Question

from the top of a 7m high building the angle of elevation of the top of a cable tower is 60 degree the angle of depression of its foot is 45 degree determine the height of the tower​

Answer 1

Answer:

the answer is 7(√3+1)..

Answer 2

Given :-

• Height of building (AB) = 7 m

• The angle of elevation of the top of a cable tower is 60°

• The angle of depression of its foot is 45°

• BC = AD , AB = CD = 7 m

To find :-

• Height of the tower (EC)

Solution :-

In right ∆ ABC

$\rm\:tan\:45\degree=\dfrac{p}{b}=\dfrac{AB}{BC}$

$\rm\rightarrow\:1=\dfrac{7}{BC}$

$\rm\rightarrow\:BC=7\:m$

Then,

In right ∆ ADE

$\rm\:tan\:60\degree=\dfrac{p}{b}=\dfrac{DE}{AD}$

$\rm\rightarrow\:\sqrt{3}=\dfrac{DE}{7}$

$\rm\rightarrow\:DE=7\times\:\sqrt{3}$

$\rm\rightarrow\:DE=7\sqrt{3}\:m$

Now,

Height of the tower (EC) = DE + CD

Height of the tower (EC) = 7√3 + 7

Height of the tower (EC) = 7(√3 +1) m

Hence,the height of the tower(EC) will be 7(√3+1) m.