Question

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be ${45}^{\circ }$ from a point $A$ on the plane. Let $B$ be the point $30$ m vertically above the point $A$. If the angle of elevation of the top of the tower from $B$ be ${30}^{\circ }$, then the distance (in m) of the foot of the tower from the point $A$ is :

• A. $15(3-\sqrt{3})$
• B. $15(3+\sqrt{3})$
• C. $15(1+\sqrt{3})$
• D. $15(5-\sqrt{3})$

Answer 1

The correct option is B $15(3+\sqrt{3})$


$AB=30=NP$
In $△ANM$
$\tan {45}^{\circ }=\frac{MN}{AN}=1\Rightarrow MN=AN$

$PM=MN-30=AN-30$
In $△BPM$
$\tan {30}^{\circ }=\frac{PM}{PB}=\frac{AN-30}{AN}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AN-30}{AN}\Rightarrow AN=\sqrt{3}AN-30\sqrt{3}\Rightarrow AN=\frac{30\sqrt{3}}{\sqrt{3}-1}=15(3+\sqrt{3})$