$ A B C $ is a triangular park with $ A B=A C=100 $ metres. A vertical tower is situated at the mid-point of $ \mathrm{BC} $. If the angles of elevation of the top of the tower at, $ A $ and $ B $ are $ \cot ^{-1}(3 \sqrt{2}) $ and $ \operatorname{cosec}^{-1}(2 \sqrt{2}) $ respectively, then the height of the tower (in metres) is

Question

$ A B C $ is a triangular park with $ A B=A C=100 $ metres. A vertical tower is situated at the mid-point of $ \mathrm{BC} $. If the angles of elevation of the top of the tower at, $ A $ and $ B $ are $ \cot ^{-1}(3 \sqrt{2}) $ and $ \operatorname{cosec}^{-1}(2 \sqrt{2}) $ respectively, then the height of the tower (in metres) is, $ \frac{100}{3 \sqrt{3}} $, $ 20 $, $ 25 $, $ 10 \sqrt{5} $

MARKS App · Accepted Answer

Let A D = y & D E = h (Tower) ∴ y 2 = 100 2 - x 2 Now cot θ = 3 2 ⇒ tan θ = 1 3 2 = h y ⇒ y = 3 2 h ⇒ y 2 = 100 2 - x 2 = 18 h 2 … 1 ⇒ tan ∅ = 1 7 = h x ⇒ x = h 7 … 2 Form 1 & 2 100 2 - 7 h 2 = 18 h 2 ⇒ 25 h 2 = 100 2 ⇒ h = 20

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