Question

The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are 30° and 45° respectively. Find the height of the tower and the distance between the tower and the building.

Answer 1

Let AB be the building and CD be the tower.

Let BE ⊥ CD. Then, AB = 8 m and CE = AB = 8 m.

AB = 8 m, CE = AB = 8 m, ∠EBD = 30° and ∠CAD = 45°. 

Let DE = x m.

From right ΔACD, we have

`(AC)/(CD) = cot 45^circ = 1`

⇒ `(AC)/((8 + x) m) = 1`

⇒ AC = (8 + x) m   ...(i)

∴ BE = AC = (8 + x) m.

From right ΔBED, we have

`(DE)/(BE) = tan 30^circ = 1/sqrt(3)`

⇒ `x/(8 + x) = 1/sqrt(3) xx sqrt(3)/sqrt(3) = sqrt(3)/3`

∴ `3x = sqrt(3)(8 + x)`

⇒ `(3 - sqrt(3))x = 8sqrt(3)`

∴ `x = (8sqrt(3))/((3 - sqrt(3))`

= `(8sqrt(3))/((3 - sqrt(3))) xx ((3 + sqrt(3)))/((3 + sqrt(3)))`

= `(24(sqrt(3) + 1))/((9 - 3))`

= 4(1.732 + 1)

= 10.928

Height of the tower = CD

= CE + ED 

= (8 + x) m 

= (8 + 10.928) m

= 18.928 m

Distance between tower and building = AC

= (8 + x) m   ...[From (i)]

= (8 + 10.928) m 

= 18.928 m