A-ladder-placed-against-a-vertical-walls-ubtends-an-angle-of-45-degree-with-thewall-The-distance-between-the-footo-f-the-ladder-and-the-wall-is-15mt-calculae-the-length-of-the-ladder-correctto-the-

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

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

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

$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{ladder}\:\mathrm{is}\:\frac{\mathrm{15}}{\mathrm{cos45}°}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{2}}/\mathrm{2}} \\ $$$$=\frac{\mathrm{30}}{\:\sqrt{\mathrm{2}}}=\mathrm{15}\sqrt{\mathrm{2}}\approx\mathrm{21}\:\mathrm{m} \\ $$

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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