Question

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 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 :

• A. $\frac{6}{5}$
• B. $\frac{9}{5}$
• C. $\frac{4}{3}$
• D. $\frac{7}{3}$

Answer 1

Answer: (A)

given $OB =15$
$\cos \beta=\frac{3}{\sqrt{13}}$

$\tan \beta=\frac{2}{3}$

$\tan \beta=\frac{ h }{15}$
$\frac{2}{3}=\frac{h}{15} $
$ 10=h$

$OA ^2+ AB ^2=225 $
$OA ^2+81=225 $
$ OA =12$

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