The height of the street light represent the opposite side.
The adjacent is 20ft
the angle is 40 degrees
using trigonometry ratio for tangent
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan 40^0=\frac{x}{20} \\ \text{where x=height of the stre}etlight \end{gathered}[/tex][tex]\begin{gathered} x=20\times\tan 40 \\ =20\times0.8390 \\ =16.78 \end{gathered}[/tex]Therefore the streetlight is approximately 16.8ft
The right option is C