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Question Number 106373 by chichi last updated on 04/Aug/20

$$$$$$\mathrm{A}\mathrm{ladder}\mathrm{placed}\mathrm{against}\mathrm{a}\mathrm{vertical}\mathrm{walls}$$$$\mathrm{ubtends}\mathrm{an}\mathrm{angle}\mathrm{of}45\mathrm{degree}\mathrm{with}$$$$\mathrm{thewall}\mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{footo}$$$$\mathrm{f}\mathrm{the}\mathrm{ladder}\mathrm{and}\mathrm{the}\mathrm{wall}\mathrm{is}15\mathrm{mt}$$$$\mathrm{calculae}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{ladder}$$$$\mathrm{correctto}\mathrm{the}\mathrm{nearest}\mathrm{whole}\mathrm{number}.$$

Answered by 1549442205PVT last updated on 05/Aug/20

$$\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{ladder}\mathrm{is}\frac{15}{\cos 45°}=\frac{15}{\sqrt{2}/2}$$$$=\frac{30}{\sqrt{2}}=15\sqrt{2}\approx 21\mathrm{m}$$

Commented by chichi last updated on 05/Aug/20

$$$$$$\mathrm{thanks}\mathrm{a}\mathrm{lot}\mathrm{for}\mathrm{your}\mathrm{help}$$

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