Question

The angle of elevation of the top of a tower from the top and foot of 100m building are ${30}^{\circ }and{45}^{\circ }$ respectively. The height of the tower is

• A. $50(3-\sqrt{3})m$
• B. $50(3+\sqrt{3})m$
• C. $50\sqrt{3}m$
• D. 150 m

Answer 1

The correct option is B

$50(3+\sqrt{3})m$

CD = BE = 100m,
AB = H metre
$\therefore$ AE = AB - BE = H - 100, CE = BD
In $\Delta ACE,$ $tan{30}^{\circ }=\frac{AE}{CE}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{H-100}{x}\Rightarrow x=\sqrt{3}(H-100)....\left(1\right)In\Delta ABD,tan{45}^{\circ }=\frac{AB}{BD}\Rightarrow 1=\frac{H}{x}\Rightarrow x=H.....\left(2\right)$
From (1) and (2), we have
$H=\sqrt{3}(H-100)\Rightarrow H=\sqrt{3}H-100\sqrt{3}\Rightarrow \sqrt{3}H-H=100\sqrt{3}\Rightarrow H=\frac{100\sqrt{3}}{\sqrt{3}-1}\Rightarrow H=\frac{100\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{100\sqrt{3}(\sqrt{3}+1)}{2}\Rightarrow H=50\sqrt{3}(\sqrt{3}+1)\Rightarrow H=50(3+\sqrt{3})$