ABCD is a trapezium such that AB and CD are parallel and `BC bot CD`. If `angleADB = theta, BC = p and CD = q`, then AB is equal to

Options

A. `((p^(2)+q^(2))sintheta)/(pcostheta+qsintheta)`
B. `(p^(2)+q^(2)costheta)/(pcosthet+qsintheta)`
C. `(p^(2)+q^(2))/(p^(2)costheta+q^(2)sintheta)`
D. `((p^(2)+q^(2))sintheta)/((pcostheta+qsintheta)^(2))`

Explanation

Let `angleBDC=alpha`. Then, `tanalpha=p/q`
Clearly, `angleBDC=angleABD=alpha`

In `DeltaABD`, using sine rule, we get
`(AB)/(sintheta)=(BD)/(sin(pi-(theta+alpha)))`
`rArr(AB)/(sintheta)=(sqrt(p^(2)+q^(2))/(sin(theta+alpha))=(sqrt(p^(2)+q^(2)))/(sinthetacosalpha+costhetasinalpha)`
`rArrAB=(sqrt(p^(2)+q^(2))sintheta)/(sintheta(q/(sqrt(p^(2)+q^(2))))+costheta(p/(sqrt(p^(2)+q^(2)))))`
`rArrAB=((p^(2)+q^(2))sintheta)/(qsintheta+pcostheta)`