The angles of depression of the top and bottom of a 8m tall building from the top of a multi storied building are 30° and 45°, respectively. Find the height of the multi storied building and the distance between the two buildings.
OR
A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°.After sometime, the angle of elevation reduces 30°.Find the distance travelled by the balloon during the interval.


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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