Question

The angle of elevation of the top of 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 $30m$ 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

Answer: (B)

$15(3+\sqrt{3})$

From $△ACP,$

$\tan {45}^{\circ }=\frac{x+30}{y}$

$\Rightarrow 1=\frac{x+30}{y}$

$\Rightarrow x+30=y$ $...\left(i\right)$

From $△BDP,$

$\tan {30}^{\circ }=\frac{x}{y}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{y}$

$\Rightarrow x=\frac{y}{\sqrt{3}}$ $...\left(ii\right)$

Substitute the value of $y$ rom equations $\left(ii\right)$ in equation $\left(i\right)$

$\Rightarrow \frac{y}{\sqrt{3}}+30=y$

$\Rightarrow y=y\sqrt{3}-30\sqrt{3}$

$\Rightarrow y(\sqrt{3}-1)=30\sqrt{3}$

$\Rightarrow y=\frac{30\sqrt{3}}{\sqrt{3}-1}$

$\Rightarrow y=\frac{30\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$\Rightarrow y=\frac{30\sqrt{3}(\sqrt{3}+1)}{3-1}$

$\Rightarrow y=\frac{30\sqrt{3}(\sqrt{3}+1)}{2}$

$\Rightarrow y=15\sqrt{3}(\sqrt{3}+1)$

$\Rightarrow y=15(3+\sqrt{3})$