Question

The angle of elevation of the top of the tower from a point on the ground, which is 25 m away from the foot of the tower is 60° find the height of the tower.​

Answer 1

Answer:

R.E.F. Image.

Height of the tower, H = (h+10)m

In △ABC,

tan60  

∘

=  

AB

h+10

​  

 

AB=  

3

​  

 

h+10

​  

⋯(i)

In △DCO,

tan45  

∘

=  

AB

h

​  

 

∴AB=h⋯(ii)

comparing eq.(i) and (ii), we get :

h=  

3

​  

 

h+10

​  

 

∴  

3

​  

h=h+10

⇒  

3

​  

h−h=10

⇒h(  

3

​  

−1)=10

∴h=  

3

​  

−1

10

​  

 

Height of tower = h + 10

=  

3

​  

−1

10

​  

+10

=  

3

​  

−1

10+10  

3

​  

−10

​  

 

H=  

3

​  

−1

10  

3

​  

 

​  

m