Applications of trigonometry

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Question

$ABC is a triangular park with AB = AC = 100 m . A clock tower is situated at the midpoint$
$of BC . The angles of elevation of the top of the tower at A and B are \cot^{- 1} 3.2, {cosec}^{- 1} 2.6$
$The height of tower is \dots mt$

Moderate

Solution

$A B = A C = 100 m (correction)$
$\begin{array}{l} Let D P = clock tower = hm \\ ∠ P A D = α = \cot^{- 1} 3.2 \Rightarrow \cot α = 3.2 \\ ∠ P B D = β = {cosec}^{- 1} 2.6 \Rightarrow \cos e c β = 2.6 \\ \cot β = 2.4 \\ In △ P A D and 9 n Δ P B D \end{array}$
$\begin{array}{l} A D = h \cot α = 3.2 h B D = h \cot B = 2.4 h \\ In △ A B D \\ A B^{2} = A D^{2} + B D^{2} \\ 100^{2} = [(3.2)^{2} + (2 . h)^{2}] h^{2} = 16 h^{2} \end{array}$
$∴ h = 25 m$

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Applications of trigonometry

Remember concepts with our Masterclasses.

ABC is a triangular park with AB=AC=100m . A clock tower is situated at the midpoint of BC . The angles of elevation of the top of the tower at A and B are cot−1⁡3.2,cosec−1⁡2.6 The height of tower is …mt

AB=AC=100m (correction) Let DP= clock tower =hm∠PAD=α=cot−1⁡3.2⇒cot⁡α=3.2∠PBD=β=cosec−1⁡2.6⇒cos⁡ecβ=2.6cot⁡β=2.4 In △PAD and 9nΔPBDAD=hcot⁡α=3.2hBD=hcot⁡B=2.4h In △ABDAB2=AD2+BD21002=(3.2)2+(2.h)2h2=16h2∴h=25m

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free study material

$ABC is a triangular park with AB = AC = 100 m . A clock tower is situated at the midpoint$
$of BC . The angles of elevation of the top of the tower at A and B are \cot^{- 1} 3.2, {cosec}^{- 1} 2.6$
$The height of tower is \dots mt$