Answer
$b\approx5.7, A\approx35^\circ, C\approx90^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$b^2=a^2+c^2-2ac\ cosB$, or $b^2=(4)^2+(7)^2-2(4)(7)cos(55^\circ)\approx 32.88$, thus $b=\sqrt {32.88}\approx 5.7$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinB}{b}$, $SinA=\frac{4}{\sqrt {32.88}}sin(55^\circ)\approx 0.5714$, thus $A=sin^{-1}(0.5714)\approx35^\circ$
Step 3. We can find the angle as
$C=180^\circ-55^\circ-35^\circ=90^\circ$
Step 4. We solved the triangle with
$b\approx5.7, A\approx35^\circ, C\approx90^\circ$
