ABCD is a trapezium such that `A B||C Da n dC B` is perpendicular to them. If `/_A D B=theta,B C=p ,a n dC D=q` , show that `A B=((p^2+q^2)sintheta)/(pcostheta+qsintheta)`

Explanation

To solve the problem, we need to show that in trapezium ABCD, with \( AB \parallel CD \), \( CB \perp AB \) and \( AD \) making an angle \( \theta \) with \( DB \), the length of \( AB \) can be expressed as:

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

where \( BC = p \) and \( CD = q \).

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