Question

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tove
60° and the angle of depression of its foot is 45º. Determine the height of the tower​

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given that the height of a building is 7m
• The angle of elevation of the top of a cable tove is 60° and the angle of depression of its foot is 45º.

To Find:

• We need to find the height of tower.

Solution:

Let BD be the cable tove and AE be the building.

Now, we need to find BD.

AE = CD = 7m

Let BC = x

Now, in ΔACD,

$\sf{ \tan 45° = \dfrac{P}{B} = \dfrac{CD}{AC} = \dfrac{7}{AC}}$

We know that the value of tan 45° = 1.

$\implies\sf{1 = \dfrac{7}{AC} }$

$\implies\sf{AC = 7m}$

Now, in ΔABC, we have

$\sf{ \tan 60° = \dfrac{x}{AC} = \dfrac{x}{7} }$

We know that the value of tan 60° = √3.

$\implies\sf{ \sqrt{3} = \dfrac{x}{7} }$

$\implies\sf{ x = 7 \sqrt{3}m}$

Now, BD = (x + 7m)

$\sf{BD = (7 \sqrt{3} + 7)}$

$\implies\sf{ BD = 7( \sqrt{3} + 1)m}$

Hence, the height of tower is

7(√3 + 1)m.