The size of angle BAJ in the given regular polygons is 84⁰.
The given parameters;
- regular hexagon = ABCDEF = 6 sides
- regular pentagon = AJFGH = 5 sides
Each interior angles of a regular hexagon is calculated as follows;
[tex]= \frac{(6-2) \times 180}{6} \\\\= 120^0[/tex]
Each interior angles of a regular pentagon is calculated as follows;
[tex]= \frac{(5 -2)\times 180}{5} \\\\= 108^0[/tex]
The size of angle BAJ is calculated as follows;
[tex]JAF + BAJ = 120 \ \ ----- (1)\\\\JAF + JFA + 108 = 180 \ (sum \ of \ angles\ in \ \Delta\ AJF)\\\\|AJ| = |JF| \\\\2JAF = 180-108\\\\2JAF = 72\\\\JAF = 36^0\\\\BAJ = 120 - JAF\\\\BAJ = 120 - 36\\\\BAJ = 84^0[/tex]
Thus, the size of angle BAJ in the given regular polygons is 84⁰.
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