Question

ABCD is a trapezium such that AB and CD are parallel and `B C_|_C D` . If `/_A D B""=theta,""B C""=""p""a n d""C D""=""q` , then AB is equal to (1) `(p^2+q^2costheta)/(pcostheta+qsintheta)` (2) `(p^2+q^2)/(p^2costheta+q^2sintheta)` (3) `((p^2+q^2)sintheta)/((pcostheta+qsintheta)^2)` (4) `((p^2+q^2)sintheta)/(pcostheta+qsintheta)`

Answer 1

`BD= sqrt(p^2 + q^2)`
using pythagoras theorem in `/_ BCD`
in `/_ABD`
`(AB)/(sin theta) = (BD)/(sin(pi-(theta + alpha))`
`(AB)/(sin theta) = (BD)/(sin ( theta + alpha))`
`AB= (sqrt(p^2 + q^2)* sin theta)/(sin theta cos alpha + cos alpha sin theta)`
`cos alpha = q/(sqrt(p^2 + q^2))`
so, `AB= (sqrt(p^2 + q^2) sin theta)/(sin theta q/sqrt(p^2 + q^2)+ cos theta p/sqrt(p^2 + q^2)`
`AB= (p^2 + q^2sin theta)/(q sin theta + p cos theta)`
option 4 is correct