Question

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer 1

Let AB be a building and CD be a cable tower. 

In ∆ABD, 

\(\frac{AB}{ BD} = tan 45^{\degree}\)

\(\frac7{ BD} = 1\)

\(BD = 7\,m\)

In ∆ACE, 
\(AC = BD = 7m\)

\(\frac{CE}{ AE} = tan 60^{\degree}\)

\(\frac{CE} 7 = \sqrt3\)

\(CE = 7\sqrt3\)

\(CD = CE + ED = (7\sqrt3 +7)m\)
\(CD= 7(\sqrt3 + 1)\,m\)

Therefore, the height of the cable tower is \(7(\sqrt3+1) \,m\).