the angle of depression of the top and the bottom of an 8m tall building from the top of a multi-storeyed building are 30° and 45° ,respectively .find the height of the multi- storyed building and the distance between the two buildings.
Answers
→ 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
AC = DE = x m [Given]
Also,
∠FBD = ∠BDE = 30° [ Corresponding angles ]
∠FBC = ∠BCA = 45° [ Corresponding angles ]
Now,
In Δ ACB,
tan 45° = AB / AC
1 = h / x ----(i)
In Δ BDE,
tan 30° = BE / ED
1 / √3 = h - 8 / x
⇒ x = √3 (x - 8). ----(ii)
From eqn. (i) and (ii),
⇒ h = √3h - 8√3
⇒ √3h - h = 8√3
⇒ h (√3 - 1) = 8√3
⇒ h = 8√3 / √3 - 1
⇒ h = 8√3 (√3 + 1) / 2
⇒ h = 4√3 (√3 + 1)
⇒ h = 12 + 4√3 m
and,
Distance between the two building :-
h = x = 12 + 4√3
Hence, The required answer is :-
Height of tower (h) = Distance between two building (x) = 12 + 4√3
Hope it helps! :)
Hi there!
→ 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
AC = DE = x m [Given]
Also,
∠FBD = ∠BDE = 30° [ Corresponding angles ]
∠FBC = ∠BCA = 45° [ Corresponding angles ]
Now,
In Δ ACB,
tan 45° = AB / AC
1 = h / x ----(i)
In Δ BDE,
tan 30° = BE / ED
1 / √3 = h - 8 / x
⇒ x = √3 (x - 8). ----(ii)
From eqn. (i) and (ii),
⇒ h = √3h - 8√3
⇒ √3h - h = 8√3
⇒ h (√3 - 1) = 8√3
⇒ h = 8√3 / √3 - 1
⇒ h = 8√3 (√3 + 1) / 2
⇒ h = 4√3 (√3 + 1)
⇒ h = 12 + 4√3 m
and,
Distance between the two building :-
h = x = 12 + 4√3
Hence, The required answer is :-
Height of tower (h) = Distance between two building (x) = 12 + 4√3
Hope it helps! :)