From the given picture we can see
ACB is a right triangle at C
AC = b
CB = a
AB = c
Since mSince a = 5 ft
Then to find b and c we will use the trigonometry ratios
[tex]\begin{gathered} \sin A=\frac{a}{c} \\ \sin 60=\frac{5}{c} \end{gathered}[/tex]Substitute the value of sin 60
[tex]\begin{gathered} \sin 60=\frac{\sqrt[]{3}}{2} \\ \frac{\sqrt[]{3}}{2}=\frac{5}{c} \end{gathered}[/tex]By using the cross multiplication
[tex]\begin{gathered} \sqrt[]{3}\times c=2\times5 \\ \sqrt[]{3}c=10 \end{gathered}[/tex]Divide both sides by root 3
[tex]\begin{gathered} \frac{\sqrt[]{3}c}{\sqrt[]{3}}=\frac{10}{\sqrt[]{3}} \\ c=\frac{10}{\sqrt[]{3}} \end{gathered}[/tex]To find b we will use the tan ratio
[tex]\begin{gathered} \tan 60=\frac{a}{b} \\ \tan 60=\frac{5}{b} \end{gathered}[/tex]Substitute the value of tan 60
[tex]\begin{gathered} \tan 60=\sqrt[]{3} \\ \sqrt[]{3}=\frac{5}{b} \end{gathered}[/tex]Switch b and root 3
[tex]b=\frac{5}{\sqrt[]{3}}[/tex]The exact values of b and c are
[tex]\begin{gathered} b=\frac{5}{\sqrt[]{3}} \\ c=\frac{10}{\sqrt[]{3}} \end{gathered}[/tex]