#### Answer

$A=31^\circ, B=122^\circ, C=27^\circ$

#### Work Step by Step

Step 1. Using the given figure in the exercise and the Law of Cosines, we have
$b^2=a^2+c^2-2ac\ cosB$
Thus
$cosB=\frac{17^2+15^2-28^2}{2(17)(15)}\approx-0.5294$
and
$B=acos(-0.5294)\approx122^\circ$
Step 2. Using the Law of Sines, we have
$\frac{sinA}{a}=\frac{sinB}{b}$
and
$sinA=\frac{17sin122^\circ}{28}\approx0.5149$
and
$A=asin(0.5149)\approx31^\circ$
Step 3. Angle $C=180^\circ-B-A\approx27^\circ$
Step 4. The solutions are $A=31^\circ, B=122^\circ, C=27^\circ$