如图所示，$OC = 4\text{ m}$，$OD = 7\text{ m}$ 且角 $CBO = \text{角 } DOC = \text{角 } DAO = 90^\circ$。
已知角 $COB = \theta$，其中 $0^\circ \le \theta \le 90^\circ$。

1. Find the angle $DOA$ in terms of $\theta$.

Since $A, O, B$ lie on a straight line, the sum of angles at $O$ is $180^\circ$.
$\angle DOA + \angle DOC + \angle COB = 180^\circ$
$\angle DOA + 90^\circ + \theta = 180^\circ$
$\angle DOA = 90^\circ - \theta$

2. Express the lengths $OA$ and $OB$ in terms of $\theta$.

In the right-angled triangle $ODA$:
$\cos(\angle DOA) = \frac{OA}{OD}$
$\cos(90^\circ - \theta) = \frac{OA}{7}$
Since $\cos(90^\circ - \theta) = \sin \theta$, we have $OA = 7 \sin \theta$.

In the right-angled triangle $OCB$:
$\cos \theta = \frac{OB}{OC}$
$\cos \theta = \frac{OB}{4}$
$OB = 4 \cos \theta$

3. Find the total length $AB$.

$AB = OA + OB$
$AB = 7 \sin \theta + 4 \cos \theta$

1. Use the result from part (a) and equate it to the expanded form of $R \sin(\theta + \alpha)$

From part (a), $AB = 7 \sin \theta + 4 \cos \theta$.
Let $7 \sin \theta + 4 \cos \theta = R \sin(\theta + \alpha) = R \sin \theta \cos \alpha + R \cos \theta \sin \alpha$.
Comparing coefficients:
$R \cos \alpha = 7$
$R \sin \alpha = 4$

2. Calculate the value of $R$

$R = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65}$

3. Calculate the value of $\alpha$

$\tan \alpha = \frac{4}{7}$
$\alpha = \tan^{-1}\left(\frac{4}{7}\right) \approx 29.7448^\circ \approx 29.7^\circ$ (to 1 d.p.)

$AB = \sqrt{65} \sin (\theta + 29.7^\circ)$

1. Determine the values of $R$ and $\alpha$ from the expression $AB = 7 \sin \theta + 4 \cos \theta$ (derived in parts a and b)

$R = \sqrt{7^2 + 4^2} = \sqrt{65}$
$\tan \alpha = \frac{4}{7} \implies \alpha = \tan^{-1}\left(\frac{4}{7}\right) \approx 29.745^\circ$
So, $AB = \sqrt{65} \sin(\theta + 29.745^\circ)$

2. State the maximum length of $AB$

The maximum value of $\sin(\theta + 29.745^\circ)$ is $1$.
Maximum length of $AB = \sqrt{65} \approx 8.06\text{ m}$

3. Find the corresponding value of $\theta$

$\sin(\theta + 29.745^\circ) = 1$
$\theta + 29.745^\circ = 90^\circ$
$\theta = 90^\circ - 29.745^\circ = 60.255^\circ \approx 60.3^\circ$

1. Set up the equation using the $R$-formula expression from part (b)

From part (b), $AB = \sqrt{65} \sin(\theta + 29.745^\circ)$.
Equating this to $\sqrt{55}$:
$\sqrt{65} \sin(\theta + 29.745^\circ) = \sqrt{55}$
$\sin(\theta + 29.745^\circ) = \sqrt{\frac{55}{65}} = \sqrt{\frac{11}{13}}$

2. Find the possible values for the angle $(\theta + \alpha)$ in the required range

The basic angle is $\sin^{-1}\left(\sqrt{\frac{11}{13}}\right) \approx 66.906^\circ$.
Since $\sin$ is positive, the angle can be in the first or second quadrant.
$\theta + 29.745^\circ = 66.906^\circ \text{ or } 180^\circ - 66.906^\circ$
$\theta + 29.745^\circ = 66.906^\circ \text{ or } 113.094^\circ$

3. Solve for $\theta$ and round to 1 decimal place

$\theta = 66.906^\circ - 29.745^\circ \text{ or } 113.094^\circ - 29.745^\circ$
$\theta = 37.161^\circ \text{ or } 83.349^\circ$
$\theta \approx 37.2^\circ \text{ or } 83.3^\circ$