The diagram shows a trapezium $ABCD$. The point $X$ lies on line $AB$ such that $DX = 12\text{ cm}$ and $CX = 9\text{ cm}$. $\angle ADX = \theta^\circ$ and $\angle DAX = \angle DXC = \angle XBC = 90^\circ$.

1. Find the length of $AX$ in terms of $\theta$

In $\triangle ADX$, $\angle DAX = 90^\circ$.
$\sin \theta = \frac{AX}{DX}$
$AX = 12 \sin \theta$

2. Find $\angle BXC$ in terms of $\theta$

In $\triangle ADX$, $\angle AXD = 180^\circ - 90^\circ - \theta = 90^\circ - \theta$.
Since $A, X, B$ lie on a straight line, the sum of angles at $X$ is $180^\circ$.
$\angle BXC = 180^\circ - \angle AXD - \angle DXC$
$\angle BXC = 180^\circ - (90^\circ - \theta) - 90^\circ = \theta$

3. Find the length of $XB$ and hence show the expression for $AB$

In $\triangle XBC$, $\angle XBC = 90^\circ$.
$\cos \angle BXC = \frac{XB}{CX}$
$XB = 9 \cos \theta$
Since $X$ lies on $AB$, $AB = AX + XB$.
$AB = 12 \sin \theta + 9 \cos \theta$
$AB = 9 \cos \theta + 12 \sin \theta$ (Shown)

1. Find the value of $R$

$R = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$

2. Find the value of $\alpha$

$\tan \alpha = \frac{12}{9} = \frac{4}{3}$
$\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.130^\circ$

3. Express $AB$ in the required form

$AB = 15 \cos(\theta - 53.1^\circ)$

4. State the maximum length of $AB$

Maximum length of $AB = R = 15\text{ cm}$

5. Find the corresponding value of $\theta$

For maximum length, $\cos(\theta - 53.130^\circ) = 1$
$\theta - 53.130^\circ = 0^\circ$
$\theta = 53.1^\circ$ (to 1 d.p.)

1. Recall the $R$-formula expression for $AB$ from part (b)

From part (b), $AB = 15 \cos(\theta - 53.13^\circ)$.
We set this equal to 11:
$15 \cos(\theta - 53.13^\circ) = 11$

2. Solve for the basic angle

$\cos(\theta - 53.13^\circ) = \frac{11}{15}$
$\theta - 53.13^\circ = \cos^{-1}\left(\frac{11}{15}\right)$
$\theta - 53.13^\circ = \pm 42.83^\circ$

3. Find the valid value of $\theta$ in the context of the geometric figure

Since $\theta$ is an angle in the right-angled triangle $ADX$, $0^\circ < \theta < 90^\circ$.
Taking the negative value:
$\theta = 53.13^\circ - 42.83^\circ = 10.3^\circ \text{ (to 1 d.p.)}$
(Taking the positive value gives $\theta = 95.96^\circ$, which is rejected as it is obtuse).