Question

At a point, the angle of elevation of a tower is such that its tangent is 5
/12
.On
walking 240m towards the tower, the tangent of the angle of elevation becomes
3
/4
. Find the height of the tower.​

Answer 1

$tan\theta = \frac{5}{12} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( eqn1)$

$and, \: tan\theta = \frac{3}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:( eqn2)$

BC AB

$let \:BC \: = x \: m\: \: \: and \: AB = y \: m$

$NOW, In \: right \: angled \: triangle \\ ABC \ \\ tan \theta = \frac{y}{x} \: \: \: \: \: \: \: \: \: \: (eqn \: 3)$

$from \: {2}^{nd \:} and \: {3}^{rd} \\ \implies \frac{3}{4} = \frac{y}{x}$

$\implies \: x = y \times \frac{4}{3} \: \: \: \: \: \: \: \: \: \: (eqn \: 4)$

$Also \: in \: right \: angled \: triangle \\ ABD, \: we \: get \\ \implies tan \theta = \frac{y}{x + 240} \: \: \: \: \: \: \:( eqn \: 5)$

$From \: 1 \: and \: 5, we \: get \\ \implies \: \frac{5}{12} = \frac{y}{x + 240}$

$\implies 5(x + 240) = 12(y)$

$\implies \: 5x + 1200 = 12y$

$\implies12y = 4 \times \frac{4}{3} y + 1200 \: \\ (using \: 4)$

$\implies12y - \frac{20}{3}y = 1200$

$\implies \: \frac{36y - 20y}{3} = 1200$

$\implies16y = 3600$

$\implies y = \frac{3600}{16}$

$\implies \: y = 225$

$Hense, \: the \: height \: of \: Tower \: is \: \\ 225 \: meter$

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