Answer
$a\approx4.3, B\approx66^\circ, C\approx14^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$a^2=b^2+c^2-2bc\ cosA$, or $a^2=(4)^2+(1)^2-2(4)(1)cos(100^\circ)\approx 18.39$, thus $c=\sqrt {18.39}\approx 4.3$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{4}{4.3}sin(100^\circ)\approx 0.9161$, thus $B=sin^{-1}(0.9161)\approx66^\circ$
Step 3. We can find the angle as
$C=180^\circ-100^\circ-66^\circ=14^\circ$
Step 4. We solved the triangle with
$a\approx4.3, B\approx66^\circ, C\approx14^\circ$
