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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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