From a point on the ground …
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Sia ? 6 years, 6 months ago
Let AP = y m and BC = xm.
{tex}\therefore {/tex} In {tex}\triangle B A P = \frac { B A } { P A } = \tan 45 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \frac { 20 } { y } = 1{/tex}
{tex}\Rightarrow {/tex} y = 20
In {tex}\triangle {/tex}CAP {tex}\frac { C A } { P A } = \tan 60 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \frac { 20 + x } { y } = \sqrt { 3 }{/tex}
{tex}\Rightarrow \quad 20 + x = y \sqrt { 3 }{/tex} (using y = 20)
{tex}\Rightarrow \quad 20 + x = 20 \sqrt { 3 }{/tex}
{tex}x = 20 \sqrt { 3 } - 20{/tex}
{tex}= 20 ( \sqrt { 3 } - 1 ){/tex}
{tex}= 20 \times ( 1 \cdot 73 - 1 ){/tex}
Hence, height of the tower = 14.64 m
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