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 $7(\sqrt3+1) \,m$.