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