Question

From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is ${60}^{\circ }$ and the angle of depression of its foot is ${45}^{\circ }$. Determine the height of the tower.

Answer 1

$EC=DE+CD$

$CD=AB=7\text{m}$ and $BC=AD$

In $△ABC$,

$\tan {45}^{\circ }=\frac{AB}{BC}=\frac{CD}{AD}$

$BC=7\times 1\text{m}$ $=AD=7\text{m}$

$AD=CD=7\text{m}$

In $△ADE$,

$\tan {60}^{\circ }=\frac{DE}{AD}=\frac{DE}{CD}$

$DE=7\sqrt{3}\text{m}$

Height of tower $=EC=DE+CD$

$=(7\sqrt{3}+7)\text{m}$

$=7(\sqrt{3}+1)\text{m}$