Answer
$A\approx35^\circ, B\approx86^\circ, C\approx59^\circ$
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 $(7)^2=(4)^2+(6)^2-2(4)(6)cos(B)$, thus $cosB\approx-0.0625$ and $B=cos^{-1}(0.0625)\approx86.42\approx86^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinB}{b}$, $SinA=\frac{4}{7}sin(86.42^\circ)\approx 0.5703$, thus $A=sin^{-1}(0.5703)\approx35^\circ$
Step 3. We can find the angle as
$C=180^\circ-86^\circ-35^\circ=59^\circ$
Step 4. We solved the triangle with
$A\approx35^\circ, B\approx86^\circ, C\approx59^\circ$
