Triangle ABCD in FIGURE 3 below is drawn such that A = 30°; B = 90°; DC splits C such that angle θ and angle α are formed, also DC splits AB such that x and angle β are formed. BD = 10 cm and BC = 40 cm. Determine the following: 2.1.1. θ, 2.1.2. x, 2.1.3. α, 2.1.4. β, 2.1.5. AC, 2.1.6. AD

Consider $\triangle BDC$. It is a right-angled triangle with $\angle B = 90^\circ$.
The side opposite to $\theta$ is $BD = 10$ cm.
The side adjacent to $\theta$ is $BC = 40$ cm.

The value of $\theta$ is \boxed{\text{14.04^\circ}}.

2.1.2 Determine $x$

The problem states that $BD = 10$ cm.
The diagram labels the length of the side $BD$ as $x$.

The value of $x$ is $\boxed{\text{10 cm}}$.

2.1.3 Determine $\alpha$

Consider $\triangle ABC$. It is a right-angled triangle with $\angle B = 90^\circ$ and $\angle A = 30^\circ$.
The sum of angles in a triangle is $180^\circ$.

From the diagram, $\angle BCA = \alpha + \theta$.
We found $\theta \approx 14.04^\circ$ in 2.1.1.

The value of $\alpha$ is \boxed{\text{45.96^\circ}}.

2.1.4 Determine $\beta$

Consider $\triangle ADC$. We know $\angle A = 30^\circ$ and $\angle ACD = \alpha \approx 45.96^\circ$.
The angle $\beta$ is $\angle ADC$.

The sum of angles in a triangle is $180^\circ$.

The value of $\beta$ is \boxed{\text{104.04^\circ}}.

2.1.5 Determine AC

Consider $\triangle ABC$. It is a right-angled triangle with $\angle B = 90^\circ$.
We know $\angle A = 30^\circ$ and the side opposite to it, $BC = 40$ cm.
AC is the hypotenuse.

The length of AC is $\boxed{\text{80 cm}}$.

2.1.6 Determine AD

Consider $\triangle ABC$. We know $\angle A = 30^\circ$, $\angle B = 90^\circ$, and $BC = 40$ cm.
We can use the tangent function.

From the diagram, $D$ is a point on $AB$, so $AD = AB - BD$.
We know $BD = x = 10$ cm from 2.1.2.

The length of AD is \boxed{\text{(40\sqrt{3} - 10) cm} \text{ or approximately *59.28 cm}}.