ABCD is a trapezium such that AB and CD are parallel and `BCbotCD` if `angleABD=theta,Bc=p and CD=q,` then AB is equal to

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. q`((p^(2)+q^(2)))/((p^(2) cos theta+q^(2) sin theta)^(2))`

Answer

A

Explanation

Applying sine rule in `DeltaABD`,

`(AB)/(sin theta)=(sqrt(p^(2)+q^(2)))/(sin{pi-(theta+alpha)})`
`rArr (AB)/(sin theta)=(sqrt(p^(2)+q^(2)))/(sin(theta+alpha))`
`rArrAB=(sqrt(p^(2)+q^(2)) sin theta)/(sin theta cos alpha +costheta sin alpha)[:. cosalpha=(q)/(sqrt(p^(2)+q^(2)))]`
`=((p^(2)+a^(2) sin theta))/(p cos 0+q sin theta)`
Alternate Solotion
Let AB=x

In `DeltaDAM, tan (pi theta-alpha)=(p)/(x-q)`
`rArr tan (theta+alpha)=(p)/(q-x)`
`rArr q-xp cot (theta+alpha)`
`rArr x=q-pcos (theta+alpha)`
`=p-q((cot theta cot alpha-1)/(cot alpha+cot theta))=[:.cot alpha=(q)/(p)]`
`=q-p(((q)/(p)cot theta-1)/((q)/(p)+cot theta))=q-p((q cot theta-p)/(q +p cot theta))`
`=q-p((q cos theta-p sin theta)/(q sin theta+p cos theta))`
`rArr x=(q^(2)sin theta+qpcos theta-pq cos theta+p^(2)sin theta)/(p cos theta+q sin theta)`
`rArr AB=((p^(2)+q^(2)))/(p cos theta+q sin theta)`