Question

5Sin19°Sin71° / Sin38°

Answer 1

$frac{5sin19cdot sin71}{sin38} =[, sin(90- alpha )=cos alpha , ]= frac{5sin19cdot cos19}{sin38} =\\=[, 2sin alpha cdot cos alpha =sin2 alpha , ]= frac{5cdot frac{1}{2}sin38}{sin38} =frac{5}{2}=2,5$