Question

In a regular polygon, outer angle is twice the inner angle.
i) How much is each of its inner angle?
ii) How many sides does it have? [4 Marks]

Answer 1

i) Let, inner angle $=x$
∴ Outer angle $=2x$
Now,
Inner angle + Outer angle $={180}^0$ [1 Mark]
$x+2x={180}^0$
$3x={180}^0$
$x={60}^0$ [1 Mark]

ii) Let us assume the polygon has $‘n\text{'}$ sides.
Each outer angle $=2x$
$\therefore 2x=(2\times 60{)}^0={120}^0$ [1 Mark]
Now,
Sum of all outer angles $={360}^0$
∴ we can say,
${120}^0\times n={360}^0$
$n=3$ [1 Mark]