Given that measure of angle A = 72 degree
Given that measure of angle B = 16 degree
We know that sum of all angles of the triangle is 180 degree
A+B+C=180
72+16+C=180
88+C=180
C=180-88
C=92
Hence measure of angle C = 92 degree
Now we can use sine formula to find the missing side "a"
[tex] \frac{a}{\sin\left(A\right)}=\frac{b}{\sin\left(B\right)}=\frac{c}{\sin\left(C\right)} [/tex]
[tex] \frac{a}{\sin\left(A\right)}=\frac{c}{\sin\left(C\right)} [/tex]
[tex] \frac{a}{\sin\left(72\right)}=\frac{61}{\sin\left(92\right)} [/tex]
[tex] \frac{a}{0.951}=\frac{61}{0.99939} [/tex]
[tex] \frac{a}{0.951}=61.037232712 [/tex]
a=61.037232712*0.951
a=58.0464083091
Hence final answer is approx a=58.05
