The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos -1((3/√13)). If the distance of the point B from the tower is 15 units, then cot α is equal to :

Question

The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos -1((3/√13)). If the distance of the point B from the tower is 15 units, then cot α is equal to : (A) (6/5) (B) (9/5) (C) (4/3) (D) (7/3)

Tardigrade Admin · Accepted Answer

given OB =15 cos β=(3/√13)

tan β=(2/3)

tan β=( h /15) (2/3)=(h/15) 10=h

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

tan α=(10/12) cot α=(12/10)=(6/5)

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

$\frac{6}{5}$

$\frac{9}{5}$

$\frac{4}{3}$

$\frac{7}{3}$

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

$tan β = \frac{2}{3}$

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

$O A^{2} + A B^{2} = 225$
$O A^{2} + 81 = 225$
$O A = 12$

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

Q. The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos−1(13​3​). If the distance of the point B from the tower is 15 units, then cotα is equal to :

56​

59​

34​

37​