The angles of depression of the top and bottom of a 12 m tall building, from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multi-storeyed building.

Question

Philoid · Accepted Answer

Given: Height of the building = DE = 12 m

Let AC be the multi-storeyed building.

Let Height of the multi-storeyed building = y m

Let distance between the two building = DC = EB = x m

Now, in right angled triangle ADC,

tan 60° = AC/DE

⇒ √3 = y/x

⇒ y = √3.x ………….(1)

Also, in the right angles triangle AEB,

tan 30° = AB/BE

⇒ 1/√3 = (y-12)/x

Put the value of y from equation (1) in the above equation:

⇒ 1/√3 = (√3x-12)/x

⇒ x = √3 × (√3x-12)

⇒ x = 3x-12√3

⇒ 2x = 12√3

⇒ x = 6√3

∴ y = √3x

⇒ y = √3(6√3)

= 18

∴ Height of the multi-storeyed building,

y = 18 m