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ABCD is a trapezium such that AB,DC.are ...

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`

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 thet)`

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))`

Text Solution

Verified by Experts

The correct Answer is:
A

`BD = sqrt(p^(2) + q^(2))`
`angleABD = angleBDC = alpha`
`rArr angle DAB = pi - (theta + alpha)`
`tan alpha = (p)/(q)`
In `DeltaABD`, using sine rule
`(AB)/(sin theta) = (BD)/(sin (pi - (theta - aplpha))) = (BD)/(sin (theta + alpha))`
`:. AB = (BD sin theta)/(sin (theta + alpha)) = (BD^(2) sin theta)/(BD sin (theta + alpha))`
`= (BD^(2) sin theta)/(BD sin theta cos alpha + BD cos theta sin alpha)`
`= ((p^(2) + q^(2)) sin theta)/(q sin theta + p cos theta)`
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