In the triangle $ABC$, $AB = 1$, $BC = 4$ and angle $ABC = \theta$ radians. Given that $\theta$ is a sufficiently small angle, show that $AC \approx \left(9 + 4\theta^2\right)^{\frac{1}{2}} \approx a + b\theta^2,$ for constants $a$ and $b$ to be determined.

1. Formulate an expression for $AC^2$ using the Cosine Rule on triangle $ABC$.

$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos \theta$
$AC^2 = 1^2 + 4^2 - 2(1)(4)\cos \theta$
$AC^2 = 17 - 8\cos \theta$

2. Apply the small angle approximation for $\cos \theta$ to show the first result.

Since $\theta$ is a small angle, $\cos \theta \approx 1 - \frac{\theta^2}{2}$.
$AC^2 \approx 17 - 8\left(1 - \frac{\theta^2}{2}\right)$
$AC^2 \approx 17 - 8 + 4\theta^2$
$AC^2 \approx 9 + 4\theta^2$
$AC \approx (9 + 4\theta^2)^{\frac{1}{2}}$

3. Factor out 9 to prepare the expression for the standard series expansion of $(1+x)^n$.

$(9 + 4\theta^2)^{\frac{1}{2}} = \left[9\left(1 + \frac{4\theta^2}{9}\right)\right]^{\frac{1}{2}}$
$= 3\left(1 + \frac{4}{9}\theta^2\right)^{\frac{1}{2}}$

4. Expand using the binomial expansion and state the constants $a$ and $b$.

Using $(1+x)^n \approx 1 + nx$ for small $x$:
$\approx 3\left[1 + \frac{1}{2}\left(\frac{4}{9}\theta^2\right)\right]$
$= 3\left(1 + \frac{2}{9}\theta^2\right)$
$= 3 + \frac{2}{3}\theta^2$
Comparing with $a + b\theta^2$, we obtain $a = 3$ and $b = \frac{2}{3}$.