The angle of elevation to the top of a tower 24√3 m high from two points on the level ground on its opposite sides are 45° and 60°. What is the distance (in m) between the two points?

65.56 m

⇒ AB = 24 \(\sqrt {3}\)

⇒ In Δ ABC:

tan 60° =   \(\sqrt {3}\)    :         1 

                   (AB)                (BC)  

                      ⇓                    ⇓

              24 \(\sqrt {3}\) (given)          24

⇒ In Δ ABD:

tan 45° =        1          :         1 

                   (AB)                (BD)  

                      ⇓                    ⇓

             24 \(\sqrt {3}\) (given)     24 \(\sqrt {3}\)

CD = BC + BD = 24 + 24\(\sqrt {3}\) = 24 (1 + \(\sqrt {3}\)) = 24 (1 + 1.732) = 24 (2.732) = 65.56