Question

the angle of elevation top of a tower 30m high from the foot of another tower in the same plane is 60 degree ,and the angle of elevation of the top of the second tower from the foot of first tower is 30 degree .find the distance between the two Towers and also height of the other tower.
please answer soon.....

Answer 1

The required answer is in the attachment.

Answer 2

Answer:

The distance between the two Towers is 17.320 m and the height of another tower is 9.9 m.

Step-by-step explanation:

Refer the attached figure

Height of first tower = DC = 30 m

Height of second tower = AB

The angle of elevation top of a tower 30m high from the foot of another tower in the same plane is 60 degree i.e. ∠DBC=60°

The angle of elevation of the top of the second tower from the foot of first tower is 30 degree .i.e.∠ACB = 30°

Now we are supposed to find  the distance between the two Towers i.e. BC and height of tower AB

In ΔDBC

Using trigonometric ratio

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

$Tan60^{\circ} = \frac{DC}{BC}$

$\sqrt{3} = \frac{30}{BC}$

$BC= \frac{30}{\sqrt{3}}$

$BC= 17.320$

So, the distance between the two Towers is 17.320 m

Now In ΔABC

Using trigonometric ratio

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

$Tan30^{\circ} = \frac{AB}{BC}$

$\frac{1}{\sqrt{3}} = \frac{AB}{ 17.320}$

$17.320 \times\frac{1}{\sqrt{3}} =AB$

$9.9 =AB$

Hence the distance between the two Towers is 17.320 m and the height of another tower is 9.9 m.