Answer
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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.
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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