A-person-stands-in-the-diagonal-produced-of-the-square-base-of-a-church-tower-at-a-distance-2a-from-it-and-observes-the-angle-of-elevation-of-each-of-the-two-outer-corners-of-the-top-of-the-tower-to

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Question Number 84651 by ajfour last updated on 14/Mar/20

$$Apersonstandsinthediagonal$$$$producedofthesquarebaseofa$$$$churchtower,atadistance2a$$$$fromit,andobservestheangle$$$$ofelevationofeachofthetwo$$$$outercornersofthetopofthe$$$$towertobe30°,whilethatofthe$$$$nearestcorneris45°.Findthe$$$$breadthofthetower.$$

Answered by mr W last updated on 14/Mar/20

$$b=breadthoftower$$$$h=heightoftower$$$$\tan 45°=\frac{h}{2a}\Rightarrow h=2a$$$$h=\sqrt{{\left(\frac{b}{\sqrt{2}}\right)}^2+{(\frac{b}{\sqrt{2}}+2a)}^2}\tan 30°$$$$2a=\sqrt{{\left(\frac{b}{\sqrt{2}}\right)}^2+{(\frac{b}{\sqrt{2}}+2a)}^2}\times \frac{1}{\sqrt{3}}$$$$12a^2=2{\left(\frac{b}{\sqrt{2}}\right)}^2+4a\left(\frac{b}{\sqrt{2}}\right)+4a^2$$$${\left(\frac{b}{\sqrt{2}}\right)}^2+2a\left(\frac{b}{\sqrt{2}}\right)-4a^2=0$$$${(\frac{b}{\sqrt{2}}+a)}^2=5a^2$$$$\frac{b}{\sqrt{2}}+a=\sqrt{5}a$$$$\Rightarrow b=(\sqrt{10}-\sqrt{2})a\approx 1.748a$$

Commented by ajfour last updated on 14/Mar/20

$$correctsir,andwellpresented,$$$$thanks.$$

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