Answer
$c\approx4.7, A\approx45^\circ, B\approx93^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$c^2=a^2+b^2-2ab\ cosC$, we have $c^2=(5)^2+(7)^2-2(5)(7)cos(42^\circ)\approx21.98$, thus $c=\sqrt {21.98}\approx4.7$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinC}{c}$, $SinA=\frac{5}{4.7}sin(42^\circ)\approx0.7118$, thus $A=sin^{-1}(0.7118)\approx45^\circ$
Step 3. We can find the angle as
$B=180^\circ-42^\circ-45^\circ=93^\circ$
Step 4. We solved the triangle with
$c\approx4.7, A\approx45^\circ, B\approx93^\circ$
