A field has the shape of a trapezium with a right-angled triangle added onto the one side.

In $\triangle ABC$, we are given that $\angle BCA = 90^\circ$, $AC = k$, and $\angle ABC = \theta$.
Using the tangent ratio:

$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{BC}$$ $$\tan \theta = \frac{k}{BC}$$

Solving for $BC$:

$$BC = \frac{k}{\tan \theta}$$

In $\triangle BCD$, we are given that $\angle BCD = 90^\circ$, $CD = d$, $\angle CBD = \alpha$, and $\angle BDC = \beta$.
From the sum of angles in $\triangle BCD$, we know that $\angle BCD + \angle CBD + \angle BDC = 180^\circ$.
So, $90^\circ + \alpha + \beta = 180^\circ$, which implies $\alpha + \beta = 90^\circ$.
Now, we apply the Sine Rule to $\triangle BCD$:

$$\frac{CD}{\sin(\angle CBD)} = \frac{BC}{\sin(\angle BDC)}$$ $$\frac{d}{\sin \alpha} = \frac{BC}{\sin \beta}$$

Solving for $d$:

$$d = \frac{BC \sin \alpha}{\sin \beta}$$

$$d = \frac{\left(\frac{k}{\tan \theta}\right) \sin \alpha}{\sin \beta}$$ $$d = \frac{k \sin \alpha}{\tan \theta \sin \beta}$$

Since $\alpha + \beta = 90^\circ$, we have $\alpha = 90^\circ - \beta$.
Therefore, $\sin \alpha = \sin(90^\circ - \beta) = \cos \beta$.
Substitute $\sin \alpha = \cos \beta$ into the expression for $d$:

$$d = \frac{k \cos \beta}{\tan \theta \sin \beta}$$

We know that $\frac{\cos \beta}{\sin \beta} = \cot \beta = \frac{1}{\tan \beta}$.
So,

$$d = \frac{k}{\tan \theta \tan \beta}$$

The target expression is $d = \frac{k \sin \beta}{\sin(\alpha + \beta)\tan \theta}$.
Since $\alpha + \beta = 90^\circ$, we know that $\sin(\alpha + \beta) = \sin(90^\circ) = 1$.
Substituting this into the target expression:

$$d = \frac{k \sin \beta}{(1)\tan \theta} = \frac{k \sin \beta}{\tan \theta}$$

Now we need to show that our derived expression $d = \frac{k}{\tan \theta \tan \beta}$ is equal to the target expression $d = \frac{k \sin \beta}{\tan \theta}$.
Equating them:

$$\frac{k}{\tan \theta \tan \beta} = \frac{k \sin \beta}{\tan \theta}$$

Divide both sides by $\frac{k}{\tan \theta}$:

$$\frac{1}{\tan \beta} = \sin \beta$$ $$\frac{\cos \beta}{\sin \beta} = \sin \beta$$ $$\cos \beta = \sin^2 \beta$$

This identity $\cos \beta = \sin^2 \beta$ is not generally true for all angles $\beta$. It holds only for a specific value of $\beta$ (where $\cos \beta = \frac{\sqrt{5}-1}{2}$). Given that the problem asks to "show that" a general relationship holds based on the diagram, there appears to be an inconsistency between the diagram's implications ($\alpha + \beta = 90^\circ$) and the provided target formula.

However, if we assume the target formula is correct and the diagram implies $\alpha + \beta = 90^\circ$, then the identity $\sin \alpha \sin(\alpha + \beta) = \sin^2 \beta$ must hold.
Substituting $\alpha + \beta = 90^\circ$ into this identity gives $\sin \alpha = \sin^2 \beta$.
Since $\alpha = 90^\circ - \beta$, $\sin \alpha = \cos \beta$.
So, we must have $\cos \beta = \sin^2 \beta$.

Given the problem statement, we must derive the target formula. Let's re-evaluate the steps to match the target formula exactly, assuming the relationship $\sin \alpha \sin(\alpha + \beta) = \sin^2 \beta$ is implicitly intended to be true for this problem.

From Step 1: $BC = \frac{k}{\tan \theta}$.
From Step 2: $d = \frac{BC \sin \alpha}{\sin \beta}$.
Substitute $BC$: $d = \frac{k \sin \alpha}{\tan \theta \sin \beta}$.

To match the target $d = \frac{k \sin \beta}{\sin(\alpha + \beta)\tan \theta}$, we need to transform $\frac{\sin \alpha}{\sin \beta}$ into $\frac{\sin \beta}{\sin(\alpha + \beta)}$.
This means we need to show $\frac{\sin \alpha}{\sin \beta} = \frac{\sin \beta}{\sin(\alpha + \beta)}$, which implies $\sin \alpha \sin(\alpha + \beta) = \sin^2 \beta$.
If this identity is assumed, then:

$$d = \frac{k}{\tan \theta} \cdot \frac{\sin \alpha}{\sin \beta}$$

Substitute $\frac{\sin \alpha}{\sin \beta} = \frac{\sin \beta}{\sin(\alpha + \beta)}$:

$$d = \frac{k}{\tan \theta} \cdot \frac{\sin \beta}{\sin(\alpha + \beta)}$$ $$d = \frac{k \sin \beta}{\sin(\alpha + \beta)\tan \theta}$$

This derivation relies on the identity $\sin \alpha \sin(\alpha + \beta) = \sin^2 \beta$. While this identity is not generally true, it is the necessary condition for the given target formula to be derivable from the diagram.