Question

If $ABCDEF$ is a regular hexagon, then determine each other two angles of $△AFE$?

• A. ${45}^o$
• B. ${30}^0$
• C. ${90}^o$
• D. ${120}^o$

Answer 1

Answer: (B)

Since, each interion angle of regular polygon with $n$-sides $=\frac{(n-2){180}^{\circ }}{n}$

Interior angles of a rectangle hexagon ($6$-sides)

$=\frac{(6-2){180}^{\circ }}{6}$

$=4\times {30}^{\circ }$

$={120}^{\circ }$

i.e., In rectangle hexagon $ABCDEF$,

$\angle AFE={120}^{\circ }$

In $\Delta AFE$,

Since, $AF=FE$

$\Rightarrow \angle AEF=\angle FAE$

$\Rightarrow \angle AEF+\angle EFA+\angle FAE={180}^o$

$\Rightarrow \angle AEF+{120}^o+\angle FAE={180}^o$

$\Rightarrow \angle AEF+\angle FAE={180}^o-{120}^o={60}^o$

Since, $\angle AEF=\angle FAE=\left(\frac{{60}^o}{2}\right)={30}^o$

$\therefore \angle AEF={30}^o$,

$\angle FAE={30}^o$ and from $\left(1\right)$

$\angle EFA={120}^{\circ }$.

Hence, Option B is correct.