Answer
$A\approx139^\circ, B\approx14^\circ, C\approx27^\circ$
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 $(66)^2=(25)^2+(45)^2-2(25)(45)cos(A)$, thus $cosA\approx-0.7582$ and $A=cos^{-1}(-0.7582)\approx139.3\approx139^\circ$
Step 2. Using the Law of Sines, we have
$\frac{SinB}{b}=\frac{SinA}{a}$, $SinB=\frac{25}{66}sin(139.3^\circ)\approx 0.247$, thus $B=sin^{-1}(0.247)\approx14^\circ$
Step 3. We can find the angle as
$C=180^\circ-139^\circ-14^\circ=27^\circ$
Step 4. We solved the triangle with
$A\approx139^\circ, B\approx14^\circ, C\approx27^\circ$
