在三角形 $ABC$ 中，$AB = 2$，$BC = \sqrt{3}$，$AC = x$ 且角 $ABC = \frac{\pi}{6} + \theta$（见图）。

1. Apply the Cosine Rule to triangle ABC to express $x^2$ in terms of $\theta$.

$x^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3})\cos\left(\frac{\pi}{6} + \theta\right)$
$x^2 = 4 + 3 - 4\sqrt{3}\cos\left(\frac{\pi}{6} + \theta\right)$
$x^2 = 7 - 4\sqrt{3}\cos\left(\frac{\pi}{6} + \theta\right)$

2. Use the compound angle formula to expand the cosine term and simplify.

Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$x^2 = 7 - 4\sqrt{3} \left[ \cos\frac{\pi}{6}\cos\theta - \sin\frac{\pi}{6}\sin\theta \right]$
Since $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$:
$x^2 = 7 - 4\sqrt{3} \left( \frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta \right)$
$x^2 = 7 - 4\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos\theta\right) + 4\sqrt{3}\left(\frac{1}{2}\sin\theta\right)$
$x^2 = 7 - 6\cos\theta + 2\sqrt{3}\sin\theta$ (shown)

1. Differentiate the result from part (a) implicitly with respect to $\theta$.

From part (a), we have:
$x^{2} = 7 - 6\cos\theta + 2\sqrt{3}\sin\theta$

Differentiating both sides implicitly with respect to $\theta$:
$2x\frac{dx}{d\theta} = 6\sin\theta + 2\sqrt{3}\cos\theta$

2. Simplify and differentiate again using the product rule to obtain the required identity.

Dividing the equation by 2 gives:
$x\frac{dx}{d\theta} = 3\sin\theta + \sqrt{3}\cos\theta$

Differentiating both sides again with respect to $\theta$ using the product rule on the left-hand side:
$\left(\frac{dx}{d\theta}\right)\left(\frac{dx}{d\theta}\right) + x\left(\frac{d^{2}x}{d\theta^{2}}\right) = 3\cos\theta - \sqrt{3}\sin\theta$

$x\frac{d^{2}x}{d\theta^{2}} + \left(\frac{dx}{d\theta}\right)^{2} = 3\cos\theta - \sqrt{3}\sin\theta \quad \text{(shown)}$

1. Evaluate $x$, $\frac{dx}{d\theta}$, and $\frac{d^2x}{d\theta^2}$ at $\theta = 0$ using results from earlier parts.

At $\theta = 0$, from the expression in part (a):
$x^2 = 7 - 6\cos 0 + 2\sqrt{3}\sin 0 = 1$
Since $x$ is a length, $x = 1$.

Differentiating the equation in part (a) gives an intermediate result needed for part (b):
$2x\frac{dx}{d\theta} = 6\sin\theta + 2\sqrt{3}\cos\theta \implies x\frac{dx}{d\theta} = 3\sin\theta + \sqrt{3}\cos\theta$
At $\theta = 0$:
$1 \cdot \frac{dx}{d\theta} = 3(0) + \sqrt{3}(1) \implies \frac{dx}{d\theta} = \sqrt{3}$

From the equation proven in part (b):
$x\frac{d^2x}{d\theta^2} + \left(\frac{dx}{d\theta}\right)^2 = 3\cos\theta - \sqrt{3}\sin\theta$
At $\theta = 0$:
$1 \cdot \frac{d^2x}{d\theta^2} + (\sqrt{3})^2 = 3(1) - 0 \implies \frac{d^2x}{d\theta^2} = 0$

2. Differentiate the equation from part (b) with respect to $\theta$ to find an expression for $\frac{d^3x}{d\theta^3}$.

$\frac{d}{d\theta}\left[x\frac{d^2x}{d\theta^2} + \left(\frac{dx}{d\theta}\right)^2\right] = \frac{d}{d\theta}\left[3\cos\theta - \sqrt{3}\sin\theta\right]$
Using the product rule and chain rule:
$x\frac{d^3x}{d\theta^3} + \frac{dx}{d\theta}\frac{d^2x}{d\theta^2} + 2\frac{dx}{d\theta}\frac{d^2x}{d\theta^2} = -3\sin\theta - \sqrt{3}\cos\theta$
$x\frac{d^3x}{d\theta^3} + 3\frac{dx}{d\theta}\frac{d^2x}{d\theta^2} = -3\sin\theta - \sqrt{3}\cos\theta$

