Question

From the top of a 32 m building, the angle of elevation of the top of a cable tower is 45o and the angle of depression of its foot is 30o, Find the height of the tower.(√3 =1.73).

Answer 1

Given:

From the top of a 32 m building, the angle of elevation of the top of a cable tower is 45° and the angle of depression of its foot is 30°.

To find:

Height of tower.

Calculation:

From the diagram , we can see that :

BD = CE = 32 m .......(1)

In ∆BDE:

$\tan( {30}^{ \circ} ) = \dfrac{32}{x}$

$= > \dfrac{1}{ \sqrt{3} } = \dfrac{32}{x}$

$= > x = 32 \sqrt{3} \: m$

So, DE = BC = x = 32√3 m ......(2)

In ∆ ABC:

$\therefore \: \tan( {45}^{ \circ} ) = \dfrac{AC}{x}$

$= > \: 1 = \dfrac{AC}{x}$

$= > \: AC = x = 32 \sqrt{3} \: m$

So , height of tower be H ;

H = AC + CE

=> H = 32√3 + 32

=> H = 32 (1+√3)

=> H = 32 ( 1 + 1.73)

=> H = 32 × 2.73

=> H = 87.36 m

So, height of tower is 87.36 m.