过C做CD垂直AB,设BC=R,角C=50度,角A=180-B-50
CD=SINB*R DB=COSB*R,DA=CD*ctgA=SINB*R*ctgA
DA+DB=SINB*R*ctgA+COSB*R=m
R=m/(SINB*ctgA+COSB)=m/[SINB*ctg(130-B)+COSB]
=m*sin(130-B)/sin(130-B+B)=m*sin(130-B)/sin130
R取得最大值,sin（130－B）＝1,130－B=90
则B=40