A polygon with $$ n $$ sides can be split into $$ n-2 $$ triangles Therefore, the sum of internal angles in an $$ n-g o n=(n-2) imes \square $$

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A polygon with $$ n $$ sides can be split into $$ n-2 $$ triangles Therefore, the sum of internal angles in an $$ n-g o n=(n-2) 	imes \square $$

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Sum of internal angles in an $$ n $$-gon = $$ (n - 2) \times 180^\circ $$ The sum of the internal angles in an $$ n $$-gon can be calculated using the formula: $$ \text{Sum of internal angles} = (n - 2) \times 180^\circ $$ Here, $$ n $$ is the number of sides of the polygon. The reason for this formula is that any polygon can be divided into $$ n - 2 $$ triangles, and since each triangle has a sum of internal angles equal to $$ 180^\circ $$, multiplying the number of triangles by $$ 180^\circ $$ gives the total sum of the internal angles of the polygon. So, the complete expression for the sum of internal angles in an $$ n $$-gon is: $$ \text{Sum of internal angles} = (n - 2) \times 180^\circ $$

A polygon with \( n \) sides can be split into \( n-2 \) triangles Therefore, the sum of internal angles in an \( n-g o n=(n-2) \times \square \)

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A polygon with \( n \) sides can be split into \( n-2 \) triangles Therefore, the sum of internal angles in an \( n-g o n=(n-2) \times \square \)

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