ABCDE is a regular pentagon. Its sides are extended as shown in the figure. The value of $$\frac{\angle ABC + 2\angle EGD + 3\angle BAJ}{6}$$ is:

Given,

ABCDE is a regular pentagon

Each angle of regualr pentagon = $$\frac{\left(n-2\right)180^{\circ\ }}{n}=\frac{\left(5-3\right)180^{\circ\ }}{5}=108^{\circ\ }$$

$$=$$>  $$\angle\ $$ABC = $$\angle\ $$BCD = $$\angle\ $$CDE = $$\angle\ $$DEA = $$\angle\ $$EAB = 108$$^{\circ\ }$$

From the figure,

$$\angle\ $$BAJ + $$\angle\ $$EAB = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$BAJ + 108$$^{\circ\ }$$ = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$BAJ = 72$$^{\circ\ }$$

$$\angle\ $$DEA + $$\angle\ $$GED = 180$$^{\circ\ }$$

$$=$$>  108$$^{\circ\ }$$ + $$\angle\ $$GED = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$GED = 72$$^{\circ\ }$$

$$\angle\ $$CDE + $$\angle\ $$GDE = 180$$^{\circ\ }$$

$$=$$>  108$$^{\circ\ }$$ + $$\angle\ $$GDE = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$GDE = 72$$^{\circ\ }$$

In $$\triangle\ $$GED,

$$\angle\ $$GDE + $$\angle\ $$GED + $$\angle\ $$EGD = 180$$^{\circ\ }$$

$$=$$>  72$$^{\circ\ }$$ + 72$$^{\circ\ }$$ + $$\angle\ $$EGD = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$EGD = 36$$^{\circ\ }$$

$$\therefore\ $$ $$\frac{\angle ABC+2\angle EGD+3\angle BAJ}{6}=\frac{108^{\circ\ }+2\left(36^{\circ\ }\right)+3\left(72^{\circ\ }\right)}{6}=\frac{108^{\circ\ }+72^{\circ\ }+216^{\circ\ }}{6}=\frac{396^{\circ\ }}{6}=66^{\circ\ }$$

Hence, the correct answer is Option C