Question
In a hexagon $ABCDEF$, side $AB$ is parallel to side $FE$ and $∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3$. Find $∠B$ and $∠D.$
✓
Answer
Given: Hexagon $\mathrm{ABCDEF}$ in which $\mathrm{AB}|| \mathrm{EF}$ and $\angle \mathrm{B}: \angle \mathrm{C} : \angle \mathrm{D}: \angle \mathrm{E}=6: 4: 2: 3$.
To find: $\angle B$ and $\angle D$
Proof: No. of. sides $n=6$
$\therefore$ Sum of interior angles $=(n-2) \times 180^{\circ}$
$=(6-2) \times 180^{\circ}$
$ =720^{\circ}$
$ \because A B \| E F \ ($Given$)$
$ \therefore \angle A+\angle F=180^{\circ}$
But $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=720^{\circ} ($Proved$)$
$\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}+180^{\circ}=720^{\circ}$
$\therefore \angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}=720^{\circ}-180^{\circ}$
Ratio $=6: 4: 2: 3$
Sum of parts $=6+4+2+3=15$
$\therefore \angle \mathrm{B}=\frac{6}{15} \times 540=216^{\circ}$
$ \angle \mathrm{D}=\frac{2}{15} \times 540^{\circ}=72^{\circ}$
Hence $\angle B=216^{\circ} ; \angle D=72^{\circ}$
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