80 metre away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height (in metres) of the tower?

AB = Tower

BC = 80 metre

∠ACB = 60°

In ’ABC,

$an 60^{circ}=frac{mathrm{AB}}{mathrm{BC}}$

$Rightarrow sqrt{3}=frac{mathrm{AB}}{80}$

$Rightarrow mathrm{AB}=80 sqrt{3}$ metre