Question

$ABCDEF$ is a regular hexagon. Find each angle of triangle $BEF$.

• A. ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$
• B. ${60}^{\circ },{60}^{\circ }$ and ${60}^{\circ }$
• C. ${45}^{\circ },{45}^{\circ }$ and ${90}^{\circ }$
• D. ${30}^{\circ },{30}^{\circ }$ and ${120}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$

ABCDEF is a regular hexagon

$\therefore \angle BAF=\angle ABC=\angle BCD=\angle CDE=\angle DEF=\angle AFE=(6-2)\times \frac{180}{6}={120}^o...eq.1$

In $△ABFAB=AF\Rightarrow \angle ABF=\angle AFB....eq.2\left[\text{angle opp. to equal sides are equal}\right]$

Using angle sum property, for $△ABF$

$\angle ABF+\angle AFB+\angle BAF={180}^o\Rightarrow \angle ABF+\angle ABF+{120}^o={180}^0\text{(using eq.2)}\Rightarrow 2\angle ABF={180}^o-{120}^o\Rightarrow \angle ABF={30}^o...eq3$

From eq.2 and eq.3

$\angle ABF=\angle AFB={30}^0...eq4$

Now, $\angle AFE={120}^o$

$\angle AFB+\angle BFE={120}^o\Rightarrow \angle BFE=120-30={90}^o...eq.5$

In $△ABF\&△CBD$

$AB=CB$ [sides of regular hexagon]

$AF=CD$ [sides of regular hexagon]

$\angle BAF=\angle BCD$ [angle of regular hexagon]

$\therefore △ABF\cong △CBD$ [By SAS congurency]

$\Rightarrow BF=BD$ [By CPCT] ....eq.6

And $\angle ABF=\angle CBD\Rightarrow \angle CBD={30}^o...eq.7$

In $△BEF\&△BED$

$EF=FD$ [sides of regular hexagon]

$BF=BD$ [From eq. 6]

$BE=EB$ [common]

$\therefore △BEF\cong △BED$ [By SSS congurency]

$\angle EBF=\angle EBD...eq.8$ [By CPCT]

Now, $\angle ABC={120}^o$ [using 1]

$\Rightarrow \angle ABF+\angle EBF+\angle EBD+\angle CBD={120}^o\Rightarrow {30}^o+2\angle EBF+{30}^o={120}^0[\text{using}3,6,7\&8]\Rightarrow 2\angle EBF={120}^o-{60}^o\Rightarrow \angle EBF={30}^o$

Now using angle sum property

$\angle EBF+\angle BFE+\angle BEF={180}^o\Rightarrow {30}^o+{90}^o+\angle BEF={180}^o\Rightarrow \angle BEF=180-120={60}^o$

$\therefore$ option A is the correct answer.