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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 + q sin theta )^(2))`

Text Solution

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The correct Answer is:
A
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