Question

Q9. ABCDEF is a regular hexagon. AJFGH is a regular pentagon. Work out the size of angle BAJ​

Answer 1

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;

$= \frac{(6-2) \times 180}{6} \\\\= 120^0$

Each interior angles of a regular pentagon is calculated as follows;

$= \frac{(5 -2)\times 180}{5} \\\\= 108^0$

The size of angle BAJ is calculated as follows;

$JAF + BAJ = 120 \ \ ----- (1)\\\\JAF + JFA + 108 = 180 \ (sum \ of \ angles\ in \ \Delta\ AJF)\\\\|AJ| = |JF| \\\\2JAF = 180-108\\\\2JAF = 72\\\\JAF = 36^0\\\\BAJ = 120 - JAF\\\\BAJ = 120 - 36\\\\BAJ = 84^0$

Thus, the size of angle BAJ in the given regular polygons is 84⁰.

Learn more here: brainly.com/question/18565704