The angle of elevation of the …
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Posted by Utkarsh Kumar Yadav 1 year, 8 months ago
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Ankit Class 1 year, 8 months ago
Ankit Class 1 year, 8 months ago
Ankit Class 1 year, 8 months ago
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Preeti Dabral 1 year, 8 months ago
Let distance between the two towers = AB = x m
and height of the other tower = PA = h m
Given that, height of the tower = QB = 30 m and {tex}\angle{/tex}QAB = 60°, {tex}\angle{/tex}PBA = 30°.
Now, in {tex}\Delta Q A B, \tan 60^{\circ}=\frac{Q B}{A B}=\frac{30}{x}{/tex}
{tex}\Rightarrow \sqrt{3}=\frac{30}{x}{/tex}
{tex}x=\frac{x^{2}}{3}-\frac{x}{3}-\frac{x-3}{3}-x+0.5 \pi{/tex}
and in {tex}\Delta P B A{/tex}
{tex}\tan 30^{\circ}=\frac{P A}{A B}=\frac{h}{x}{/tex}
{tex}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{10 \sqrt{3}}{/tex} {tex}[\because x=10 \sqrt{3} \mathrm{m}]{/tex}
{tex}\Rightarrow h = 10 m{/tex}
Hence, the requied distance and height are {tex}10\sqrt3{/tex} and 10m, respectively.
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