Answer
$A \approx39.4^\circ$,
$B \approx54.7^\circ$,
$C \approx85.9^\circ$.
Work Step by Step
1. Use the Law of Cosines, we have $A=cos^{-1}(\frac{11^2+9^2-7^2}{2(11)(9)})\approx39.4^\circ$,
2. Use the Law of Cosines, we have $B=cos^{-1}(\frac{11^2+7^2-9^2}{2(11)(7)})\approx54.7^\circ$,
3. Find the third angle $C=180-A-B\approx85.9^\circ$.
