Question

The angle of elevation of the top of 36 m tall tower from the initial position of a person on ground was 60degree she walked away in a manner that the foot of the her initial position and final position are in straight . the angle of of the top of the tower from her final position was 30 degree how much did she walked from her

Answer 1

here is ur answer....

Answer 2

Image

Let us consider CD as x

From the fig, consider $tan60 = \frac { AB }{ BC }$

$\tan { 60\quad =\quad \frac { AB }{ BC } \quad =\quad \sqrt { 3 } } \quad =\quad \frac { 36 }{ BC }$

By simplifying

$BC\quad =\quad 12\sqrt { 3 }$

$\tan { 30\quad =\quad \frac { 36 }{ BC+x } \quad =\quad \frac { 1 }{ \sqrt { 3 } } } \quad =\quad \frac { 36 }{ 12\sqrt { 3+x }}$

By simplifying

=41.568 m.

She walked nearly 41.568 m.