Question

the angle of elevation of the top of a tower from a point on the ground is 45 degree for a while walking 3 m towards the tower the angle of elevation become 60 degree find the height of the tower and the original distance from the foot of the tower

Answer 1

this is the correct answer

Answer 2

Answer:

The height of the tower and the original distance from the foot of the tower is 7.09 m and 4.09 m respectively .

Step-by-step explanation:

Refer the attached figure

Height of the tower = AB

In ΔABC

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

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

$\sqrt{3}BC= AB$

In ΔABD

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

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

$1 = \frac{AB}{BC+3}$

$BC+3=AB$

So, $BC+3=\sqrt{3}BC$

$BC =4.09$

AB = BC+3 = 4.09+3=7.09 m

Hence the height of the tower and the original distance from the foot of the tower is 7.09 m and 4.09 m respectively .