Q. $ABC$ is a triangular park with $AB = AC =100\, m$. A TV tower stands at the mid-point of $BC$. The angles of elevation of the top of the tower at $A, B, C$ are $45^{\circ}, 60^{\circ}, 60^{\circ}$ respectively. The height of the tower is
BITSATBITSAT 2020
Solution:
$\tan 45^{\circ}=\frac{P Q}{A Q}=\frac{h}{A Q} \Rightarrow h=A Q$
Where $P Q$ is tower and $A B C$ is the park, with $Q$ being mid point of the side
$BC$ and $PQ =h$
Also,$ AQ ^{2}+ BQ ^{2}=100^{2} $
$\Rightarrow h^{2}+h^{2} \cot ^{2} 60^{\circ}=100^{2} $
$\Rightarrow h^{2}\left[1+\frac{1}{3}\right]=100^{2} $
$\Rightarrow h^{2}=\frac{3 \times 100^{2}}{4}$
$ \Rightarrow h=50 \sqrt{3}$
