Question

from the top of 7m building the angle of elevation of the top of the cable tower is 60° and angle of depression of the foot is 45° find the height of tower 6mark question​

Answer 1

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

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

$\sf1=\dfrac{7}{BC}$

$\bf{BC=7\:m</p><p>}$

Then,

In right ∆ ADE

$\blue{\rm\:tan\:60\degree=\dfrac{p}{b}=\dfrac{DE}{AD}}</p><p>$

$\sf\sqrt{3}=\dfrac{DE}{7}$

$\sf{DE=7\times\:\sqrt{3}}$

$\sf{DE=7\sqrt{3}\:m}</p><p>$

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.

Answer 2

Answer:

Given :-

Given :-Height of building (AB) = 7 m

Given :-Height of building (AB) = 7 mThe angle of elevation of the top of a cable tower is 60°

Given :-Height of building (AB) = 7 mThe angle of elevation of the top of a cable tower is 60°The angle of depression of its foot is 45°

Given :-Height of building (AB) = 7 mThe 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

Given :-Height of building (AB) = 7 mThe 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 mTo find :-

Given :-Height of building (AB) = 7 mThe 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 mTo find :-Height of the tower (EC)

Given :-Height of building (AB) = 7 mThe 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 mTo find :-Height of the tower (EC)Solution :tan\:45\degree=\dfrac{p}{b}=\dfrac{AB