Question : 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 7.5 m from Subhas and a distance of 45 m from the building. Further, the eyes of Subhas, the top of the tree, and the top of the building fall in one line. Find the height (in m ) from the ground at which Subhas's eyes are situated.
Option 1: 1.75
Option 2: 1.8
Option 3: 1.6
Option 4: 1.5
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Correct Answer: 1.8
Solution :
Let XA be Subhas, YB be the tree, and ZC be the building.
YB = 3.15 m
ZC = 11.25 m
AB = 7.5 m
BC = 45 m
XO = AB = 7.5 m
OP = BC = 45 m
⇒ XP = XO + OP = 7.5 + 45 = 52.5 m
Let the height from the ground at which Subhas's eyes are situated be $h$.
OY = BY – OB = 3.15 – $h$
PZ = ZC – PC = 11.25 – $h$
Let the angle of elevation be $\theta$.
In $\triangle$XPZ, $\tan \theta = \frac{\text{PZ}}{\text{PX}} = \frac{11.25-h}{52.5}$ -----------(i)
In $\triangle$XOY, $\tan \theta = \frac{\text{OY}}{\text{OX}} = \frac{3.15-h}{7.5}$ ------------(ii)
$⇒ \frac{11.25-h}{52.5} = \frac{3.15-h}{7.5}$
$⇒ \frac{11.25-h}{7} = \frac{3.15-h}{1}$
$⇒11.25-h = 7(3.15-h)$
$⇒11.25-h = 22.05-7h$
$⇒6h = 10.8$
$\therefore h = \frac{10.8}{6} = 1.8$ m
Hence, the correct answer is 1.8.
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