Subhas, a 3.15 m tall tree, and an 11.25 m tall building are positioned such that their feet on the ground are collinear and the tree is located between Subhas and the building. The tree is located at a distance of 45 m from the building. Further, the eyes of Subhas at the top of the tree and the top of the building fall in one line, and the eyes of Subhas are at a height of 1.8 m from the ground. At what distance ( in m ) from Subhas is the tree located?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 7.5.
Given that
Subhas, a 3.15 m tall tree, and an 11.25 m tall building are positioned such that their feet on the ground are collinear.
A tall tree PD = 3.15 m
A building QF = 11.25 m
Distance between tree and building CE = 45 m
'A' is a Subhas eye that makes an angle θ at the top of the tree and the top of building.
Let the distance between Subhas and tree be AC
Calculation
AB = CD = EF = 1.8 m
So, QE = QF - EF = 11.25 - 1.8 = 9.45 m
PC = PD - CD = 3.15 - 1.8 = 1.35 m
In triangle AQE, and In triangle PAC
⇒ tan θ = QE/AE = PC/AC
⇒ 9.45/AE = 1.35/AC
⇒ 9.45 AC = 1.35 AE = 1.35 (AC + CE) = 1.35 (AC + 45) = 1.35 AC + 1.35 × 45
⇒ 9.45 AC - 1.35 AC = 60.75
⇒ 8.1 AC = 60.75
⇒ AC = 60.75/8.1 = 7.5 m