[tex]a=25,m\angle A=111^{\circ},b=24,\angle B=63.67,c=2.49,\angle C=5.33^{\circ}[/tex]
1) Let's begin by sketching out this triangle to beter gras p it:
2) So now, we can write out the Law of Sines formula:
[tex]\begin{gathered} \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)} \\ \\ \frac{a}{\sin(A)}=\frac{b}{\sin(B)} \\ \\ \frac{25}{\sin(111)}=\frac{24}{\sin(B)} \\ 25\sin(B)=24\sin(111) \\ \frac{25\sin \left(B\right)}{25}=\frac{\sin \left(111^{\circ \:}\right)\cdot \:24}{25} \\ \sin \left(B\right)=\frac{\sin \left(111^{\circ \:}\right)\cdot \:24}{25} \\ B=\arcsin \left(\frac{\sin \left(111^{\circ \:}\right)\cdot \:24}{25}\right) \\ m\angle B=63.67 \end{gathered}[/tex]
Note that we picked two ratios to find one variable. So now, let's find the measure of the angle C:
Since we know angle A, and angle B let's find the angle C by using the Triangle Sum Theorem:
[tex]\begin{gathered} m\angle A+m\angle B+m\angle C=180 \\ 111+63.67+m\angle C=180 \\ 174.67+m\angle C=180 \\ m\angle C=180-174.67 \\ m\angle C=5.33 \end{gathered}[/tex]
3) Now, let's find the one missing the leg c:
[tex]\begin{gathered} \frac{a}{\sin(A)}=\frac{c}{\sin(C)} \\ \frac{25}{\sin(111)}=\frac{c}{\sin(5.33)} \\ \frac{c}{\sin \left(5.33\right)}=\frac{25}{\sin \left(111^{\circ \:}\right)} \\ \frac{c\sin \left(5.33\right)}{\sin \left(5.33\right)}=\frac{25\sin \left(5.33\right)}{\sin \left(111^{\circ \:}\right)} \\ c=\frac{25\sin \left(5.33\right)}{\sin \left(111^{\circ \:}\right)} \\ c=2.49 \end{gathered}[/tex]
Note that we rounded off to the nearest hundredth.
