I confused on this one

The top t appears to be the hypotenuse of a right triangle with angle 14.5 degrees amd opposite side 6.
[tex] \sin 14.5^\circ = \dfrac{6}{t} [/tex]
[tex] t = \dfrac{6}{\sin 14.5} \approx 23.96 \textrm{ inches} \approx 24 \textrm{ in}[/tex]