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

Last updated date: 13th Jun 2024
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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 $AB$.

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.