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.
Solution and Explanation
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 \(7(\sqrt3+1) \,m\).
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