3. Evaluate $\frac{d^3x}{d\theta^3}$ at $\theta = 0$.

Substitute $\theta = 0$ into the differentiated equation:
$1 \cdot \frac{d^3x}{d\theta^3} + 3(\sqrt{3})(0) = -3(0) - \sqrt{3}(1)$
$\frac{d^3x}{d\theta^3} = -\sqrt{3}$

4. Construct the Maclaurin expansion for $x$ up to the term in $\theta^3$.

The Maclaurin series is given by:
$x(\theta) \approx x(0) + x'(0)\theta + \frac{x''(0)}{2!}\theta^2 + \frac{x'''(0)}{3!}\theta^3$
Substituting the values obtained:
$x \approx 1 + \sqrt{3}\theta + \frac{0}{2!}\theta^2 + \frac{-\sqrt{3}}{3!}\theta^3$
$x \approx 1 + \sqrt{3}\theta - \frac{\sqrt{3}}{6}\theta^3$

1. Find the Maclaurin expansion for $x$ up to $\theta^3$ as required from part (c).

Evaluate $x$ and its derivatives at $\theta = 0$:
From (a), $x^2 = 7 - 6\cos(0) + 2\sqrt{3}\sin(0) = 1 \implies x(0) = 1$ (since length $x > 0$).
Differentiating (a) implicitly: $2x \frac{dx}{d\theta} = 6\sin\theta + 2\sqrt{3}\cos\theta$.
At $\theta = 0$: $2(1)\frac{dx}{d\theta} = 2\sqrt{3} \implies x'(0) = \sqrt{3}$.
From (b), $x\frac{d^{2}x}{d\theta^{2}} + \left(\frac{dx}{d\theta}\right)^{2} = 3\cos\theta - \sqrt{3}\sin\theta$.
At $\theta = 0$: $1 \cdot x''(0) + (\sqrt{3})^2 = 3(1) - 0 \implies x''(0) = 0$.
Differentiating (b): $x\frac{d^{3}x}{d\theta^{3}} + 3\frac{dx}{d\theta}\frac{d^{2}x}{d\theta^{2}} = -3\sin\theta - \sqrt{3}\cos\theta$.
At $\theta = 0$: $1 \cdot x'''(0) + 3(\sqrt{3})(0) = -\sqrt{3} \implies x'''(0) = -\sqrt{3}$.

Thus, the Maclaurin series for $x$ is:
$x \approx x(0) + x'(0)\theta + \frac{x''(0)}{2!}\theta^2 + \frac{x'''(0)}{3!}\theta^3$
$x \approx 1 + \sqrt{3}\theta - \frac{\sqrt{3}}{6}\theta^3$

2. Determine the value of $\theta$ in radians corresponding to angle $ABC = 35^{\circ}$.

Given angle $ABC = \frac{\pi}{6} + \theta$.
Since angle $ABC = 35^{\circ}$ and $\frac{\pi}{6}$ radians is equivalent to $30^{\circ}$:
$30^{\circ} + \theta = 35^{\circ} \implies \theta = 5^{\circ}$
Convert $\theta$ to radians (required for Maclaurin expansions of trigonometric functions):
$\theta = \frac{5\pi}{180} = \frac{\pi}{36}$ radians.

3. Approximate $x$ by substituting the value of $\theta$ into the Maclaurin series.

Substituting $\theta = \frac{\pi}{36}$ into the expansion:
$x \approx 1 + \sqrt{3}\left(\frac{\pi}{36}\right) - \frac{\sqrt{3}}{6}\left(\frac{\pi}{36}\right)^3$
$x \approx 1 + 0.15114995 - 0.00019186$
$x \approx 1.15095809$
$\therefore x \approx 1.150958 \text{ (correct to 6 d.p.)}$