Question

$ABC$ is a triangular park with $AB=AC=100m$. 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

• A. $50m$
• B. $50\sqrt{3}m$
• C. $50\sqrt{2}m$
• D. $50(3-\sqrt{3})m$

Answer 1

Answer: (B)

$\tan 45^{\circ }=\frac{PQ}{AQ}=\frac{h}{AQ}\Rightarrow h=AQ$
Where $PQ$ is tower and $ABC$ is the park, with $Q$ being mid point of the side
$BC$ and $PQ=h$
Also,$A{Q}^2+B{Q}^2=100^2$
$\Rightarrow h^2+h^2{\cot }^{2}60^{\circ }=100^2$
$\Rightarrow h^2\left[1+\frac{1}{3}\right]=100^2$
$\Rightarrow h^2=\frac{3\times 100^2}{4}$
$\Rightarrow h=50\sqrt{3}$