Find sin B.
Triangle ADC is a Pythagorean Triple 6 - 8 -10 right triangle ....so AD = 6
And in right triangle BDA , hypotenuse BA = sqrt (6^2 + 12^2) = sqrt ( 180) = 6sqrt (5)
So....sin B = AD / BA = 6 / (6sqrt (5)) = 1 /sqrt (5) = sqrt (5) / 5