Question

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of a tower from the foot of the building is 60°. If the tower is 50 m high, then find the height of the building.

Answer 1

Let AB be the building and CD be the tower.

In ΔCDB,

`"CB"/"BD"` = tan 60°

`50/("BD") = sqrt3`

`"BD" = 50/sqrt3`

In ΔABD,

`("AB")/("BD") = tan 30°`

AB = `50/sqrt3 xx 1/sqrt3`

= `50/3`

= `16 2/3`

Therefore, the height of the building is `16 2/3` m.

Answer 2

Let AD be the building of height h m. and an angle of elevation of the top of building from the foot of the tower is 30° and an angle of the top of the tower from the foot of building is 60°.

Let AD = h, AB = x and BC = 50 and ∠DBA = 30°, ∠CAB = 60°

So we use trigonometric ratios.

In a triangle ABC

⇒ `tan 60° = 50/x`

⇒ `sqrt3 = 50/x`

⇒ `x = 50/sqrt3`

Again in a triangle ABD

⇒ `tan 30° = ("AD")/("AB")`

⇒ `1/sqrt3 xx h/x`

⇒ `h = x/sqrt3`

⇒ `h = 50/(sqrt3 xx sqrt3)`

⇒ `h = 50/3`

⇒ `h =16 2/3`

Therefore, the height of the building is `16 2/3` m.

Answer 3

According to the question,

In ΔABD

`tan 60^circ = "AB"/"BD"`

⇒ `sqrt(3) = 50/"BD"`

⇒ `"BD" = 50/sqrt(3)`

Now, in ΔBDC,

⇒ `tan 30^circ = "CD"/"BD"`

⇒ `1/sqrt(3) = h/(50/sqrt(3))`

⇒ `1/sqrt(3) = (hsqrt(3))/50`

⇒ 3h = 50

⇒ `h = 50/3`

⇒ h = 16.67

Hence, the height of the building is 16.67 m.