Question

Two men standing on either side of a tower of 60m high observe the angle of elevation of the top of tower to be 45degree and 60degree respectively. Find the distance between the two men

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{\mathcal{Solution}}}}}$

$Given \begin{cases}</p><p></p><p>\small{\textbf{1. height(h) of the tower=60 metre}} \\</p><p></p><p>\small{\textbf{2. Angle of elevation from one side is }}\: 45\degree\\ </p><p>\small{\textbf{3. Angle of elevation from other side is }}\: 60\degree</p><p></p><p>\end{cases}$

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$\underline{\large{\red{\textbf{Let's Solve The Problem}}}}$

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Now for the Triangle ∆ABC & ∆ABD ,

$\bf\Longrightarrow\angle ABC=90\degree\\</p><p>\bf\Longrightarrow\angle ABD=90\degree$

$\therefore$∆ABC & ∆ABD are Right angled triangle.

$\underline{\red\textbf{From the Fig:-}}$

$\bf\Longrightarrow height of the tower=AB=60m\\</p><p>\bf\Longrightarrow\angle ACB=45\degree\\</p><p>\bf\Longrightarrow\angle ADB=60\degree\\</p><p>\bf\Longrightarrow BC=x_1 \:(let)\\</p><p>\bf\Longrightarrow BD=x_2\:(let)\\</p><p>\bf\therefore\footnotesize{ the\: distance\: between\: the \:two\: men=CD=BC+BD=(x_1+x_2) \: metres}$

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Now Applying Trigonometric Knowledge

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1)To the ∆ABC:-

$\bf\Longrightarrow tan\angle ACB=\frac{AB}{BC}\\ \bf\Longrightarrow tan45\degree=\frac{60}{x_1}\\ \bf\Longrightarrow</p><p>1=\frac{60}{x_1}\\ \bf\Longrightarrow\boxed{ \bf x_1=60\:metre}$

2)To the ∆ABD:-

$\bf\Longrightarrow tan\angle ADB=\frac{AB}{BD}$

$\bf\Longrightarrow tan60\degree=\frac{60}{x_2}$

$\bf\Longrightarrow</p><p>\sqrt{3}=\frac{60}{x_2}\\ \bf\Longrightarrow \bf x_2=\frac{60}{\sqrt{3}}\:\\</p><p> \bf\Longrightarrow \bf x_2=\frac{20\times\sqrt{3}\times\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}}$

$\bf\Longrightarrow\boxed{ \bf x_2=20\sqrt{3}\:metre}$

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$\bf\therefore\normalsize{ the\: distance\: between\: the \:two\: men}\\</p><p>\bf\Longrightarrow(x_1+x_2)\\</p><p> \bf\Longrightarrow(60+20\sqrt{3})\: metres\\</p><p>\bf\Longrightarrow( 60+34.641)\:metres\\</p><p>\bf\Longrightarrow\boxed{\large{\mathfrak\red{\bf 94.641\:metres}}}\:(approx)$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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