The angle of elevation of the …
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Sia ? 6 years, 4 months ago
Given,
Let the height of the tower be h cm.
Now, In {tex}\triangle{/tex}PAB,
{tex}\tan 60^{\circ}=\frac{A P}{A B}{/tex}
{tex}\Rightarrow \sqrt{3}=\frac{h}{A B}{/tex}
{tex}\Rightarrow A B=\frac{h}{\sqrt{3}}{/tex} .....(i)
And, In {tex}\triangle{/tex}PCD,
{tex}\tan 30^{\circ}=\frac{P D}{C D}{/tex}
{tex}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{C D}{/tex}
{tex}\Rightarrow{/tex} CD = {tex}\sqrt{3}(h-10){/tex} ...(ii)
Since, AB = CD, so equation (ii) becomes,
{tex}A B=\sqrt{3}(h-10){/tex} ....(iii)
Equating equation (i) and (iii), we get,
{tex}\frac{h}{\sqrt{3}}=\sqrt{3}(h-10){/tex}
{tex}\Rightarrow h=3(h-10){/tex}
{tex}\Rightarrow{/tex} {tex}2h = 30 {/tex}
{tex}\Rightarrow{/tex} {tex}h = 15 cm{/tex}
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