Question -
Answer -
We know that the sum of the interior anglesof a polygon = (n тАУ 2) ╧А
And each angle ofpolygon = sum of interior angles of polygon / number of sides┬а
Now, let uscalculate the magnitude of
(i)┬аPentagon
Number of sides inpentagon = 5
Sum of interiorangles of pentagon = (5 тАУ 2) ╧А = 3╧А
тИ┤┬аEachangle of pentagon = 3╧А/5 ├Ч 180o/ ╧А =108o
(ii)┬аOctagon
Number of sides inoctagon = 8
Sum of interiorangles of octagon = (8 тАУ 2) ╧А = 6╧А
тИ┤┬аEachangle of octagon = 6╧А/8 ├Ч 180o/ ╧А =135o┬а
(iii)┬аHeptagon
Number of sides inheptagon = 7
Sum of interiorangles of heptagon = (7 тАУ 2) ╧А = 5╧А
тИ┤┬аEachangle of heptagon = 5╧А/7 ├Ч 180o/ ╧А =900o/7 = 128o┬а34тА▓17тАЭ ┬а
(iv)┬аDuodecagon
Number of sides induodecagon = 12
Sum of interiorangles of duodecagon = (12 тАУ 2) ╧А = 10╧А
тИ┤┬аEachangle of duodecagon = 10╧А/12 ├Ч 180o/╧А = 150o