The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

We have

Let height of building be AB = 50 m as given & height of the tower be CD.

The angle of depression of the top of the building = 30°

The angle of depression of the bottom of the building = 60°

Or ∠QDB = 30° & ∠QDA = 60°

Draw BE parallel to AC. Also, AB = 50 m = CE.

Since BE ∥ QD, therefore ∠QDB = ∠DBE = 30° as they are alternate angles.

Also since BE ∥ AC, therefore ∠QDA = ∠DAC = 60° as they are alternate angles.

Now we can apply trigonometric identities in ∆DBE,

tan 30° =

[∵ tan θ = ]

⇒

[∵ tan 30° = 1/√3]

⇒ BE = √3 DE …(i)

In ∆DAC,

tan 60° =

⇒

[∵ tan 60° = √3]

⇒ DC = √3 AC …(ii)

Now since AC = BE, we can re-write the equation (ii) as

DC = √3 BE

⇒ BE = DC/√3 …(iii)

Comparing equations (i) & (iii), we get

√3 DE = DC/√3

⇒ 3 DE = DC

⇒ 3 DE = DE + EC [∵ DC = DE + EC]

⇒ 3 DE – DE = EC

⇒ 2 DE = EC

⇒ 2 DE = 50 [∵ EC = 50 m]

⇒ DE = 50/2

⇒ DE = 25

DC = DE + EC = 25 + 50 = 75 m

Thus, the height of the tower is 75 m.

Substituting DC = 75 m in equation (ii), we get

75 = √3 AC

⇒ AC = 75/√3

⇒ AC =

[Multiplying and dividing by √3]

⇒ AC = 75√3/3

⇒ AC = 25√3

Hence, the distance between two building is 25√3 m.