Let AB and PQ be two vertical poles,160 m ap | vedclass

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(C) Let $BC = CQ = x$,$AB = h$,and $PQ = 2h$.From the right-angled triangles $	riangle ABC$ and $	riangle PQC$:$	an 	heta = \frac{AB}{BC} = \frac{

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$\frac{3-2 \sqrt{2}}{4}$

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(C) Let $BC = CQ = x$,$AB = h$,and $PQ = 2h$.From the right-angled triangles $	riangle ABC$ and $	riangle PQC$:$	an 	heta = \frac{AB}{BC} = \frac{h}{x}$$	an \left(\frac{\pi}{8}ight) = \frac{PQ}{CQ} = \frac{2h}{x}$Dividing the two equations:$\frac{	an 	heta}{	an \left(\frac{\pi}{8}ight)} = \frac{h/x}{2h/x} = \frac{1}{2}$Thus,$	an 	heta = \frac{1}{2} 	an \left(\frac{\pi}{8}ight)$.We know that $	an \left(\frac{\pi}{8}ight) = \sqrt{2} - 1$.Therefore,$	an 	heta = \frac{1}{2}(\sqrt{2} - 1)$.Squaring both sides:$	an^{2} 	heta = \frac{1}{4}(\sqrt{2} - 1)^{2} = \frac{1}{4}(2 + 1 - 2\sqrt{2}) = \frac{3 - 2\sqrt{2}}{4}$.

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