Question

The angle of elevation of a tower at a point is 45 degree.after going 40m towards the foot of the tower,the angle of elevation of the tower becomes 60 degree.find the height of the tower

Answer 1

Plz refer attachment below

Answer 2

Answer:

94.641 m

Step-by-step explanation:

Refer the attached figure

AB is the height of the tower

The angle of elevation of a tower at a point is 45 degree i.e. ∠ADC = 45°

After going 40 m towards the foot of the tower,the angle of elevation of the tower becomes 60 degree.i.e. ∠ACB = 60°

CD = 40 m

Let CB be x

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 60^{\circ} = \frac{AB}{BC}$

$\sqrt{3}= \frac{AB}{x}$

$\sqrt{3}x=AB$   ---1

In ΔADB

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 45^{\circ} = \frac{AB}{BD}$

$1= \frac{AB}{x+40}$

$x+40=AB$    -----2

With 1 and 2

$\sqrt{3}x=x+40$  

$\sqrt{3}x-x=40$  

$(\sqrt{3}-1)x=40$  

$x=\frac{40}{(\sqrt{3}-1)}$  

$x=54.641$  

AB = x+40=54.641+40=94.641 m

Hence the height of the tower is 94.641 m