Answer :

Let AB and CD be the multi-storeyed building and the building respectively. Let the height of the multi-storeyed building = h m and the distance between the two buildings = x m.

AE = CD = 8 m [Given]

BE = AB – AE = (h – 8) m

and AC = DE = x m [Given]

Also, ∠FBD = ∠BDE = 30° (Alternate angles)

∠FBC = ∠BCA = 45° (Alternate angles) 1/2

Now,

In Δ ACB, In Δ BDE,

⇒ tan 45° =

⇒

⇒ x = h ….. (i) 1

In ΔBDE,

⇒ tan30° =

⇒

⇒ ….. (ii)

From (i) and (ii), we get,

h = √3 – 8√3

√3h – h =8√3

h (√3 – 1) = 8√3

h =

h =

h – = 4√3 (√3 + 1)

h = 12 + 4√3 m

Distance between the two building

**OR**

From the figure, the angle of elevation for the first position of the balloon

∟EAD = 60° and for second position ∟BAC = 30°.

The vertical distance 1

ED = CB = 88.2 – 1.2 = 87 m.

Let AD = x m and AB = y m.

Then in right Δ ADE, tan 60° =

…….(i) 1

In right ΔABC, tan 30° =

y = 87√3 ..….(ii)

Subtracting (i) and (ii) 1

y – x = 87√3 –

y – x = 1

y – x = 58√3 m

Hence, the distance travelled by the balloon is equal to BD

y – x = 58√3 m.

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