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

1. Apply the cosine rule to triangle $ABC$ to find an initial expression for $x^2$.

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

2. Expand the trigonometric expression using the addition formula and simplify to the target form.

Using the addition formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$x^2 = 7 - 4\sqrt{3}\left(\cos\frac{\pi}{6}\cos\theta - \sin\frac{\pi}{6}\sin\theta\right)$
Substituting the exact trigonometric ratios $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$:
$x^2 = 7 - 4\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta\right)$
$x^2 = 7 - 4\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\cos\theta + 4\sqrt{3}\left(\frac{1}{2}\right)\sin\theta$
$x^2 = 7 - 6\cos\theta + 2\sqrt{3}\sin\theta$

1. Find the values of $x$ and its first derivative at $\theta = 0$.

From part (a), $x^{2} = 7 - 6\cos\theta + 2\sqrt{3}\sin\theta$.
At $\theta = 0$:
$x(0)^2 = 7 - 6(1) + 0 = 1$
Since $x$ is a side length of a triangle, $x > 0$, so $x(0) = 1$.

Differentiating implicitly with respect to $\theta$:
$2x \frac{dx}{d\theta} = 6\sin\theta + 2\sqrt{3}\cos\theta$
Substitute $\theta = 0$ and $x(0) = 1$:
$2(1) \frac{dx}{d\theta} = 6(0) + 2\sqrt{3}(1) \implies \frac{dx}{d\theta}\bigg|_{\theta=0} = \sqrt{3}$

2. Use the equation from part (b) to evaluate the second derivative at $\theta = 0$.

From part (b):
$x\frac{d^{2}x}{d\theta^{2}} + \left(\frac{dx}{d\theta}\right)^{2} = 3\cos\theta - \sqrt{3}\sin\theta$
Substitute $\theta = 0$, $x(0) = 1$, and $\frac{dx}{d\theta}\Big|_{\theta=0} = \sqrt{3}$:
$1 \cdot \frac{d^{2}x}{d\theta^{2}}\bigg|_{\theta=0} + (\sqrt{3})^2 = 3(1) - 0$
$\frac{d^{2}x}{d\theta^{2}}\bigg|_{\theta=0} + 3 = 3 \implies \frac{d^{2}x}{d\theta^{2}}\bigg|_{\theta=0} = 0$

3. Differentiate the equation from part (b) to find the third derivative at $\theta = 0$.

Differentiate the equation from part (b) implicitly with respect to $\theta$:
$\frac{dx}{d\theta}\frac{d^{2}x}{d\theta^{2}} + x\frac{d^{3}x}{d\theta^{3}} + 2\left(\frac{dx}{d\theta}\right)\frac{d^{2}x}{d\theta^{2}} = -3\sin\theta - \sqrt{3}\cos\theta$
$x\frac{d^{3}x}{d\theta^{3}} + 3\frac{dx}{d\theta}\frac{d^{2}x}{d\theta^{2}} = -3\sin\theta - \sqrt{3}\cos\theta$
Substitute $\theta = 0$, along with our known evaluated derivatives:
$1 \cdot \frac{d^{3}x}{d\theta^{3}}\bigg|_{\theta=0} + 3(\sqrt{3})(0) = -3(0) - \sqrt{3}(1)$
$\frac{d^{3}x}{d\theta^{3}}\bigg|_{\theta=0} = -\sqrt{3}$

4. Substitute the evaluated derivatives into the Maclaurin series formula.

The Maclaurin expansion for $x$ is:
$x(\theta) = x(0) + x'(0)\theta + \frac{x''(0)}{2!}\theta^2 + \frac{x'''(0)}{3!}\theta^3 + \dots$
Substituting the evaluated constants:
$x(\theta) = 1 + \sqrt{3}\theta + \frac{0}{2}\theta^2 + \frac{-\sqrt{3}}{6}\theta^3$
$x(\theta) = 1 + \sqrt{3}\theta - \frac{\sqrt{3}}{6}\theta^3$

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

From part (a), at $\theta = 0$, $x^2 = 7 - 6(1) + 0 = 1 \implies x = 1$ (since length $x > 0$).

Differentiating the equation from (a) with respect to $\theta$:
$2x\frac{dx}{d\theta} = 6\sin\theta + 2\sqrt{3}\cos\theta$
At $\theta = 0$: $2(1)\frac{dx}{d\theta} = 2\sqrt{3} \implies \frac{dx}{d\theta} = \sqrt{3}$.

Using the given equation from (b) at $\theta = 0$:
$(1)\frac{d^2x}{d\theta^2} + (\sqrt{3})^2 = 3(1) - 0 \implies \frac{d^2x}{d\theta^2} = 0$

Differentiating the equation from (b) with respect to $\theta$ using the Product and Chain rules:
$x\frac{d^3x}{d\theta^3} + \frac{dx}{d\theta}\frac{d^2x}{d\theta^2} + 2\frac{dx}{d\theta}\frac{d^2x}{d\theta^2} = -3\sin\theta - \sqrt{3}\cos\theta$
At $\theta = 0$:
$(1)\frac{d^3x}{d\theta^3} + 3(\sqrt{3})(0) = -\sqrt{3} \implies \frac{d^3x}{d\theta^3} = -\sqrt{3}$

Substituting these derivatives into the Maclaurin series formula:
$x \approx 1 + (\sqrt{3})\theta + \frac{0}{2!}\theta^2 + \frac{-\sqrt{3}}{3!}\theta^3$
$x \approx 1 + \sqrt{3}\theta - \frac{\sqrt{3}}{6}\theta^3$

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

We are given that angle $ABC = \frac{\pi}{6} + \theta$.
Since $\frac{\pi}{6}$ radians is equivalent to $30^\circ$, when the angle is $35^\circ$, we have:
$30^\circ + \theta^\circ = 35^\circ \implies \theta^\circ = 5^\circ$
To use the Maclaurin expansion, the angle $\theta$ must be in radians:
$\theta = 5^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{36} \text{ radians}$

3. Substitute $\theta = \frac{\pi}{36}$ into the Maclaurin expansion to approximate $x$.

Substitute $\theta = \frac{\pi}{36}$ into the Maclaurin expansion obtained from part (c):
$x \approx 1 + \sqrt{3}\left(\frac{\pi}{36}\right) - \frac{\sqrt{3}}{6}\left(\frac{\pi}{36}\right)^3$
Using a calculator to compute the value:
$x \approx 1 + 0.15115002 - 0.00019185$
$x \approx 1.15095817$
Rounding the result to 6 decimal places, we get:
$x \approx 1.150958$