Answer
$A\approx117^\circ, B\approx18^\circ, C\approx45^\circ$
Work Step by Step
Step 1. Based on the given conditions, using the Law of Cosines, we have
$a^2=b^2+c^2-2bc\ cosA$, or $(63)^2=(22)^2+(50)^2-2(22)(50)cos(A)$, thus $cosA\approx-0.4477$ and $A=cos^{-1}(-0.4477)\approx116.6\approx117^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{22}{63}sin(116.6^\circ)\approx 0.3122$, thus $B=sin^{-1}(0.3122)\approx18^\circ$
Step 3. We can find the angle as
$C=180^\circ-117^\circ-18^\circ=45^\circ$
Step 4. We solved the triangle with
$A\approx117^\circ, B\approx18^\circ, C\approx45^\circ$
