in a triangle ABC, AD i the altitude from A givven b>c, angle C =2 degree and AD= (abc)/(b^2-c^2)find angle b

AD=c sinB=abc/(b² - c²)

             or sinB=ab/(b² - c²)

              or sinB=sinA sinB/sin(B+C) sin(B-C)

                or sinB=sinAsinB/sinAsin(B-C)

                or sin(B-C)=1=sin90°

                     or B-C=90°

but C = 2

so angle B = 92 degree