50 ft from the tower
angle=60°
since we know the angle and the Adj leg
we want to solve the Opp leg
then
[tex]tan(\theta)=\frac{Opp}{Adj}[/tex][tex]tan(60)=\frac{Opp}{60}[/tex][tex]60tan(60)=Opp[/tex][tex]60\sqrt{3}=Opp[/tex][tex]Opp=103.92304[/tex]
Rounded to the nearest tenth
the tower is 103.9 ft tall
