Question

from the top of the building the angle elevation of the top of the tower is 60degree and the amgle of depresion of its foot is 45 degree find the height of tower ​

Answer 1

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

$\implies AB= height\: of \:the\: building=h_b\:(let)$

$\implies CD=height \:of \:the\: tower=h_t\:(let)$

$\implies \angle CAE=angle\: of\: elevation \:of\: the\: top \\\:of \:the\: tower\: from\: the \:top\: of\: the \:building.$

$\implies \angle DAE=angle\: of\: depression \:of\: the\\ feet \:of \:the \:tower\: from\: the\: top\: of\: the \:building.$

$\implies BD=distance\: between \:the\: building\: and\:\\ the \:tower\: from\: feet\: to\: feet.$

Now.......

$\rightarrow AE=BD=x\:(let)\\ \rightarrow</p><p>AB=DE=h_b \:(\because\: according\: to\: the\: figure)\\ \rightarrow CD=h_t=DE+CE=h_b+CE\\</p><p>\rightarrow \boxed{\bf\red{h_t=h_b+CE}}$

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$\star$ NOw from the Right angled triangle ADE :-

$\implies \bf tan 45\degree=\frac{AE}{DE}\\</p><p>\implies\bf 1=\frac{BD}{h_b}\\ \implies\boxed{ \bf\blue{\bf x=h_b}}\:\:(\because BD=x)$

& again

$\implies \bf tan 60\degree=\frac{CE}{AE}\\</p><p>\implies\bf \sqrt{3}=\frac{CE}{x}\\ \implies \bf \sqrt{3}=\frac{CE}{h_b}\\ \implies\boxed{ \blue{CE=\sqrt{3}h_b}}$

$\bf\therefore h_t=h_b+CE\\ \bf\implies h_t=h_b+\sqrt{3}h_b\\ \bf\implies\boxed{ \bf\red{h_t=(1+\sqrt{3})h_b}}$

$\bf\therefore Height \:of \:the\: tower\\=(1+\sqrt{3})\times Height \:of \:the\: building$

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

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