Question

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Answer 1

$\sf In \: ∆ADE,$

$\sf tan \: 60° = \frac{60}{DE}$ = DE = $\sf \frac{60}{ \sqrt{3} }$ = $\sf 20 \sqrt{3}$

$\sf and \: we \: can \: see \: that, \: BCDE \: rec/tan \: gle$

$\sf So, \: BC = DE ⇒BC = 20 \sqrt{3}$

$\sf and \: BD = CE \: ......(1)$

$\sf and \: in \: ∆ABC,$

$\sf tan \: 30° =$ $\sf \frac{AB}{20\sqrt{3} }$ $\sf ⇒AB = 20\sqrt{3} \times \frac{1}{ \sqrt{3} } = 20$

$\sf Now, \: as \: AD = AB + BD \\ \sf⇒ 60 = 20 + BD \\ \sf⇒ BD = 40$

$\sf And \: from \: (1),$

$\sf BD = CE = 40 \: (which \: is \: the \: hieght \: of \: the \: building)$

$\sf Therefore, \: answer \: is \: 40.$