Question

From the top of a 7 m high building, the angle of elevation of the top of a cable tower measures ${60}^0$ and the angle of depression of its foot measures ${45}^0.$ Determine the height of the tower. $(\sqrt{3}=1.732)$

Answer 1

The diagram will be as shown

$AB$ =building

$CD=$ Cable tower

$CE=AB$

In $△ABC$

$\frac{AB}{BC}=\tan {45}^o$

$\Rightarrow$ $AB=BC=7m$

$AE=BC=7m$

In $△AED$

$\frac{DE}{AE}=\tan {60}^o$

$\Rightarrow$ $DE=AE\left(\sqrt{3}\right)=7\sqrt{3}$

Heignt of cable tower $=CD$

$=CE+DE$

$=7+7\sqrt{3}$

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

$=7\left(2.732\right)$

$=19.124m$