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The angle of elevation of the top of a tower from a point A due north of it is $\alpha$ and from a point B at a distance of 9 units due west of $\mathrm{A \: is \cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)}$ . If the distance of the point B from the tower is 15 units, then $\cot \alpha$ is equal to :
Option: 1
$\frac{6}{5}$

Option: 2
$\frac{9}{5}$

Option: 3
$\frac{4}{3}$

Option: 4
$\frac{7}{3}$

Answers (1)

best_answer

$\text { Giren } \mathrm{O B=15}$

$\mathrm{\cos B =\frac{3}{\sqrt{13}} }$

$\mathrm{\tan B =\frac{h}{\sqrt{15}} }$
$\mathrm{\frac{2}{\sqrt{3}}=\frac{h}{15} }$
$\mathrm{h=10 }$

$\begin{gathered} \mathrm{O A^2+A B^2=225} \\ \mathrm{O A^2+81=225} \\ \mathrm{O A=12} \end{gathered}$

$\mathrm{\tan 2=\frac{10}{12} }\\ \mathrm{\cot \alpha=\frac{12}{10}=\frac{6}{5}}$

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seema garhwal

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