Question

from a point on ground the angle of elevation of the top of a tower is observed to be 60 degrees from a point 40 m vertically above the first point of observation the angle of elevation of the top of the tower is 30 degrees find the height the height of the tower and its horizontal distance from the point of observation

Answer 1

Let the height of the tower be h cm. 

Now, In




And, In



Since, AB = CD, SO, equation (2) becomes,



Equating equation (1) and (3), we get,



Hence, height of tower will be 15 cm.

Answer 2

Answer:

Height of tower is  60 m and Distance between tower & point is 20√3 m.

Step-by-step explanation:

Given: Angle of elevation form C = 60°

           Angle of elevation from D = 30°

           CD  = BE = 40 m

            CB = DE

To find: height of tower , AB and Distance between point & tower , CB

let AE be x

In ΔABC,

$tan\,60^{\circ}=\frac{AB}{CB}$

$\sqrt{3}=\frac{40+x}{CB}$

$CB=\frac{40+x}{\sqrt{3}}$

In ΔADE,

$tan\,30^{\circ}=\frac{AE}{DE}$

$\frac{1}{\sqrt{3}}=\frac{x}{CB}$

$CB=x\sqrt{3}$

from above,

$\sqrt{3}x=\frac{40+x}{\sqrt{3}}$

$3x=40+x$

$2x=40$

$x=20$

AB = 40 + 20 = 60 m

⇒ CB = 20√3 m

Therefore, Height of tower is  60 m and Distance between tower & point is 20√3 m