ)$

1. Set up the equation for when the claw closes

When the claw closes, $A$ and $F$ touch, so $AF = 0$.
$k - 2k(\sin\theta - \cos\theta) = 0$
$1 - 2(\sin\theta - \cos\theta) = 0$
$\sin\theta - \cos\theta = \frac{1}{2}$

2. Substitute the R-formula expression from part (ii)

From part (ii), $\sin\theta - \cos\theta = \sqrt{2}\sin(\theta - 45^\circ)$.
$\sqrt{2}\sin(\theta - 45^\circ) = \frac{1}{2}$
$\sin(\theta - 45^\circ) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

3. Solve for theta

Basic angle $\alpha = \sin^{-1}\left(\frac{1}{2\sqrt{2}}\right) \approx 20.7048^\circ$
Since $\theta > 45^\circ$, $\theta - 45^\circ > 0$.
$\theta - 45^\circ = 20.7048^\circ$
$\theta = 65.7048^\circ \approx 65.7^\circ$