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**Hint:**Sine rule of triangle:

Let us consider, sides of a triangle \[\Delta ABC\] are \[AB = c,BC = a, CA = b\], then we get,

\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]

Using this formula, we can find the length of \[AB\].

**Complete step-by-step answer:**

It is given that, \[ABCD\] is a trapezium such that \[AB\] and \[CD\] are parallel and \[BC \bot CD\].

Also it is given that,

\[\angle ADB = \theta , BC = p, CD = q\].

Then we have to find the value of \[AB\].

Let us consider, \[\angle BDC = \alpha \]

Since, \[AB\] and \[CD\] are parallel and \[BD\] is the transversal.

So,

\[\angle BDC = \angle ABD = \theta \]

Since, they are opposite interior angles.

From, \[\Delta BCD\] using the Pythagoras theorem, we get,

\[BD = \sqrt {{p^2} + {q^2}} \]

Where BD is the hypotenuse and p, q are adjacent and opposite sides of the triangle.

And using the relation between angles and sides of triangle we get,

\[\sin \theta = \dfrac{p}{{\sqrt {{p^2} + {q^2}} }}\] and \[\cos \theta = \dfrac{q}{{\sqrt {{p^2} + {q^2}} }}\]

Let us consider \[\angle ADB = \alpha \], also we know that total angle of a triangle is \[\pi \]

Now, in \[\Delta ABD\], we have \[\angle ABD = \alpha \] and \[\angle ADB = \theta \] using the above condition we get, \[\angle BAD = \pi - (\theta + \alpha )\]

Now by using rule of sine we get,

\[\dfrac{{AB}}{{\sin \theta }} = \dfrac{{BD}}{{\sin [\pi - (\theta + \alpha )]}}\]

We know that \[\sin [\pi - a] = \sin (a)\] using this we get,

\[\dfrac{{AB}}{{\sin \theta }} = \dfrac{{BD}}{{\sin (\theta + \alpha )}}\]

Let us substitute the value of \[BD\] in the above formula and applying the formula of \[\sin (\theta + \alpha ) = \sin \theta \cos \alpha + \cos \theta \sin \alpha \]we get,

\[AB = \dfrac{{\sqrt {{p^2} + {q^2}} \sin \theta }}{{\sin \theta \cos \alpha + \cos \theta \sin \alpha }}\]

Substitute the values of \[\cos {\rm{ }}\& \sin \] we get,

\[AB = \dfrac{{\sqrt {{p^2} + {q^2}} \sin \theta }}{{\sin \theta \dfrac{q}{{\sqrt {{p^2} + {q^2}} }} + \cos \theta \dfrac{p}{{\sqrt {{p^2} + {q^2}} }}}}\]

Let us multiply the above equation by \[\sqrt {{p^2} + {q^2}} \] in numerator and denominator we get,

\[AB = \dfrac{{({p^2} + {q^2})\sin \theta }}{{p\cos \theta + q\sin \theta }}\]

Hence, the value \[AB = \dfrac{{({p^2} + {q^2})\sin \theta }}{{p\cos \theta + q\sin \theta }}\]

**Additional Information:**A quadrilateral is called a trapezium when two opposite sides are parallel and the other two opposite sides are non-parallel.

Pythagoras theorem states that, for a right-angle triangle, the square of the hypotenuse is equal to the sum of the square of base and the square of perpendicular.

**Note:**The relation between side and angles in the triangle is given as,

\(\begin{array}{l}\sin a = \dfrac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}\\\cos a = \dfrac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}}\\\tan a = \dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}\end{array}\)

Here while using the relation we should be careful as the values of sine and cosine may vary with respect to the angle made by the side.

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