Question

Each exterior angle of a regular polygon measures 72°. Find the sum of the measures of the polygon's interior angles.

Answer 1

`\mathrm{Let\ }n\ \mathrm{be\ the\ number\ of\ sides\ of\ the\ polygon.}\\\mathrm{Then\ measure\ of\ each\ interior\ angle}=(180(n-2))/(n)\\\\\mathrm{Measure\ of\ each\ exterior\ angle=}180-(180(n-2))/(n)\\\mathrm{or,\ }72=180-(180(n-2))/(n)\\\mathrm{or,\ }(180(n-2))/(n)=180-72=108\\\mathrm{or,\ }180(n-2)=108n\\\mathrm{or,\ }180n-360=108n\\\mathrm{or,\ }72n=360\\\mathrm{or,\ }n=5\\`

`\mathrm{Now,}\\\mathrm{Sum\ of\ interior\ angles\ of\ polygon=}(n-2)*180^o=(5-2)* 180^o=540^o`