Question : Suhas, a 3.15 m tall tree, and a building are positioned such that their feet on the ground are collinear and the tree is located between Suhas and the building. The tree is located at a distance of 7.5 m from Suhas and a distance of 45 m from the building. Further, the eyes of Suhas, the top of the tree, and the top of the building fall in one line, and the eyes of Suhas are at a height of 1.8 m from the ground. Find the height (in m) of the building.
Option 1: 11.25
Option 2: 11.50
Option 3: 10.25
Option 4: 10.75
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Correct Answer: 11.25
Solution :
The height of the tree, BY = 3.15 m
Let the height of the building, CZ be $h$.
Height of suhas, XA = 1.8 m
The distance between Suhas and the tree, AB = 7.5 m
The distance between the tree and the building, BC = 45 m
So, AC = 7.5 + 45 = 52.5 = PX
OY = BY – BO = 3.15 – 1.8 = 1.35
It is given that the eyes of Subhas, the top of the tree, and the top of the building fall in one line.
The angle made between this line and the straight line of sight of Subhas is θ.
In $\triangle$XOY,
$\tan\theta=\frac{OY}{OX}=\frac{1.35}{7.5}$ [As AB = OX]
$\tan\theta=\frac{9}{50}$ -----(1)
Similarly, $\triangle$XPZ,
$\tan\theta=\frac{PZ}{PX}=\frac{h-1.8}{52.5}$ [As PZ = CZ – CP, CP = XA]-----(2)
From equations (1) and (2), we get,
$\frac{9}{50}=\frac{h-1.8}{52.5}$
$⇒50h-90=472.5$
$⇒50h=562.5$
$\therefore h = 11.25$ m
Hence, the correct answer is 11.25 m.
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