ABCD is a trapezium such that AB,DC.are parallel and BC is perpendicular to them. If `angleADB=theta, BC= p and CD=q` , show that AB=`((p^2+q^2)sintheta)/(pcostheta+qsintheta`

Explanation

Let `angle ABD = angle BDC = alpha`
`:. Angle BAD = 180^(@) - (theta + alpha)`
By the sine formula, in `Delta ABD`, we have
`(AB)/(sin theta) = (BD)/(sin (180^(@) - (theta + alpha)))`
or `AB = (BD sin theta)/(sin (theta + alpha)) = (BD sin theta)/(sin theta cos alpha + cos theta sin alpha)` (i)

In `Delta BCD , sin alpha = p//BD`
and `cos alpha = q//BD`
Also `BD^(2) = p^(2) + q^(2)`
Therefore, from Eq. (i), we have
`AB = (BD sin theta)/(sin theta (q//BD) + cos theta (p//BD))`
`= (BD^(2) sin theta)/(q sin theta + p cos theta) = ((p^(2) + q^(2)) sin theta)/(p cos theta + q sin theta)`