Question 211794
Draw CE such that CE is perpendicular to line AB.
Triangle AEC is a right triangle.
Let h=CE
h=b sin A =12 sin 47 = 8.776
{{{AE=sqrt(b^2 - h^2)=sqrt(12^2 - 8.776^2)= sqrt(66.982)=8.184}}}
Let d=BE
BE=AE-Be=8.184-8=0.184
{{{ a= sqrt(BE^2 + h^2)=sqrt(0.184^2 + 8.776^2)=sqrt(77.052)=8.778 }}}
So, a=8.778