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Sia ? 6 years, 5 months ago
In right triangle ABD,
{tex}\tan {45^0} = \frac{{AB}}{{BD}}{/tex}
{tex} \Rightarrow {/tex} {tex}1 = \frac { 7 } { B D }{/tex}
{tex} \Rightarrow {/tex} BD = 7 m
{tex} \Rightarrow {/tex} AE = 7 m
In right triangle AEC,
{tex}\tan 60 ^ { \circ } = \frac { C E } { A E }{/tex}
{tex}\Rightarrow \sqrt { 3 } = \frac { \mathrm { CE } } { 7 } \Rightarrow \mathrm { CE } = 7 \sqrt { 3 } \mathrm { m }{/tex}
{tex}\therefore{/tex} CD = CE + ED
= CE + AB
{tex}= 7 \sqrt { 3 } + 7 = 7 ( \sqrt { 3 } + 1 ) { \mathrm { m } }{/tex}
Hence height of the tower is {tex}7 ( \sqrt { 3 } + 1 ){/tex} m.
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