tan α = [tex] \frac{6.67}{25} [/tex]
α = 14 degree 56' (sorry, there's no degree symbol here)
Let x be the distance of the basketball player from the light
tan 14 degree 56' = [tex] \frac{15}{25 + x} [/tex]
25 + x = [tex] \frac{15}{tan 14 degree 56'} [/tex]
25 + x = 56.22
x = 56.22 - 25
x = 31.22 ft --> answer
