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

Solution (By Examveda Team)

∠ABC = 108° {Internal Angle of regular pentagon}
∠EGD = 108° - (∠E + ∠D)
= 108° - (72° + 72°)
= 36°
∠BAJ = 72°
$$\eqalign{ & \frac{{\angle {\text{ABC}} + 2\angle {\text{EGD}} + 3\angle {\text{BAJ}}}}{6} \cr & = \frac{{{{72}^ \circ } + 2 \times {{36}^ \circ } + 3 \times {{72}^ \circ }}}{6} \cr & = {18^ \circ } + {12^ \circ } + {36^ \circ } \cr & = {66^ \circ } \cr} $$