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
(a)`((p^(2)+q^(2))sin theta)/(p cos theta +q sin theta)`
(b)`(p ^(2) + q ^(2)cos theta)/(p cos theta +q sin theta)`
(c)`(p ^(2)+ q^(2))/(p^(2)cos theta +q^(2) sin theta)`
(d)`((p^(2) +q^(2))sin theta)/((p cos theta + sin theta)^(2))`

Options

A. `((p^(2)+q^(2))sin theta)/(p cos theta +q sin theta)`
B. `(p ^(2) + q ^(2)cos theta)/(p cos theta +q sin theta)`
C. `(p ^(2)+ q^(2))/(p^(2)cos theta +q^(2) sin theta)`
D. `((p^(2) +q^(2))sin theta)/((p cos theta + sin theta)^(2))`

Answer

A