From the top of a 7 m high building, the angle of the elevation of the top of a tower is $60^{o}$ and the angle of the depression of the foot of the tower is $30^{o}$. Find the height of the tower.
Given: Height of the building$=7$ m, angle of the elevation off the top of the tower$=60^{o}$ and the angle of depression of the foot of the tower$=30^{o}$
To do: To find out the height of the tower.
Solution:
Let AB is the building and CD is tower
As given
$\angle EAD=60^{o}$
$\because AE\parallel BC$
$\Rightarrow \angle ACB=\angle EAC=45^{o}$
In $\vartriangle ABC$, $tan45^{o} =\frac{AB}{BC} =\frac{7}{BC} =1$ $( \because tan45^{o} =1)$
$BC=7$
$BC=AE=7$
Now in $\vartriangle AED$
$tan60^{o} =\sqrt{3} =\frac{DE}{AE} =\frac{DE}{7}$
$\Rightarrow DE=7\sqrt{3}$
We know that the height of the tower is $CD=CE+ED=7+7\sqrt{3} =7\left( 1+\sqrt{3}\right)$
$=19.13$ m.
$\therefore$ Height of the tower is 19.13 m
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