Question

the+
angle+of+elevation+of+a+top+of+tower+from+a+point+was+observed+to+be+45+degree+on+walking+30+meter+away+from+the+point+it+was+found+to+be+30+degree+find+the+height+of+tower

Answer 1

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