Question

The angle of elevation of the top of a tower from an observer on the ground is 60 .if the tower is 40m high how far is the observer from the foot the tower

Answer 1

Answer:

$\frac{40\sqrt{3} }{3}$  or 23.1

Step-by-step explanation:

By the given information, we know that,

perpendicular height p(let) = 40 m

Angle of elevation Θ(let) = 60°

We know that,

tan Θ = $\frac{p}{b}$                      [where b is the base]

∴ tan 60° = $\frac{40}{b}$

⇒b = $\frac{40}{\sqrt{3} }$

⇒b = $\frac{40\sqrt{3} }{3}$                       [multiplying numerator and denominator by $\sqrt{3}$]

⇒b = 23.0940107676 ≈ 23.1