InterviewSolution
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InterviewSolution
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ABCD is a trapezium such that AB and CD are paralleland `B C_|_C D`. If `/_A D B""=theta,""B C""=""p""a n d""C D""=""q`, then AB isequal 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)` |
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Answer» `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 |
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