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

Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is ${60}^{\circ }$ and the angle of depression of its foot is ${45}^{\circ }$.
$EC=DE+CD$
Also, $CD=AB=7m$ and $BC=AD$
According to question,
In right $\Delta ABC,$
tan ${45}^{\circ }=\frac{AB}{BC}$
$\Rightarrow 1=\frac{7}{BC}$
$\Rightarrow BC=7m=AD$
Also,
In right $\Delta ADE,$
tan ${60}^{\circ }=\frac{DE}{AD}$
$\Rightarrow \sqrt{3}=\frac{DE}{7}$
$\Rightarrow DE=7\sqrt{3}m$
Height of the tower $=EC=DE+CD$ $=(7\sqrt{3}+7)m=7(\sqrt{3}+1)m$