$$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \perp CD$$. If $$\angle ADB = \theta$$, $$BC = p$$ and $$CD = q$$, then $$AB$$ is equal to

In the trapezium $$ABCD$$, since $$AB \parallel CD$$ and $$BC \perp CD$$, it follows that $$BC \perp AB$$. Thus, $$\angle BCD = 90^\circ$$ and $$\angle ABC = 90^\circ$$.

In the right-angled triangle $$BCD$$:

$$BD = \sqrt{BC^2 + CD^2} = \sqrt{p^2 + q^2}$$. Let $$\angle BDC = \alpha$$.
$$\sin \alpha = \frac{BC}{BD} = \frac{p}{\sqrt{p^2 + q^2}}$$

$$\cos \alpha = \frac{CD}{BD} = \frac{q}{\sqrt{p^2 + q^2}}$$

Since $$AB \parallel CD$$, the alternate interior angles are equal. $$\angle ABD = \angle BDC = \alpha$$

In $$\triangle ABD$$, $$\angle ADB = \theta$$, $$\angle ABD = \alpha$$

$$\angle DAB = 180^\circ - (\theta + \alpha)$$

Using the Sine Rule in $$\triangle ABD$$:

$$\frac{AB}{\sin \theta} = \frac{BD}{\sin(\angle DAB)}$$

$$\frac{AB}{\sin \theta} = \frac{\sqrt{p^2 + q^2}}{\sin(180^\circ - (\theta + \alpha))}$$

$$\frac{AB}{\sin \theta} = \frac{\sqrt{p^2 + q^2}}{\sin \theta \cos \alpha + \cos \theta \sin \alpha}$$

Substitute the values of $$\sin \alpha$$ and $$\cos \alpha$$:

$$AB = \frac{\sqrt{p^2 + q^2} \sin \theta}{\sin \theta \left( \frac{q}{\sqrt{p^2 + q^2}} \right) + \cos \theta \left( \frac{p}{\sqrt{p^2 + q^2}} \right)}$$

$$AB = \frac{(p^2 + q^2) \sin \theta}{q \sin \theta + p \cos \theta}$$