Question

The angle of elevation of the top of a tower is observed to be $60^\circ $ ; at a point $40\,m$ above the first point of observation, the elevation is found to be $45^\circ $ . Find the height of the tower is $hm$ then the value of $\left[h\right]$ is where $\left[\cdot \right]$ is the greatest integer function.

Answer 1

Suppose $AC$ is the tower. The angle of elevation of $A$ at $B$ is $60^\circ $ and $BD=40\,m$ . The angle of elevation at $D$ is $45^\circ $ .
$\tan 60^\circ =\frac{A C}{B C}=\frac{A E + E C}{B C}$
$\Rightarrow \sqrt{3}=\frac{A E + 40}{B C}$
$\Rightarrow AE+40=\sqrt{3}BC...\left(i\right)$
Now, $\tan 45^\circ =\frac{A E}{D E}=1$
$\therefore AE=DE=BC...\left(ii\right)$
From Equations $\left(i\right)\&\left(ii\right)$ ,
$AE+40=\sqrt{3}AE$
$\Rightarrow AE=\frac{40}{\sqrt{3} - 1}=54.641\,m$ .
And $AC=\left(40 + 54 . 641\right)m$ .
$\therefore $ Height of the tower $=94.641\,m$ .
And, horizontal distance of the tower
$BC=AE=54.641\,m$ .