Answer
$A\approx28^\circ, B\approx40^\circ, C\approx112^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$c^2=a^2+b^2-2ab\ cosC$, or $(10)^2=(5)^2+(7)^2-2(5)(7)cos(C)$, thus $cosC\approx-0.3714$ and $C=cos^{-1}(-0.3714)\approx112^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinA}{a}=\frac{SinC}{c}$, $SinA=\frac{5}{10}sin(112^\circ)\approx 0.4636$, thus $A=sin^{-1}(0.4636)\approx28^\circ$
Step 3. We can find the angle as
$B=180^\circ-112^\circ-28^\circ=40^\circ$
Step 4. We solved the triangle with
$A\approx28^\circ, B\approx40^\circ, C\approx112^\circ$
