在 $\Delta ABC$ 中，$AC = 12\text{ cm}$，$\angle ABC$ 为直角，$\angle ACB = \frac{\pi}{6}$ 弧度，且 $M$ 是 $BC$ 的中点。在不使用计算器的情况下，求整数 $k$ 的值，使得 $\angle CAM = \sin^{-1}\left(\frac{\sqrt{k}}{14}\right)$。

1. Find the lengths of AB and BC using exact trigonometric ratios

In $\Delta ABC$, $\angle ABC = \frac{\pi}{2}$ and $\angle ACB = \frac{\pi}{6}$.
$AB = AC \sin\left(\frac{\pi}{6}\right) = 12 \times \frac{1}{2} = 6\text{ cm}$.
$BC = AC \cos\left(\frac{\pi}{6}\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\text{ cm}$.

2. Find the length of CM and BM

Since $M$ is the mid-point of $BC$,
$CM = BM = \frac{1}{2} BC = \frac{1}{2}(6\sqrt{3}) = 3\sqrt{3}\text{ cm}$.

3. Find the length of AM using Pythagoras' theorem in triangle ABM

In right-angled $\Delta ABM$,
$AM^2 = AB^2 + BM^2$
$AM^2 = 6^2 + (3\sqrt{3})^2 = 36 + 27 = 63$
$AM = \sqrt{63} = 3\sqrt{7}\text{ cm}$.

4. Use the sine rule in triangle ACM to find the sine of angle CAM

In $\Delta ACM$, using the sine rule:
$\frac{\sin(\angle CAM)}{CM} = \frac{\sin(\angle ACM)}{AM}$
$\frac{\sin(\angle CAM)}{3\sqrt{3}} = \frac{\sin(\frac{\pi}{6})}{3\sqrt{7}}$
$\sin(\angle CAM) = \frac{3\sqrt{3} \times \frac{1}{2}}{3\sqrt{7}} = \frac{\sqrt{3}}{2\sqrt{7}}$

5. Rationalise the denominator and find k

$\sin(\angle CAM) = \frac{\sqrt{3}}{2\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{21}}{14}$
Since $\angle CAM = \sin^{-1}\left(\frac{\sqrt{k}}{14}\right)$, we have $\frac{\sqrt{k}}{14} = \frac{\sqrt{21}}{14}$.
Therefore, $k = 21$.