$tan\alpha=\dfrac{2}{5}↔\dfrac{sin\alpha}{cos\alpha}=\dfrac{2}{5}$
$→sin\alpha=\dfrac{2}{5}cos\alpha$
$B=\dfrac{5sin\alpha+2cos\alpha}{2sin\alpha-5cos\alpha}$
$=\dfrac{5.\dfrac{2}{5}cos\alpha+2cos\alpha}{2.\dfrac{2}{5}cos\alpha-5cos\alpha}$
$=\dfrac{2cos\alpha+2cos\alpha}{\dfrac{4}{5}cos\alpha-5cos\alpha}$
$=\dfrac{4cos\alpha}{\dfrac{-21}{5}cos\alpha}$
$=\dfrac{-20}{21}$