Question

A man stands on the ground at a point A, Which is on the same horizontal plane as B, the foot of a vertical Pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground He observes the angle of clevation at C, the top of the pole as $x^{\circ }$,
where tan $x^{\circ }=\frac{2}{5}.$ The distance AB ( in meters ) is

• A. 15 m
• B. 18 m
• C. 20 m
• D. 16 m

Answer 1

The correct option is C 20 m

Height of the pole = 10 m
height of the man's eye above the ground = 2 m
Let the man's eye be at point A' and similar point on the pole be B'
Now, In $△A^{\prime }{B}^{\prime }C$
$\tan x=\frac{B^{\prime }C}{A^{\prime }{B}^{\prime }}$
$\frac{2}{5}=\frac{8}{A^{\prime }{B}^{\prime }}$
$A^{\prime }{B}^{\prime }=20$ m
Now, $AB=A^{\prime }{B}^{\prime }=20$ m