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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 \[AB\].

Complete step-by-step answer:
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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.
\[\angle BDC = \angle ABD = \theta \]
 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.