Answer
$a=6.0,B=29^\circ,C=105^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$a^2=b^2+c^2-2bc\ cosA=(4)^2+(8)^2-2(4)(8)cos(46)\approx35.54$, thus $a\approx\sqrt {35.54}\approx6.0$
Step 2. To find the unknown angles, using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{4}{6}sin(46)\approx0.4827$, thus $B=sin^{-1}(0.4827)\approx29^\circ$
Step 3. We can find the angle as
$C=180^\circ-46^\circ-29^\circ=105^\circ$
Step 4. We solved the triangle with $a=6.0,B=29^\circ,C=105^\circ$
