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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Question : Suhas, 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 Suhas and the building. The tree is located at a distance of 45 m from the building. Further. the eyes of Suhas, the top of
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